Reducing Engine Speed to Allow Smoke Unit to work

[chieffan]

Has anyone come up with a good way to reduce the engine motor speed in order to have the smoke unit operate most of the time?  I did away with the standard bulb headlamp and put in a large bright white LED powered by 2 AA batteries in the belly hold on my F unit.  A switch on the side turns the light on and off.  The 2 batteries should last the year with no problem.

Now I would like to have smoke about all the time.  Not sure which way to go.  A resistor in line to the motors, or a 9V battery to power the smoke unit.  Not sure 9V would be enough power or how long it would last.

Would appreciate any thoughts or a solution to this.  Once the F unit is taken care of will move on the Heavy Hauler steam unit.

Roger

Thanks much.

[NarrowMinded]

Its simple you just need to add a resister inline with the motor

This will slow your top speed but the run too fast anyway.

NM-Jeff

[Chuck N]

I may be wrong, but most modern engines have a voltage regulator on the circuit board for the smoke unit.  I don't think that anything you do will put more volts into the smoke unit.  If you should succeed, you will probably burn out the unit.  

Most smoke units are anemic.  I always turn off the smoke unit if it is possible, if not I cut one of the wires.

Chuck

[chieffan]

Chuck, your right in that adding more power to smoke unit would probably burn it out.  I don't need more power to the smoke unit, I need less power to the motors while maintaining the power to the smoke unit.

Bachmann smoke units are relatively tough as far as burn out is concerned, as long as one stays in the proper power range.  If you race the engine with no fluid in the smoke unit, yep it will go up in smoke.

I would guess that a 1/2 watt resistor in line with the motor feed would be heavy enough.  What resistance (ohm) would be a good starting point?  The shell isd a real pain to remove and install for to much trial and error.

Thanks much for your comments and suggestions.

Roger

[NarrowMinded]

I would start low like 330 ohm if you have low ohm resistors you can jus add them in series till your happy.

[Loco Bill Canelos]

Trying to think through this, I am thinking a resistor would help but not that much. A better way would be some kind of electronic inverter which could up the power to the smoke unit rather than reduce the power to the motor.

Considering the short time it takes to use up the smoke fluid, raising  power to the  unit would use fluid up even faster.  Seems to me that you might end up constantly adding fluid.

No doubt smoke users could use a better smoke unit design which might provide for more voltage electronically while building in some kind of fluid feed so that you wouldn't have to keep adding drops so often.   

Any of you electronics guys have any ideas?

Bill

[Joe Satnik]

Dear All,

Diode bridge (plus single diodes) to drop motor voltage:

(Diodes work better than resistors.) 

http://www.dallee.com/PDFs/MotorDiodeDrop.pdf

Works for DC or AC (universal) motors.

Hope this helps.

Sincerely,

Joe Satnik

[chieffan]

I am not an electrical engineer, but diodes regulate the direction of flow of electricity.  That is how they regulate if the front or rear light is illuminated on a loco.  I fail to see how a diode bridge or any other diode configuration will lower voltage any noticeable amount.  That is not what diodes are designed for.

As for running out of fluid, the loco I am working with has a rather large fluid tank, about 3" long, 1/5" wide and 3/4" deep so it would run for some time before it ran out.  Now some of the Bachmann smoke units would only hold few drops and are a pain to try and keep fluid in them.

I am going to try running the smoke unit with 9V from a battery and see what happens?

Thanks for all your input.

Rog

[671]

Hi chieffan,
         
           671 here. Joe Satnik's answer is the best way to reduce voltage ( Electrical pressure ) to your motor/motors.
           My scale is 'O'. I have several of the W by B Steam locos. The problem with these locos is that their motors require relatively low voltage to obtain scale like speed.
           Now the challenge is this...The smoke unit needs a track voltage of approximately 9 + volts in order for the smoke unit's voltage regulator to provide the design output voltage of 7.9 volts for the 8 volt Seuth Smoke unit. Where did that voltage difference go? It was lost to the electronics within the loco's electronics.
           Now, a diode helps in controlling the directional flow of the current. But diodes are slightly parasitic. What I mean is that they consume some of the voltage that is flowing through them. They 'steal' about .6 to .7 volts for each diode the current flows through.
           Using this very constant voltage drain, allows us to drop voltage to any circuit within our locos.
           I use the Dalllee bridge rectifier circuit along with two extra diodes. This drops my motor voltage by almost three volts. This allows me to increase track voltage by approx 3 volts to run at the same speed. The smoke circuit now receives almost three more volts. My smoke units smoke like they are on fire, yet my loco runs at a closer to scale speed.
           Thank you, Joe Satnik for the slower speed and tons of smoke.

                              Smok'em if you got'em...671

[chieffan]

Well, we live and learn don't we.  At least part of the time.  Thanks for that explanation 671.  I didn't realize that a diode would drop the voltage that much.  Going to order some diodes and get the soldering iron hot.  Cheaper to pay the shipping on them than it is to drive 70 miles to a RS store to get them.

Roger

[NarrowMinded]

I still dont see the  benefits in using the diodes over a resistor, once you determine the ohms needed you simply add a single resistor.

If your worried about heat all you have to do is use one rated for higher wattage.

This has the advantage of less parts to fail and provides the exact same results, lower motor speed and  higher voltage to the smoke generator.

Just my opinion...
Nm-Jeff

[671]

Hi Narrowminded,

                 671 here.

The voltage drop across a resister is dependant upon the voltage and the current.

The voltage drop across the diode is always constant .6-.7 voltage drop.

So in conclusion if the loco is not pulling a load, the voltage drop across the resister is quite different than if it were pulling 20-30 cars up an incline.

The diode voltage reduction remains the same no matter how the voltage (throttle) changes or current (load).

The resister setup would 'pull' more power from the transformer than the diode setup for the same results.

Equal load and speed, diode reduction uses less current (watts) with a constant voltage reduction between input and output.

                                                          671

[Chuck N]

I'm not an EE, if there is a resistor or diode in the line to to the motor it will drop the voltage going to the motor slowing down the locomotive for a given voltage coming out of the power supply.  That voltage drop, I think will be dissipated as heat and will not add more electrons to the smoke generator.

I think that Chieffan just wants to slow the engine down, but keep the smoke output as is, not add watts to the smoke generator.

Chuck

[NarrowMinded]

Chuck,

The idea behind both uses is to get the voltage to the smoke generator high enough to produce the max amount of smoke like is produced at full throttle but not having the loco motive move at full throttle speed.

So putting a resistor or diodes inline between the motor and pickups will slow the motor,
The smoke generator will not be affected so it will see all the voltage an  thus produce more smoke.

Nm-jeff

[Chuck N]

Jeff:

We are now in agreement and probably have been for a while.  It is just that I didn't say things as clearly as I thought I had.

Chuck