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I have three 1122 switches on my layout. All 3 work fine with lights, and switching mechanisms.
The one problem I have is that the straight stretch on one of them has no power at the end, thus the engine hits the dead spot and stops.
- I can run the engine into the turnout and take the curved section and all is fine, cuz power is there.
- If I go straight, I have power going in, but after the mid point, there is no power and the Engine stops.

I put the meter on the curved section and it has power, but the end of the straight section has no power.
There is power on the actual switch plate, but nothing going out from there to the straight section.

This 1122 has the outside rail, of the section where the problem is, with a gap in the rail and then 1" piece of rail that goes to the end.

Insulation pins are in proper position.

Any ideas on why or fixes??
Thanks.

Mr. Bill
 

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Bill, welcome to the forum! I'm not familiar with your 1122, but basics are basics. Unless you are using fiber track pins somewhere in your setup, your switch should receive power from tracks going into and and coming out of it. Yours seems to be only receiving power going into the turnout, but nothing at the other end of the straight segment. That suggests that your turnout is not providing power on the exit end of the straight section. It also suggests, however, that you are insulated at the far end from the next section of track.
My first suggestion would be to see if the track you exit to is receiving power: if it is, check the pins connecting it to the (exit) rails of the turnout. They should be metal and they should be clean and tight. If you have no power on the track immediately after the straight section of your turnout, you've identified one solution to your problem: get some power to those tracks! It doesn't repair what is wrong with your turnout, but it delivers the power from the other end and circumvents the problem.
 

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OK, guys, here we go on 1122 switches. I recently repaired a bunch of them and found problems inside all of them. Some of them had high resistance connections that actually melted the plastic.

What is necessary to fix them is to remove the bottom plate and solder all the electrical connections. To remove the bottom plate, you must drill out the rivets. The V in the center of the frog is riveted to the bottom plate, and to allow it to be held in place, you must drill and tap a hole in the frog. I use a 4-40 tap and a 4-40 filister head machine screw. Use WD-40 as cutting oil. Otherwise you may break the bit or tap. Don't ask me how I know these things. The frog is die cast zinc. Drill out the rivet for the ground terminal and for the other rivet that is inside the switch motor housing. Bend the tabs for the outside rails up so you can remove the bottom cover. Inside you will find some crimped electrical connections, and if you have a meter, you can discover which one is bad. However, solder all of the connections. Otherwise you will be taking your switch apart again in the future. The plastic that the switch is made from is soft and cold flows over time and so the crimps become loose. Everything inside the switch that is connected to the center rail should be soldered together. The center rivet in the fat center rail should be soldered to the rail. This rivet is the electrical path for power to the other center rails and the switch motor. Make this solder connection as smooth as possible because a mound of solder will push the pick up roller to the side and short it against the movable points. Check your work with a meter before you put the bottom cover back on.

When you put the bottom cover in place, the first thing to do is solder the ground terminal to the bottom cover. This needs to be tight so it makes a good connection with the ground tab coming out of the switch motor housing. I used a pair of channel locks to clamp the screw in the hole so I could solder it. This can be done with 2 hands, but it is easier with 2 people unless you have a 3rd arm. You need to polish everything before you solder it so the solder will stick. Solder the tabs for the outer rails to the lower plate at one end of the rail. I didn't replace the rivet inside the switch motor, but you can replace it with a #4 machine screw if you want to.

The terminals use a 4-36 thread, so if you use a 4-40 machine screw to replace the ground terminal, the thumb nut won't screw on. You could buy some 4-36 machine screws and nuts to use for this terminal rather than soldering the terminal in place.

There are two types of 1122 switches. The earlier type uses fiber pins, and the later type does not need fiber pins. The early type switches use fiber pins in the short outside rails. These rails are used for the non-derailing feature. On the new switches, the long outside rails are used for the non-derailing feature. There is a cut in the long outside rails about an inch from the end. The short end of each rail is connected to the lower plate and they are also connected to the outside rails on the end where the movable points and are connected to the short outside rails that go to the frog. You need to make 6 solder connections to solder all of the outside rail pieces to the lower plate.

Bob Nelson on trains.com has a modification to these switches so that you can supply constant voltage to them and the switch motor is operated by a capacitive discharge circuit. IMHO, this is a very good modification and should be considered for a permanent layout. He went through the modification in great detail for one of the other members including indicator lights on the layout control panel.
Bruce Baker
 

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Spectacular, servoguy. Thanks for taking the time to give a detailed description.:thumbsup::thumbsup::thumbsup:
 

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I have found the 1122 to be good inexpensive switch. So inexpensive that the obvious solution is to switch it out.
Once out give it the Bruce makeover, nice reply, it is appreciated.

I have collected boxes of these for around 20, 25 each. Some I have gotten for as little as 3 bucks. Every one had a good coil.
 

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I have 2 marx 034 and had one that didn't switch all the way one direction so cleaned and lubed it it works fine now it has a screw in bulb thats burned out what voltage would that be?
 

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TJ, I think your advice about using an 18 volt bulb is good. For my 022 switches, I put a diode in series with the lamp and run the switches off the 20 volt fixed voltage tap of a KW transformer and I use 18 volt bulbs. The diode reduces the 20 volts to about 14 volts, and the lamps look a lot more realistic. Running an 18 volt bulb on 20 volts makes the lamp too bright, reduces the life of the bulb, and melts the lantern. It is unlikely that the track voltage is going to get above 14 volts.

Bruce Baker
 

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Bruce,

Thanks... but educate me a bit, if you have an extra moment ...

We're talking AC power here, right? You say "diode" ... So is that essentially eliminating half of an AC wave? I would have thought a resistor to impose a voltage drop, yet still keep a full AC wave.

I ask all this very naively on my part ...

TJ
 

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TJ,
I have an MSEE and 47 years of engineering experience, so if go a little too fast, don't hesitate to ask a question.

You can drop the voltage with a resistor, but the resistor is going to be big and dissipate a lot of power. It may also get rather hot.

A diode will rectify the voltage, supplying half wave rectified DC power to the lamp. The diode doesn't dissipate much power since it is either on or off. It is superior to a resistor in this respect. I installed half of the diodes one way and half the other way to minimize the imbalance that would have occurred if I had installed them all the same way.

Even though the diode eliminates half of the AC wave form (only the positive or negative peaks get through the diode), the voltage is not cut in half. The power is essentially cut in half (not quite true, but close enough for this discussion), and so the voltage is reduce to 70% of what it is without the diode.

While I am at it, reducing the voltage on the lamp extends its life significantly. Using the diode extends the lamp life by about a factor of 3000. Bob Nelson on trains.com has some information about this and a formula for calculating the lamp life. Find the thread 022/711 switch operating pblms. This is where my long post is on fixing 022 switches.

Bruce Baker
 

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TJ,
I have an MSEE and 47 years of engineering experience, so if go a little too fast, don't hesitate to ask a question.

You can drop the voltage with a resistor, but the resistor is going to be big and dissipate a lot of power. It may also get rather hot.

A diode will rectify the voltage, supplying half wave rectified DC power to the lamp. The diode doesn't dissipate much power since it is either on or off. It is superior to a resistor in this respect. I installed half of the diodes one way and half the other way to minimize the imbalance that would have occurred if I had installed them all the same way.

Even though the diode eliminates half of the AC wave form (only the positive or negative peaks get through the diode), the voltage is not cut in half. The power is essentially cut in half (not quite true, but close enough for this discussion), and so the voltage is reduce to 70% of what it is without the diode.

While I am at it, reducing the voltage on the lamp extends its life significantly. Using the diode extends the lamp life by about a factor of 3000. Bob Nelson on trains.com has some information about this and a formula for calculating the lamp life. Find the thread 022/711 switch operating pblms. This is where my long post is on fixing 022 switches.

Bruce Baker
HUH?:D
Sounds good to me Bruce?:laugh:

I know little about what your talking about.::D
But find your posts interesting.:thumbsup:

Question, Can you under-power your bulbs instead.
Say take an 18v and replace it with a 12v? Or will it blow the bulbs?

What would a 12v PS bulb be? What does the PS stand for?
I read that somewhere else about replacing a 16v with a 12v PS bulb.
 

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If you want to reduce the power for a light bulb, increase the bulb voltage, don't reduce it. If put 14 volts into a 12 volt bulb, the bulb will get very hot and will have a very short life.

I don't know what the PS stands for.

Here is a good discussion of light bulbs: http://www.three-rail.com/BULBS.HTM

The wattage for the bulbs is at the nominal voltage. If you reduce the voltage, the wattage will go down approximately as the square of the voltage.

Bruce Baker
 

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I usually buy bulbs at a swap meet. I have a friend in LA that has a huge selection of parts, and he has all types of bulbs, and his prices are good.

For a 022 switch running on 20 volts, even with the diode in series with the lamp, a 14.4 volt bulb is going to be a little too much. That is why I use the 18 volt bulbs.

Mike's Trains and Hobbies http://www.mikestrainsandhobbies.com/ sells light bulbs but you will have to email him to find prices as they are not on his web site. I am sure other parts suppliers also have light bulbs.

Bruce Baker
 

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Does anyone have the Marx 034 switches. I hooked these up to C and D KW190 14 volts I put the meter to the posts I had 12.5 volts. The switches work good at that voltage so back to the bulb maybe a 12 or 14 volt would work?
 

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I assume the O34 switch motors are isolated from the switch. The O27 Marx switches are isolated. If they are not isolated, then you cannot use the C & D terminals on the KW. I have never seen a Marx O34 switch. I didn't know until recently that they even made them.

I would suggest either a 14 or 18 volt bulb. The 12 volt bulb would definitely get too hot.

When I put the diodes in series with the lamps on the 022 switches, the lamps stayed cool enough that I could touch them without burning my fingers. They looked more realistic since they weren't so bright.

Bruce Baker
 

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TJ,
I have an MSEE and 47 years of engineering experience, so if go a little too fast, don't hesitate to ask a question.

You can drop the voltage with a resistor, but the resistor is going to be big and dissipate a lot of power. It may also get rather hot.

A diode will rectify the voltage, supplying half wave rectified DC power to the lamp. The diode doesn't dissipate much power since it is either on or off. It is superior to a resistor in this respect. I installed half of the diodes one way and half the other way to minimize the imbalance that would have occurred if I had installed them all the same way.

Even though the diode eliminates half of the AC wave form (only the positive or negative peaks get through the diode), the voltage is not cut in half. The power is essentially cut in half (not quite true, but close enough for this discussion), and so the voltage is reduce to 70% of what it is without the diode.

While I am at it, reducing the voltage on the lamp extends its life significantly. Using the diode extends the lamp life by about a factor of 3000. Bob Nelson on trains.com has some information about this and a formula for calculating the lamp life. Find the thread 022/711 switch operating pblms. This is where my long post is on fixing 022 switches.

Bruce Baker
OK, Bruce ... you've continued to perk my curiosity ...

Are you using one diode, two, or more? T-Man has explained to me how a bridge rectifier works, but that uses 4 diodes.

As I read your post, it sounds like you're using just one diode in series, which would allow only 1/2 of the AC sine wave to pass, yielding a sine voltage pulse for a short time, then no voltage for a short time, with that pattern repeating. So the voltage on the 1/2 sine wave that passes through remains as it was in the original AC circuit (the peak doesn't shift). But the quasi-DC "average" voltage of this half sine wave is about 70% of the peak voltage (i.e., area under the 1/2 sine curve). The net power is cut in half (as you said) because we're throwing away the other (not used) half of the sine wave.

Do I have that right?

And if so, and we've synthesized a quasi (albiet "pulsey") DC signal, are you saying that the bulb will operate OK and rather efficiently (from a lifespan standpoint) with our one-sided, pulsey power?

Sorry to chew up your time on this. No reason other than curiosity on my part.

Thanks!

TJ
 
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