As I mentally start my first (and hopefully last) multi-train layout, I thought about using a constant-voltage method for controlling train speed on grades. I like to run trains pretty slow, so I thought I could remove the track pins and electrically isolate the grades from the level mainline.
Has anyone tried this?
I have heard how you do it though. You have sperate controls for that steep grade on a seperate viarable voltage tap. And i think you just set that section to a speed or volts you want that train to climb that grade. using insulated pins obviously to seperate it from normal track power on a seperate control tap. I read it somewhere too. Not sure the month of CCT.
That is how people do it, but they shouldn’t. When you connect two outputs of a transformer together, you create a short circuit that is not protected against by the transformer’s circuit breaker. Unless the two outputs are set to exactly the same voltage, a large current can flow between the outputs. It can easily be enough to melt wire and damage your transformer.
As the train makes the transitions from level to incline and back again, the locomotive’s pickups and the pickups of any lighted cars with two pickups will create such a short circuit. The voltages will be different–otherwise, there would be no point to the arrangement. If the short circuit is brief, there will probably be no damage to the wiring nor transformer; but there will be arcing at the rail joint and pickup rollers. If the train stops at the wrong place or moves slowly, however, you could melt something.
Lionel actually advised users to do this, unwisely I believe. CTT has too, in the mistaken belief that Lionel transformers are indestructable. Lionel however did print a warning against it in the fine print of the KW transformer’s service manual, which you can find at Olsen’s.
There is a safe way to do it. Use a single transformer output and put a passive voltage-dropping element in series with the center rail where you want the train to get less voltage. That would be on the level and downhill sections. Give the center rail a direct connection for the uphill. For the voltage-dropping element, you can use a fixed power resistor, a rheostat like those that Lionel used to sell for controlling trains, or semiconductor rectifiers. The rectifiers have the advantage that the voltage drop is constant and doesn’t increase with heavier trains.
An easy way to do it is with a bridge rectifier like this:
Thanks Bob,
Your valuable information not only saves model trains, but also lives and property damage.
Ralph
Thanks so much, Bob. I’ll be heading over to Radio Shack to pick up a bunch of those rectifiers. Looks like I’ll also need some to get my 671, 681, and 2046 to smoke better.
I just finished adding insulation, freeing up the pistons, and opening up clogged vent holes, but at the low speeds I run them they don’t smoke much at all. In neutral with ~15 V on them for a few minutes they smoke real well for ~1 min after I reduce the KW to ~8 V for running-then nothing.
Is there some other bridge rectifier that is small and would drop the V by ~7? With 15 V on the smoke coil it should do real well.
You’ll need half a dozen or so to get 7 volts. Folks who have added them to their locomotives seem to get satisfactory smoke improvement with just a few. For that application, where the rectifier carries only the motor current, with very little chance of a fault current, I recommend something smaller and cheaper, like this 4-ampere rectifier: http://www.radioshack.com/product/index.jsp?productId=2062580
Question on the grade deal: if you use two different transformers for your grade section and your level section would you have any problems? You shouldn’t have more current going to less current in the same transformer. A derailment should just throw the breaker in the transformer controlling that track and that should be it.
Even with separate transformers, the fault current still flows when the train’s pickups cross the gap; and the arcing is the same. However, the transformers’ circuit breakers should protect the wiring and the transformers from the danger of fire and damage if the train stops in the wrong place, assuming that the wiring is properly sized for the circuit breakers’ ratings.
You can get the same result with a single transformer by installing extra circuit breakers or fuses in series with its multiple outputs. However, those who favor particularly fast circuit breakers or fuses may find them tripping or blowing. I have installed thermal (automotive) circuit breakers on the outputs of my transformers, for the inevitable case when I accidentally run between separately powered blocks.
The problem isn’t the same as with a derailment. The fault current flows every time a pickup straddles one of the center-rail gaps, even though the train remains on the track.
Thanks, lionelsoni. A little arcing doesn’t sound too bad, but the concern was frying a transformer. The thing about a rheostat or resistor in series with the wire from the transformer to the track to reduce current there is the fact that the locomotive would still have current of, say 14 volts at one roller and 12 volts at the other roller at the same time. The resistor would work fine going DOWNhill. UPhill you probably want 16 volts to help the train go up,and setting the voltage for the uphill grade and resistors for most of the rest of the layout would generate a lot of heat and make the transformer work harder.
I think that’s why Bob likes the bridge rectifier method better.
Two transformer outputs feeding the center rails on each side of the gap at 14 volts and 12 volts (to continue your example) will try mightily to maintain those voltages. Of course they can’t when the pickups connect the rails together. They will respond by delivering a lot of current through the pickups. But the 12-volt rail fed by a resistor or other passive dropping element from a 14-volt transformer doesn’t mind at all if the rails are connected together. The pickup assembly simply becomes a shunt around the resistor; and both rails get 14 volts as long as the pickups connect them together. Since both ends of the resistor are at 14 volts, no current at all flows through the resistor.
Using rectifiers instead of resistors as the dropping elements allows the voltage to be regulated as well as if the train were run directly from the transformer. The rectifiers, like the resistors, will waste power.
While it’s true that the resistor, or whatever, will waste power, it’s not really all that much. A train run at 14 volts dropped from a 16-volt transformer will increase the power taken from the transformer by 14 percent. There is a way to do the job safely without dropping elements: Use relays with SPDT contacts to transfer an ungapped center rail from one transformer output to another, with long control rails to operate the relays.
Here’s how I did it on my layout. Use separate blocks for level, up and down. Then use bridge rectifiers to adjust the speed on each block. In this way, up gets full power, level gets a little less and down gets even less than level. Add or subtract the rectifiers to get the speed that you want.
Thanks for posting that. I’ll make two further suggestions:
Tap the level voltage off the same string as the downhill voltage, but closer to the transformer, to save components.
Use heavier rectifiers, like this one: http://www.radioshack.com/product/index.jsp?productId=2062583
What would happen if one were to impress a slightly higher constant DC voltage, say from a battery, to the uphill section? How would the transformer react?
The RMS voltage across the gap would then be the root-sum-square of the battery and transformer voltages, rather than simply the difference between the individual voltages. For example, a transformer at 14 volts and a battery at 16 volts would result in a gap voltage of 21.3 volts rather than the 2 volts that a 14-volt and a 16-volt transformer would produce. The fault current would therefore be much heavier. Furthermore, the DC component from the battery would greatly exacerbate the arcing problem. The zero-voltage axis crossing that occurs in AC circuits 120 times each second tends to extinguish the arc quickly; but the arc tends to persist in DC circuits.
If you’re thinking instead of putting the battery in series with the transformer, just to boost the transformer voltage a couple of volts, you will need a much bigger battery than you may imagine. For example, to add 2 volts to a 14-volt transformer output, you need to use a 7.7-volt battery. In this case, the fault current will not flow through the transformer, only through the battery. But a dead short across a 7.7-volt battery when the train passes may not be a pretty sight.
If it helps anyone, here’s what I did.
I have four insulated blocks. Two level portions, with one at the top of a grade and one on the bottom. There is one up-grade and one down-grade. I bought two ancient Lionel rheostats. From the transformer, the power wires run straight to the up-grade. That is where the highest speed needs to be. From the main power wire, one rheostat steps the power down and powers the two level sections. A separate rheostat then runs to the down-grade, stepping the power down even further. the whole layout runs on just a single tap to the transformer. No arching concerns.
This works best for me because it allows me to tweak each section if I change trains or consists. And it was cheap. You can find those old rheostats for about $5 each.
Right on!
I use two different transformers from level to up hill using the “mountain section” as a block. What size resistor should I use to connect the two center rails together? I have not experienced , as yet, any problems with the transition but reading your response I wish to protect the equiptment.
Thanks,
Cliff
jjm’s suggestion to use a Lionel rheostat is a good one. You can adjust it as needed.
I would use a bridge-rectifier string myself (if I had any grades). It is also adjustable, in half-volt steps, and produces the same voltage drop no matter how much current the train draws.