smoking caboose -command environment

Install  3 amp diodes in series,one from each pickup roller pointing in the same direction. Dropping voltage with diodes is explained here

LINK 

For ease of mounting you can also use 3 leads of a 4 amp bridge rectifier. Shown here last 2 diagrams in a different application. Do not add a capacitor. 

LINK2

You can also convert to LED lighting and put a single diode 3 amp in series to the smoke unit. 

Caboose LED lighting is shown here

LINK3

Dale H

In addition to the diode method(s), to your original question, yes you can insert a power resistor in series.  The "what" depends on how much voltage/current you need to reduce.  For example, do you know the heater resistance in your smoke unit?  It's typically something in the few tens of ohms.  And have you experimented to see what track voltage is low enough to prevent overheating while still producing sufficient smoke?  We then pick a suitable valued power resistor.  It might be, for example, a 10 ohm 5 Watt resistor.  Many options for "where" - $1 on eBay free shipping from Asia, or less than $1 but $3-4 total with shipping US mail order (like DigiKey.com or Mouser.com), or even Radio Shack has a small selection of power resistors for $2-3.

Obviously inserting a resistor or diode "in series" reduces the amount of smoke you generate.  So if you actually need/want the volume of smoke that 22V (or whatever) generates, then consider cycling the smoke unit on and off.  There are many ways to do this such as a thermal cut-off switch ($1-2) or cycle-timer module that you set the on-time to X minutes and off-time to Y minutes.

A series resistor works,but it is not as efficient as a single diode. A resistor equal in value in series to the smoke resistor would generate an equal amount of heat which you would still have to be shed in the caboose. The single diode blocks half the AC wave and generates little heat. The only heat shed would be the forward drop of about .6 volts,so wattage used would be .6 times the amperage drawn by the smoke unit. This would be much less than a series resistor.  In command the smoke unit would get about the equivalent of 9 volts RMS.

Dale H

I have a real nice, at one time, Kline NYC Bay Window caboose,  Just loved the color and marker lights,  got this about 1999.  Was new to TMCC and didn't realize that there was a voltage limit to safely run with out damage,  short story  TOO  LATE..

Wow Martin, I feel for you.. that was a nice caboose..

rockstars1989 can you show us which way you went with this problem.

I would like to be able to run my caboose with smoke on and not melt it like Martin did.

Dale put links up but the amount of wires in the photos blows my mind...

thanks

Something the manufacturers should be addressing what with constant voltage trackage. Looking at the 5 manuals that were included with the smoking cabooses, not one gives a warning when used with command control.

Many were made before command control was popular. K-Line later put on command/conventional slide switches. The fault was with companies introducing command and not warning users about retrofitting cabooses and passenger cars. K-line passenger cars with stream lighting can get very hot with 18 volts applied. A simple diode or bridge rectifier in series fixes the problem in both passenger cars and cabooses.  K-Line was also the only company to address voltage spikes,offering a lock-on with a TVS.

Dale H

Dale H could you show a simple drawing of how to fit the diode to make our cabooses safer.

I did look at one of the photos you posted but there was more wires than I could follow.

thanks

Cut one wire to the smoke unit ,splice in a diode in series, polarity unimportant, insulate and you are done.

Or, Install a 4 amp bridge rectifier. You use 3 of the 4 leads. Each roller wire goes to one of the AC leads. The + or the minus lead goes to the lighting (center rail power). On passenger cars with bulbs,alternate use of the + and - between cars. In the link,just look at the bridge and 3 wires,forget the rest. If this is not clear,I will draw something up later,I am not at home now.

Dale H

The diode is the easiest and safest way to go, using a power resistor will work but then you will have to deal with the heat generated by the resistor. Remember, a resistor is doing work so while it doesn't have any moving parts to it has to do something with the energy so it converts it to heat. A diode technically isn't doing any work it is blocking the negative half of the wave form cutting the voltage in half

Amazingly, the lights marker and smoke unit still work.  Lookes like I will be scrapping this out...

Marty

If starting from 18V command voltage the single-diode method applies the equivalent of about 12V RMS.  That is, the single-diode method applies 18V AC half of the time which (for AC) is not the same as half of the voltage.

Perhaps semantics, but as above, the single-diode method does not cut the voltage in half.  It cuts the power in half.  If you cut the effective RMS voltage to a smoke unit in half you only get a quarter of the power.  This is because heating is proportional to voltage-squared.

However, all equations aside, if the single-diode method effects a suitable level of smoke that does not melt the frame then that's a good starting point.

When the OP used the value of 22V, I'm thinking worst-case where a transformer throttle got bumped applying more than 18V to track.  In such a case, the single-diode method would still apply some 15V RMS into the smoke unit. I have no idea if 15V would melt a caboose - though 15V is arguably a high conventional control voltage and may have over-heating issues to the smoke unit if sustained.

The resistor method would require additional measurements/work to identify a suitable value...but it provides settability that the single-diode method does not. The added resistor would indeed dissipate some heat (say, a few Watts) but could be located away from the roof/walls and perhaps even fastened to the steel chassis/floor if the caboose is so constructed.

The diode cuts off half the sine wave,thus it cuts effective RMS in half plus the .6 volts forward drop of the diode. The peak voltage is the same,minus the forward drop of the diode. Here is a useful link which explains

LINK

Just to add

If voltage is still to high, each additional diode in series will drop voltage.

If voltage is to low,with the shell off and unit sitting on track at 18 volts,place a good volt meter across the 2 leads to the smoke unit. Then add different small value capacitors, try 1 uf to 10 uf, 35 volt or more in proper polarity (if they are polarized) to start. This will remove some of the ripple and raise RMS.

Dale H