mondotrainsAnd I guess if I want to absolutely sure I can run 100-200 lights, I could buy this "Pyramid" 10-AMP power supply....anybody have any opinions on this one?
I suggest you read the (limited) customer reviews. The 5 AMP supply seems to have a power limitation of 3 amps or so, or it overheats.
The 10amp supply (Pyramid PS12KX or PS15KS) seem to have much better reviews.
I bought a couple of inexpensive PC/monitor power supplies off of e-bay (about $12) which I am going to distribute around the layout as needed. While they are rated at 5 amps, I don't plan on using them at full capacity.
You will need a voltage regulator to reduce the output to 12 volts (or slightly less). There was a recent thread on that here, search for LM317 voltage regulator.
Alan
Co-owner of the proposed CT River Valley RR (HO scale) http://home.comcast.net/~docinct/CTRiverValleyRR/
rrinker Keep in mind the output of that power supply is 13.8 volts - do not drive 12 volt bulbs directly with it. What I would suggest is two things. About 4 diodes of a 5+ amp rating, wired in series, which will drop the power supply output by about 2.4 volts. Second, run the power supply outout to a terminal strip. Add 5 fuses of no more than 1 amp each, and break you light circuits into 5 sections, one to each fuse. That was, there is no single high current line on the layout, and there is no need for heavy duty wiring. Plus with multiple circuits yo can switch things on and off independently, and a single short won't put a full 5 amps in one single spot of the wiring. --Randy
Keep in mind the output of that power supply is 13.8 volts - do not drive 12 volt bulbs directly with it. What I would suggest is two things. About 4 diodes of a 5+ amp rating, wired in series, which will drop the power supply output by about 2.4 volts. Second, run the power supply outout to a terminal strip. Add 5 fuses of no more than 1 amp each, and break you light circuits into 5 sections, one to each fuse. That was, there is no single high current line on the layout, and there is no need for heavy duty wiring. Plus with multiple circuits yo can switch things on and off independently, and a single short won't put a full 5 amps in one single spot of the wiring.
--Randy
Mondo
Randy has given you very good advice.
Perhaps the only change I would make is to use circuit breakers rather than fuses. But fuses are simpler and cheaper as long as you do things carefully so as not to blow them often.
Picture your electrical distribution from a single power supply as a large tree (from the ground up). Large currents and large wires are at the trunk, with small currents out at the individual light bulbs. If you depend on a single large fuse or circuit breaker, all the wiring must be robust enough to pass just under the trip current for the breaker. Yes, 18 gauge wire is capable of handling more than 5 amps continuously in open air or with reasonable ventilation. With 10 amps, you need 14 gauge wire (house wiring size).
But - if your breaker is set at 5 amps, every part of your circuit must be able to handle 4.9 amps continuously. The wire must not have enough resistance to limit the current to 4.9 amps if a short circuit occurs at the far end of your distribution. I like the quarter test used in DCC to test the wiring sufficiency. Cause a deliberate short circuit at the far end of the power distribution - if your breaker trips immediately, your wiring is good enough. If it doesn't, than you have the potential for the 4.9 amp disaster. 4.9 amps continuous through some small wire or around our models can cause damage and/or overheat things like foam or paper to the point or melting or fire. This paragraph is why Randy's idea of 5 individually fused paths of 1 amp each is the far better route.
Even with the highly recommended five 1 amp (or even 3 1.5 amp) paths, your power supply should have it's own fuse or circuit breaker for 5 amps to protect against a short upstream to the distribution terminal strip. Use 18 gauge wire or bigger from the power supply to the terminal strip.
Finally, light bulbs last much longer when operated 10%-25% below their rated voltage. Conversely, bulb life is shortened drastically by running at higher than rated voltage. Use 16 volt bulbs or use diodes as Randy described to get the voltage down to 10-11 volts for 12 volt bulbs. Each diode drops the voltage by 0.6 volts. The bulbs will a give a softer, more pleasing glow at the lower voltage.
Fred W
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
Hi Guys,
I've been looking at the prices of second-hand MRC power packs on Ebay and it looks like I'm going to spend quite a bit of money, if I buy 6 power packs to run the lighting on my layout.
I'm thinking of buying the "Pyramid" 5-amp power supply in this link:
http://www.amazon.com/Pyramid-PS9KX-Supply-Cigarette-Lighter/dp/B0002BA570/ref=sr_1_5?ie=UTF8&s=electronics&qid=1263602150&sr=8-5
For $29.99, with free shipping, this power supply could probably run the same number of lights as 5 MRC power packs, with each rated at 12VA.
And I guess if I want to absolutely sure I can run 100-200 lights, I could buy this "Pyramid" 10-AMP power supply....anybody have any opinions on this one?
http://www.amazon.com/Pyramid-PS12KX-10-amp-13-8-volt-Supply/dp/B0002JTD1Q/ref=sr_1_4?ie=UTF8&s=musical-instruments&qid=1263609067&sr=8-4
In one of my previous postings, "Fred" suggested that he's not a fan of 5amp power supplies, unless 18 gauge wire or bigger is used for the "bus" feeding all the lights. Is there anything else I should be concerned about if I go to this larger single power supply?
Thanks,
If you're talking about basic train set type power packs, you need to read the wording on the rating. VA is "volt-amps" or basically watts. If only one numebr is given, it is the total you can draw from botht he variable DC and the AC terminals combined. 12VA would be about 1 amp at 12 volts, although if the AC terminals are a more common 16V then it's more like 3/4 amp max.
Now you see why a train set power pack works fine with the single loco and 3 cars that come with the set but fall flat with double headed trains with 4 or 5 lighted cars. They just don't put out that much power.
Thanks so much Fred and Randy,
I think I'm finally getting this issue understood. I hope you realize my confusion was caused by what I reported reading in Sporandeo's book.
Now that I'm convinced that my 5 power packs running trains, along with 6 or more providing power to my lights, will in no way come close to using the 12amps left in my train room's house circuit, I will concentrate on getting the right amount of lights on each of the power packs, based on the power pack's individual ratings. Some are 12VA and others are 17VA. I'm going to break up my first "district" to include only 20 lights and then see what I get for a reading on my meter, regarding ma draw. I had my meter set to "DC10A", when I got the reading of 1.3amps. Per the instruction manual, this setting on the meter allows me to determine anywhere from 0 to 10amps....I happened to get 1.3amps.
Thanks again for you guys taking the time to respond to my questions.
At least now, I can rest assured that I have enough power in the room.
mondotrains Randy, You said "But a transformer producing 1 amp at 12 volts will not draw anywhere near 1 amp from the 120 AC input side. Probably even less than .2 of an amp - so you need not fear plugging in 5 or 6 1 amp power supplies with a single 15 amp household circuit". Getting back to what I said above, I'm not talking about only five or six 1-amp power supplies. I'm talking about 5 power packs that are hooked up to my Aristo-Craft walk-around throttles for running trains and another 6 for lighting up the 6 "districts" on my layout. Why is it that you're saying that each of the lighting power packs will only draw about .2 amps? Is it really possible that I'm powering those 25 lights in my first district with only .2 amps? Do you agree with what Andy Sporandeo says in his book, when he's discussing what size power pack to buy, that a passenger train, with engine, its headlight and its lit-up cars could pull 1.8 amps? I'd appreciate you commenting on what you would guess my 5 trains might be pulling for amps, again realizing that during an operating session, trains are not barreling around at full throttle. Thanks, Mondo
Randy,
You said "But a transformer producing 1 amp at 12 volts will not draw anywhere near 1 amp from the 120 AC input side. Probably even less than .2 of an amp - so you need not fear plugging in 5 or 6 1 amp power supplies with a single 15 amp household circuit".
Getting back to what I said above, I'm not talking about only five or six 1-amp power supplies. I'm talking about 5 power packs that are hooked up to my Aristo-Craft walk-around throttles for running trains and another 6 for lighting up the 6 "districts" on my layout. Why is it that you're saying that each of the lighting power packs will only draw about .2 amps? Is it really possible that I'm powering those 25 lights in my first district with only .2 amps?
Do you agree with what Andy Sporandeo says in his book, when he's discussing what size power pack to buy, that a passenger train, with engine, its headlight and its lit-up cars could pull 1.8 amps?
I'd appreciate you commenting on what you would guess my 5 trains might be pulling for amps, again realizing that during an operating session, trains are not barreling around at full throttle.
In a word, YES. Again, you are still confusing the current draw on the low voltage size with the current draw on your AC wall outlets. Yes, a given train as in you exampel could draw 1.8 amps, depending on the make and how many lights are in each lightes cars - many model models draw a half amp or less, even with sound. The old rule of thumbd for HO was 1 amp per locomotive. But that train drawing 1.8 amps at 12 volts on one of tyour Aristo power supplies IS NOT drawing 1.8 amps from your wall outlet. Without testing, my guess would be about .3 of an amp, allowing for power supply inefficiency plus overhead of the controller. Voltage and current are inversely proportional - reduce one and you increase the other, for the same amount of power.
I would stop worrying about the amount of pwoer you are drawing from your wall outlets. You aren't goign to come CLOSE to 12 amps with all your power supplies. You probably don;t have a 12 amp requirement on the 12 volt side, let alone the AC input. Power supplies drawing 12 amps from the AC input could theoretically supply over 96 amps at 12 volts on the output (assuming 80% efficiency). Your only real issue is figuring out how many power supplies you need to power all your lights.
As for the meter - you probably blew the fuse. Most low cost multimeters either measure in milliamps (too low for this exercise) or have a 10 amp setting, which also usuallyuses a different terminal for the positive probe. Use that range to measure the draw on your lighting circuits.
You have basically 2 options: lots of small power supplies to break up the lighting into sections and provide enough total current, or a single large supply with individual fused circuits to each section of lighting.
One other thing - jest because a powerr supply CAN put out 1 amp, if you only apply half an amp worth of load to it, it will only put out half an amp - and consume approximtely half the power it would consume putting out its full load
mondotrains Hi TomDiehl, I did what you suggested and hooked up my meter in series to the power pack on the DC circuit with the 25 lights. I got a reading of 1.3amps. Now what I don't understand is that my power pack says it is rated at max. DC 12VA. Again, if I divide the 12VA by 12, like Andy Sporandeo suggested, I would get 1amp. If my meter reads 1.3 amp draw on the circuit, how come the pack didn't get overloaded if it can only produce a max. of 1 amp? Mondo
Hi TomDiehl,
I did what you suggested and hooked up my meter in series to the power pack on the DC circuit with the 25 lights. I got a reading of 1.3amps. Now what I don't understand is that my power pack says it is rated at max. DC 12VA. Again, if I divide the 12VA by 12, like Andy Sporandeo suggested, I would get 1amp. If my meter reads 1.3 amp draw on the circuit, how come the pack didn't get overloaded if it can only produce a max. of 1 amp?
Maybe I can help without confusing the issues any more.
First, drawing 1.3 amps from a power pack rated for only 1 amp. Electrical and electronic ratings usually are based on the sustained heat load, and are usually on the conservative side. Your 12VA power pack has diodes and a transformer sized to provide 1 amp at 12 volts without overheating. Can they do more, especially for a few minutes? Almost always yes, unless the circuit breaker prevents it. That's probably the scariest part (but not all that unusual) - that your circuit breaker is not kicking out at 1.3 amps. The circuit breaker is not protecting the rest of the power pack from overheating the way that it should. The case of the power pack is probably getting warm supplying 1.3 amps. Another confirmation of the power pack struggling to provide above its rating is that the lights got dimmer as you added more - an indication that the voltage is dropping noticeably under full load - which is a natural characteristic of a highly loaded transformer.
Electrical and electronic components have very high variability between individual components of the same model and rating. If you want less variability, you pay a lot more per component. Standard resistors vary in actual resistance by +/- 20% from their rated value. Transistor gains have similar variability. This variability is why one computer processor will overclock by 20%, and the next one won't overclock at all. This is why I wouldn't have been surprised if your circuit breaker didn't trip until 1.2 amps. 1.3 amps is a little high, and I would encourage you to reduce the measured load back to 1.0 amps or less for safety. Or install an external circuit breaker rated at 1 amp - the existing circuit breaker isn't doing the job.
Second issue is the passenger train drawing 1.8 amps. Disbelief reigns because most of us never see trains with current draws that high. But let's work the numbers. Two Athearn F units at the head of the train, drawing 0.45 amps each. 10 passenger cars, each with 3 30ma light bulbs for lighting, gives 900ma for the passenger car lights. A total of 1.8 amps for a double-headed, 10 car lighted passenger train is within the realm of possibility.
The issue does beg the question - how many of us own 10 lighted passenger cars with 3 bulbs each? And how many of us have a layout big enough to run 10 full length passenger cars with double-headed power (that's an 11.5 foot long train in HO)? Most of us just don't see those kind of loads every day. And certainly the trend is to locomotives that draw only .25 or .3 amps pulling a train, and using LEDs for passenger car lighting.
With my small engines and unlit 6 car trains, I don't see current draws above 0.6 amps, even going up 4% grades with old open frame motors in the engines.
Third issue is VA (or watts) is the measure of power that is independent of voltage. A 12VA output power pack is less than a 15VA (watt) load on house power. Your 15 amp house circuit can handle 1800 watts on the circuit before the breaker trips. It's a good idea to never run circuits and power supplies at more than 80% of capacity (an added longevity and safety measure), so 1400 watts continuous is available on the house circuit. That's about 100 of your 12VA power packs tied into the single circuit.
hope these explanations make sense
In parallel circuits (like house wiring or train track) the current is additive. If you have 20 lamps each drawing 50ma then the total current of all the lamps would be be 1amp (as pointed out above by Jeff). The same is true of locomotive is they are all running on the same track.
The current is equal to the voltage divided by the resistance. If you had a 12 volt source and a lamp with a 12 ohm resistance the current draw would be 1 amp. If you reduced the voltage to 6 volts the current would be 1/2 amp. So dropping the voltage drops the current flow if the resistance stays the same.
While LEDs tend to use less power, many of the common types draw 30 to 40 ma, so one has to pay attention to them as well for calculating power needs.
Help!!!!!!!
I think the biggest flaw in the calculation you've been doing is that a train will draw 1.8A. That's a lot. A modern loco draws well under .5A. Looking at the Feb. MRR review of the Bachmann DDA40X, which has two motors, the most it can draw when stalled is 1A. The BLI SW7 draws 0.18A when stalled! (Says something about the difference between B'mann and BLI, perhaps.) Remember, if the train is moving, the motor isn't stalled, and will be drawing significantly less current. As far as car lighting goes, if it is newer cars with LED lighting the current will be very small. Any lighting isn't going to draw a lot, or you'd have a melted shell on your hand.
The importatnt thing to remember in your musings is that 1A of 12V is the same power as 0.1A of 12V. So even with the inefficiencies in the power pack taken into account, the power to run your trains is barely a concern with respect to your house wiring's ability to supply it.
Layout lighting can be another issue, if you don't watch it.
Jeff But it's a dry heat!
Thanks again to everybody who responded to my questions!
Jeff,
You said:
"The current draw of your power packs is going to be enough to not matter, in the scheme of things. Assuming 5 trains, 2A each, which is probably high by at least a factor of 2, at 15 Volts (also a high number), they are using about 30VA total. The 12A you have available on your circuit, can provide 1440 VA. At the very most, your trains will be drawing about 2% of your available power."
How did you calculate the 30VA total for the 5 trains? I don't understand the relationship between the 2A for each train, at 15 volts, resulting in 30VA total. It looks like you're multiplying the 2A times 15 volts, getting 30VA, but isn't that for one train....not all five? If each train is 2A, don't you need to multiply that by 5 for the 5 trains? I guess I need you to spell it out with an equation.
I thought that if a power pack says "12VA", that it meant it would put out 1 amp......because Andy Sporandeo's book says to determine the amp capacity of a power pack, you divide the 12VA by 12, resulting in 1 AMP. And when he says that a train draws 1.5 amps between the engine and headlight, then I thought I needed a power pack rated for at least 18VA's because 18VA divided by 12 = 1.5 amps. Based on what you're telling me, I can see where either Andy is wrong or I misunderstood what he is saying. Actually, I think he forgot a step in between the 2 calculations. I think you can see where my confusion is. Andy has a graphic where he says a passenger train with lighting will require 1.8AMPS. It really is disappointing to think that in his chapter explaining "Choosing a Power Pack", he shows you how to calculate a power pack's amp capacity (the VA's divided by 12) and then tells you how much a train draws, but from what you're telling me, there isn't a driect correlation between the two. In fact, he goes on to show that a power pack rated at 7VA "would provide .58 amps at is full voltage of 12 volts". A novice like me would then guess that this power pack couldn't even handle a freight engine with a headlight because earlier in the paragraph, he showed that the freight engine would require 1.5amps.
Come to think about it, when he says the passenger train pulls 1.8 amps and if you use his calculation to determine a pack's capacity, you'd need a power pack of 21.6VA's. None of the MRC power packs, even the more expensive ones I have, are rated more than 17VA's.
So much for reading books.
mondotrainsYou said that ...."You have to add up your individual current ratings for EACH bulb, that will give you your load. Remember not to go over %80 of max load for a pack. " Unfortunately, I cannot add up the current ratings for each bulb. For example, the Walther's lights that I bought (lamposts and gree-shaded loading dock lights don't even mention on their website or the box they came in what the current rating is. Many of my other bulbs, like the Life-Like ones to light up the inside of a building, only say that they are 12 or 16 volts.
You said that ...."You have to add up your individual current ratings for EACH bulb, that will give you your load. Remember not to go over %80 of max load for a pack. "
Unfortunately, I cannot add up the current ratings for each bulb. For example, the Walther's lights that I bought (lamposts and gree-shaded loading dock lights don't even mention on their website or the box they came in what the current rating is. Many of my other bulbs, like the Life-Like ones to light up the inside of a building, only say that they are 12 or 16 volts.
Hey Mondo,
Yes, unfortunately many manufacturers do not provide the information they should and sometimes we have to guess. If you were to assume that your 12-16 Volt bulbs draw 50ma .050 X 25 = 1.25 Amps
If your pack had a 1 Amp load capacity that would have been overloading it, so likely the bulbs are more in the 30ma range, .030 X 25 = .75 Amps. When you added 7 more bulbs, .030 X 32 =.96 Amps this load could very well cause a pack rated for 1 Amp to overload.
Unfortunately, if you do not know the loads of the devices you are connecting the only way to find out the load is to connect them with an Ammeter in series, as Tom described. I'm not sure if there is a clamp-on ammeter for inductive load testing for loads this small. I have an Amprobe clamp-on meter but it does not range that low.
If the pack you are using has overload protection, (most do I think) it will kick out if you overload it. Beware though, that the "inexpensive" packs may be a one time affair. Overload it once and it's done for good. I think most MRR power packs have some sort of auto resetting overload protection but some may not.
I hope some of this is starting to make sense. It's not the easiest part of the hobby to get a good grasp on. Keep asking questions and we'll keep trying to help you out.
Good luck!
mondotrains Okay guys, I had a "duh" moment. I took out my volt meter and turned the dial to where it said "DC/ma". When I touched the red/black testers to the wires going to my district 1 lights, it read 0.10....Does anybody know what that means? Does it mean 100 milliamps which would actually be 1/10th of an amp? Or, should I be setting the dial on my meter to something else, like "DC10A" to get a reading? Thanks, Mondo
Okay guys,
I had a "duh" moment. I took out my volt meter and turned the dial to where it said "DC/ma". When I touched the red/black testers to the wires going to my district 1 lights, it read 0.10....Does anybody know what that means? Does it mean 100 milliamps which would actually be 1/10th of an amp? Or, should I be setting the dial on my meter to something else, like "DC10A" to get a reading?
To read current with any meter (not counting clamp on ones), it needs to be connected in series to the part of the circuit being tested. In other words, you need to disconnect one wire from your power pack, connect one lead from the meter to the power pack terminal and the other lead to the wire just disconnected. Then energize the circuit (turn it on). ALWAYS start with the highest amp setting on the meter and scale down for a reading to protect your meter.
The current draw of a power pack should be listed on its ratings. Look for the "Input" ratings. For a simple transformer, the formula is volts times amps in the primary equals volts times amps in the secondary. Of course this is not an exact reading since transformers are not 100% efficient.
Going a bit farther, 5 12VA supplies would be outputting 60VA, they will be drawing more than that, but even if it is double it's only 10% of your available power.
The current draw of your power packs is going to be enough to not matter, in the scheme of things. Assuming 5 trains, 2A each, which is probably high by at least a factor of 2, at 15 Volts (also a high number), they are using about 30VA total. The 12A you have available on your circuit, can provide 1440 VA. At the very most, your trains will be drawing about 2% of your available power.
(OK, it could be a little more to allow for some inefficiencies, but it is barely a blip on the radar.)
If you connected your meter in amp mode to a voltage source, it may mean you fried your meter like I did, so the .1 is meaningless. Better check the meters other functions.
Thanks guys for your replies.
Now here's a couple of questions regarding Jsperan's input:
You say I should not go over 80% of the max load for the pack. Without knowing the load and only knowing the pack is 12VA, I'm not sure where I go from here.
I can tell you that when I hooked up the 25 lights I have connected to district 1 on my layout to the DC output from the pack, with the throttle set to maximum, the lights seemed just about right.....not too bright and not too dim. When I turned down the throttle a little, the lights were okay but honestly, they looked best at full throttle. If I'm overloading the pack, will the internal breaker kick in?
By the way, all my lights are hooked up in "parallel".
The part you are missing is that when you step down the voltage, you increase the current. Think about a computer power supply - they can put out 40 ore more amps at 12 volts - obviously they don't draw 40 amps from the wall socket yor you'd need a dedicate circuit for a single basic computer.
It's not 1 for 1, ie reduce the voltage by 10 (from 120 to 12) does not change the amps by a factor of 10, there are internal losses in the transformer that eat some of that up. But a transformer producing 1 amp at 12 volts will not draw anywhere near 1 amp from the 120 AC input side. Probably even less than .2 of an amp - so you need not fear plugging in 5 or 6 1 amp power supplies with a single 15 amp household circuit.
mondotrainsNow, here's my question: When the power pack says 12VA output, is that the total output from the DC AND the AC output terminals combined? In other words, when I hooked up this first pack with 12VA output, does that mean I am actualy drawing 1 full amp to light up this section, if the throttle is set at maximum or am is drawing less because I am not using the AC terminals?
Now, here's my question:
When the power pack says 12VA output, is that the total output from the DC AND the AC output terminals combined? In other words, when I hooked up this first pack with 12VA output, does that mean I am actualy drawing 1 full amp to light up this section, if the throttle is set at maximum or am is drawing less because I am not using the AC terminals?
Until you connect a load to the pack there is no current draw. Setting the throttle at MAX does not create a load. The train it is running or the lamps it is powering are the load. The 12VA available is likely for the entire pack, AC and DC. If you are not using the AC terminals you have %80 of the 12VA available on the DC terminals for your lamps.
You have to add up your individual current ratings for EACH bulb, that will give you your load. Remember not to go over %80 of max load for a pack.
I recently did a post regarding lighting up my layout and now have another question. As a result of input I received, I've broken up my layout into 6 districts, and plan to use a single power pack to light up each district....thus, a total of 6 power packs. Now I have another consideration. My layout room only has 12amps available for running the trains and lighting up the layout. In general, even though I have the track designed for 5 operators to be running trains, usually, I'll be running a freight and passenger train on the mainlines, with the freight at slow speed and the passenger train, with 4 cars lit up, at a higher rate of speed. Guys dropping off cars and making up trains in the yard will obviously be running at very low speed, thus drawing very little amperage.
Based on a book I have about model railroad electricity (written by Andy Sparandeo), it looks like my passenger train will draw about 1.8 amps between the engine, its headlight and 4 lit up passenger cars. The freight train, running at prototypical speed, will probably draw less than 1.5 amp. So, I'm going to estimate that my 5 train power packs during a typical operating session in total will draw 5 amps of electricity or less.
Now for the lighting. Based on input I received before, I'm going to hook up my 6 power packs, with each one feeding power to a district on my layout with around 20-25 lights per district. I was told to use the DC output of each power pack so I could "dial the lighting down", if necessary, to reduce brightness. I just installed one power pack for my first district. The power pack says it has 12VA output. My understanding is that in order to calculate the actual amps it draws, you divide the 12VA by 12...resulting in 1 amp. That's the formula Sparandeo's book gave me.
When the power pack says 12VA output, is that the total output from the DC AND the AC output terminals combined? In other words, when I hooked up this first pack with 12VA output, does that mean I am actualy drawing 1 full amp to light up this section, if the throttle is set at maximum or am I drawing less because I am not using the AC terminals?
I know I sound a little overanalytical, but again, I know I only have 12 amps available for my layout room wall plugs. Trust me, I know what is already being drawn on the 20 amp room circuit, leaving me 12 amps for the trains and lights. Unfortunately, there is no way for an electrician to bring more power into my layout room because it's on the 2nd floor (we are on a slab, with no basement) and the circuit board is in our garage, with no way, now that the room is "buttoned up" with insulation and sheetrock, to string an additional line....so, I have to plan carefully.
Any help would be appreciated.