Is one ZW transformer enough for 2 blocks?

joeymanko

New Member
I am relatively new to Command Control. I have 2 blocks which are both 10 feet long ovals and are insulated. Could I put one of the ZW's outputs to one block, and another output block? Would this be sufficient power to run 2 trains?:cry:

It is very confusing to mewall1. I know that I could always put a powerhouse 180 to the inside block, but then why did I spend all that money on a ZW? After all, a ZW has 4 outputs!

Joey
 

electric130

New Member
What is a ZW? If you give us more detail, we might be able to help.
everyone knows what a ZW is. :mrgreen:

Lionel, Product Finder

6-32930_891.jpg
 

joeymanko

New Member
The ZW

What is a ZW? If you give us more detail, we might be able to help.


Hello Railway Bob!

Thank you for taking the time to at least anser my quest. The ZW is the big Lionel transformer. They still make it today, but it is more 21st century. It is 250 volts, and has controllers for 4 trains. It kind of looks like a football with a box in the middle of it.

It seems if I put one output to one block and another ouput to another block, there just isn't enough power. I would think that it would act as 4 transformers -- that is why I have one:confused::confused:. Maybe i'm wrong.

Thanks,

Joeymanko
 
Hello Joeymanko. Sorry I can't help you on this one as I'm only into DCC.

From what I have seen on some layouts with the ZW, it seems to me that it should produce enough power to power at least two trains. And it Lionel has designed it to power 4 trains, then it should do this. Usually if there's a lack of power, it may have something to do with the transformers not being able to produce the current that is required (or it may be a problem with the locos).

Another place to look is the wiring to the tracks and the connections between the tracks. Use a volt meter to test the voltage to make sure it is up to spec.

Sorry I can't help you any more than this.

Bob M.
 

electric130

New Member
i have a loop that's 7 feet on one side and 11 feet on the other, so probably 33-34 linear feet around the track. i'm currently powering it with only a single 30w transformer. i have power fed into it at 3 different places. based on that, i doubt your ZW should have any issue running that small of a block. i'd check your connections. maybe take some resistance readings from the track.
 

joeymanko

New Member
Encouraging

i have a loop that's 7 feet on one side and 11 feet on the other, so probably 33-34 linear feet around the track. i'm currently powering it with only a single 30w transformer. i have power fed into it at 3 different places. based on that, i doubt your ZW should have any issue running that small of a block. i'd check your connections. maybe take some resistance readings from the track.

WOW!:thumb: if you are running 34 linear feet with only 30 watts, then there should be no problem with one ZW. I have heard that the newer TMCC engines draw much more power than the older ones. Maybe with two TMCC engines on the blocks, it is just not enough power. I will take some resistance readings and make sure there is power fed into at least 3 places.

Thanks, :wave:

JOEYMANKO
 

60103

Pooh Bah
The wattage has no connection with the number of feet of track. Watts will limit how many operating items you can have going at once: locos + lighted cars + operating cars + lighted/remote switches and then any accessories powered from the tracks.
The ZW at 250 watts (not volts!) has controls for 4 trains (If it's like the old ones) but remember that Lionel used to put 80 or 90 watt transformers in single train sets, so the ZW is only 3 times that.
If there is a problem with power related to distance, it is usually power loss at the track joints. For this, take some large wire and add an extra feed at the far end (or a couple of extra feeds).
A train with a single simple locomotive should take only 30 or 40 watts, but each lighted passenger car will add to the load; a 2-unit diesel with a long string of cars may pull quite a bit.
 
Some basics in electricity
Ohm's Law: Volts = Resistance x Amps (V=I*R)

Watts = Volts x Amps(W=V*I

By taking some simple voltage, resistance or amperage measurements, one can calculate watts, or visa versa.

Labels on electrical devices can be stated in a number of the above variables - watts, volts, amps - so that sometimes we end up comparing apples and oranges - eg is it 250 volts or 250 watts?

As a simple example, consider 10 volts going down the tracks from a 250 watt transformer. What is the amperage? Ohm's law tells us that W=V*I so a little bit of algebra will give us V=W/I. In this example 250 watts/ 10 volts = 25 amps

Now, take a look at the ratings of the circuit breakers in your house electrical panel. You will find that most circuit breakers are rated at 15 (amps).

That 250 watt transformer produces the power of two 100 watt lightbulbs and one 50 watt lightbulb. Now here's the question. Would you hold those three lightbulbs in your hand?

I doubt it.

One of the reasons for using such thick wire in DCC (14 AWG or 16 AWG) and using a track power buss (and it's true even of analog systems) is the high wattage that goes down the tracks when we short the tracks (derailment, running a switch, etc) Typically, we have 16 volts continuous in the tracks with a 5 amp output. This will produce 80 instantaneous watts if we short the tracks - enough to do some welding of plastic, rails, metal wheels, etc. We want the command station to detect this surge in wattage so that it will shut down immediately. This is done, in some ways, by using this heavy 14 - 16 AWG buss wire and 20 - 22 AWG track feeders placed every 3 feet or so.

Otherwise, if we are using 24 AWG telephone wire throughout, the command station might think that we are legitimately demanding more power to the locos and will keep on pumping out the extra watts. In which case, we could end up with some serious damage to our decoders.

Which is why, regardless of the size of wire throughout your layout, you should always test every foot of track on the layout with the coin test - short a coin across the rails and make sure the command station shuts down.

Anyhoo............ Just some thoughts on the idea.

Bob M.
 

electric130

New Member
As a simple example, consider 10 volts going down the tracks from a 250 watt transformer. What is the amperage? Ohm's law tells us that W=V*I so a little bit of algebra will give us V=W/I. In this example 250 watts/ 10 volts = 25 amps

250w is the max capacity, not what it puts out constant at all times. if a given locomotive takes 40w to pull a given load, and you're running it at 10v, then the transformer is only going to be putting out 4A at that time, not 25A. now with AC, it's not just simple P=VI. there's other things in there too, but for our concerns, P=VI is usually fine.
 

joeymanko

New Member
Some basics in electricity
Ohm's Law: Volts = Resistance x Amps (V=I*R)

Watts = Volts x Amps(W=V*I

By taking some simple voltage, resistance or amperage measurements, one can calculate watts, or visa versa.

Labels on electrical devices can be stated in a number of the above variables - watts, volts, amps - so that sometimes we end up comparing apples and oranges - eg is it 250 volts or 250 watts?

As a simple example, consider 10 volts going down the tracks from a 250 watt transformer. What is the amperage? Ohm's law tells us that W=V*I so a little bit of algebra will give us V=W/I. In this example 250 watts/ 10 volts = 25 amps

Now, take a look at the ratings of the circuit breakers in your house electrical panel. You will find that most circuit breakers are rated at 15 (amps).

That 250 watt transformer produces the power of two 100 watt lightbulbs and one 50 watt lightbulb. Now here's the question. Would you hold those three lightbulbs in your hand?

I doubt it.

One of the reasons for using such thick wire in DCC (14 AWG or 16 AWG) and using a track power buss (and it's true even of analog systems) is the high wattage that goes down the tracks when we short the tracks (derailment, running a switch, etc) Typically, we have 16 volts continuous in the tracks with a 5 amp output. This will produce 80 instantaneous watts if we short the tracks - enough to do some welding of plastic, rails, metal wheels, etc. We want the command station to detect this surge in wattage so that it will shut down immediately. This is done, in some ways, by using this heavy 14 - 16 AWG buss wire and 20 - 22 AWG track feeders placed every 3 feet or so.

Otherwise, if we are using 24 AWG telephone wire throughout, the command station might think that we are legitimately demanding more power to the locos and will keep on pumping out the extra watts. In which case, we could end up with some serious damage to our decoders.

Which is why, regardless of the size of wire throughout your layout, you should always test every foot of track on the layout with the coin test - short a coin across the rails and make sure the command station shuts down.

Anyhoo............ Just some thoughts on the idea.

Bob M.


Dear Bob:

You are "THE MAN"! Simple but genius. I guess I should be taking more readings with my new multimeter. In addition, I need to recheck all the wiring. I do have 3 feeders on every block, but it is possible that the gauge of the wire might be too thin.

You are a real pro.... Thank you so much

Joeymanko
 
Hey Joeymanko. No pro, and definitely not a genius. I picked this up from other discussion forums over a long period of time. Which is why it's sometimes a good idea to participate in a discussion forum such as The Gauge.

It's surprising what one can learn. Just remember to "pass it on" so that everyone else has the chance to learn.

Bob M.
 
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