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figuring out milli amps (mA)...?

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  • Member since
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  • From: Dearborn Station
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Posted by richhotrain on Friday, March 9, 2012 5:52 PM

Randy,

Thanks for that great explanation.  I appreciate it.

Rich

Alton Junction

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Posted by rrinker on Friday, March 9, 2012 2:49 PM

 AN amp is an amp is an amp (or milliamp, 1/1000 of an amp). Doesn;t matter if it's a light bulb, LED, Tortoise motor, etc. For DC.

The sine wave nature of AC introduces a lead or lag in inductive loads such as motors, comapred to a purely resitive load like an incandescent bulb. If working with incandescent bulbs, AC and DC are exactly the same. With motors and other inductive loads such as ballasts in flourescent lamps, you have to take this lag into account, the actual power drawn from the circuit is not the same as rated load. This is known as the power factor.

Really this is beyond what we need for model railroad applciations (other than planning perhaps to light the train room with flourescent lights - the actual number of fixtures you can safely install on a single branch circuit is not, for example, 15 amps worth on a 20 amp circuit (electrical codes usually go with the 75% rule too). You need to know the power factor. You may have seen some of this mentioned in computer power supplies. There are some rating for achieving certain efficiencies, like 80+ Bronze, 80+  SIlver, etc. The basic rules require the power supply to be 80% or greater efficient across a range of loads. This i susually achieved via active power factor correction circuitry so that the power suppyl is runnign with a fairly constant power factor regardless of the load. The close to 1, the better, with a power factor less than 1, more energy has to go in to ge tthe same work out. 

Back to powering layout lights - an amp is an amp is an amp. Mr Beasley's 4 amp power supply is 5 times more powerful than the 800ma one, and can run 5 times the number of lights.

Voltage and current are related through resistence - Ohm's Law. V=IR, voltage = current time resistence. So if the resistence remains constant, if voltage goes up, current goes up. Or, les current flowing throught he same resistence generates less voltage. This is how dropping resistors are calcualted for LEDs and low voltage bulbs. This formula plus a few other rules known as Kirchoff's Laws: voltages in parallel are equal, voltages in series add, current in parallel adds, current in series are equal. Mainly the last - if you imagine the circuit is power supply---resistor---LED---power supply, the current throught eh resistor and the LED will be identical. If you set the current to a safe level for the LED, say 10ma, the exact same 10ma will flow through the resistor. The LED will drop a set amount of voltage - this is int he spec for the LED. SO by applying the second in the list of laws, if the pwoer supply is 12 volts, and the LED drops 3 volts, the resistor MUST drop 9 volts. Now we know two parts of the Ohm's Law equation - volts and amps. SO we can calculator what value resistor is needed. V=IR, so R=V/I, or R=9/.010 or 900 ohms.

OK, there is the free first semester electrical engineering course...Smile, Wink & Grin

             --Randy

 


Modeling the Reading Railroad in the 1950's

 

Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.

  • Member since
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  • From: Dearborn Station
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Posted by richhotrain on Friday, March 9, 2012 4:46 AM

rrinker

 Exactly. Milliamps is a measure of current. The more current, the more load you can place on the power supply.

 If you had (to make it easy) lights that draw 10ma each, the 500ma supply could run 50 of them, the 800ma could run 80. But you don;t want to run the power supply at 100% load, 75% is much better, so you could do about 37 of the 10ma lights with the 500ma supply, and 60 on the 800ma one.

          --Randy

Randy, can you clarify and expand upon an issue that still perplexes me?

The OP has two 12 volt power packs, one rated at 500 ma (1/2 amp) and one rated at 800 ma (3/4 amp). 

Mister Beasley has a 12 volt power pack rated at 4 amps (4,000 ma) that he uses to power 16 volt incandescent bulbs rated at 30 ma each.  So, enough power for 133 bulbs at 100 percent, or 100 bulbs at 75 percent of maximum.

As far as power packs go, what is the relationship between volts and amps? 

In other words how do you decide which power pack to choose to power lights, both incandescent and LED as well as switch machines such as Tortoises? 

Are we essentially talking about DC as opposed to AC for this purpose?

Thanks.

Rich

Alton Junction

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Posted by richg1998 on Tuesday, March 6, 2012 2:01 PM

narrow gauge nuclear

The soft, warm, yellow-orange glow of an old fashion filament lamp is nice, but they are current hogs. You are forcing a thin wire to glow nearly white hot by using it to short out a power supply if you think about it.  This means heat, also.

LEDs are catching up rapidly in the all important warm, pleasing color ranges, but also surpassing filament lamps in ever reduced current drain and running cold. Many of the more efficient LEDs are running 1/10th the current of a filament lamp for the same illumination.  Unfortunately, the current ultra-high efficiency LEDs are brilliant snow white or blue white.  Keep your eyes open in future to the various lighting capabilities of LEDs. 

I am curently on a quest for a thin plastic film or yellow wash to apply to brilliant white LEDs to make them sub for tungsten filment color temperatures.

Richard 

The below link, "might" help with color. One fellow in Oz, Lauie has done a lot with LED's and has some You Tube pages also. You shoul be able to follow the discussions in Yaho

http://groups.yahoo.com/group/soundtraxx/message/32060

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.

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Posted by narrow gauge nuclear on Tuesday, March 6, 2012 1:40 PM

The soft, warm, yellow-orange glow of an old fashion filament lamp is nice, but they are current hogs. You are forcing a thin wire to glow nearly white hot by using it to short out a power supply if you think about it.  This means heat, also.

LEDs are catching up rapidly in the all important warm, pleasing color ranges, but also surpassing filament lamps in ever reduced current drain and running cold. Many of the more efficient LEDs are running 1/10th the power of a filament lamp for the same illumination.  Unfortunately, the current ultra-high efficiency LEDs are brilliant snow white or blue white.  Keep your eyes open in future to the various lighting capabilities of LEDs. 

I am curently on a quest for a thin plastic film or yellow wash to apply to brilliant white LEDs to make them sub for tungsten filment color temperatures.

Richard 

Richard

If I can't fix it, I can fix it so it can't be fixed

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Posted by MisterBeasley on Tuesday, March 6, 2012 6:36 AM

I use 16-volt bulbs for most of my layout lighting.  I run them at 12 volts, though, because it greatly extends bulb life and gives a much warmer-looking yellow glow to the lights, more appropriate for the bygone days that I model.

Miniatronics grain-of-wheat bulbs like the ones I use are rated at 30 milliamps each.  It's surprising how fast the numbers add up.  I have two 4-amp supplies for lighting, and I'll likely need more in another year or so.

Most of these power supplies do not have circuit breakers.  Instead, they are protected by a one-time fuse built into the unit.  When it goes, either because of a dead short or an overload caused by too many light bulbs, the supply is useful only for recycling.  So, it's a good idea to get a fuse holder and a supply of fuses, rated lower than your supply.

It takes an iron man to play with a toy iron horse. 

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Posted by gandydancer19 on Monday, March 5, 2012 3:32 PM

Also, 1 Amp = 1000 ma.

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.

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Posted by rrinker on Sunday, March 4, 2012 7:39 PM

 Exactly. Milliamps is a measure of current. The more current, the more load you can place on the power supply.

 If you had (to make it easy) lights that draw 10ma each, the 500ma supply could run 50 of them, the 800ma could run 80. But you don;t want to run the power supply at 100% load, 75% is much better, so you could do about 37 of the 10ma lights with the 500ma supply, and 60 on the 800ma one.

          --Randy

 


Modeling the Reading Railroad in the 1950's

 

Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.

  • Member since
    September 2011
  • 182 posts
figuring out milli amps (mA)...?
Posted by irishRR on Sunday, March 4, 2012 1:09 PM

i am wiring street lights and could use some help. my lights are 12 volt. i have two power sources and each one works just fine. however one 12 volt power source is 500 mA and the other is 800 mA.  what does that mean? is one capable of running more lights? my lights are wired in parallel. should i use one power source instead of the other? ive put a lot of time into this and i want to do it the right way. more importantly, i dont want any shorts or fire hazards. any advice is greatly appreciated. thanx.

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