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semaphore from overheating

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  • Member since
    January 2010
  • 8 posts
semaphore from overheating
Posted by LIONSAM on Monday, December 20, 2010 10:27 AM

Hello again how can I prevent an 

ORIGINAL 

LIONEL #151 AUTOMATIC SEMAPHORE 

 from being overheated? Its wired on lionel fastrack.

  • Member since
    December 2001
  • From: Austin, TX
  • 10,096 posts
Posted by lionelsoni on Monday, December 20, 2010 11:21 AM

It shouldn't overheat if it operates only intermittently, that is, when the train passes.  You don't have it wired to operate all the time, do you?

Bob Nelson

  • Member since
    November 2010
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Posted by gunrunnerjohn on Tuesday, December 21, 2010 9:38 AM

This accessory is designed to operate on 12V.  If your accessory voltage is greater than that, it can overheat.  You can also consider using an 18V bulb that will burn dimmer and also generate less heat.

  • Member since
    December 2001
  • From: Austin, TX
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Posted by lionelsoni on Tuesday, December 21, 2010 11:16 AM

A lamp with a higher voltage rating is not necessarily cooler than one with a lower voltage rating.  For example, the number 1445 is hotter than the number 53 (the 14.4-volt lamp that Lionel used in the 151) at any voltage.  A simple way to dim a lamp is to put a diode, like a 1N4001, in series with it.  This reduces the voltage by about 30 percent.

Bob Nelson

  • Member since
    November 2010
  • 993 posts
Posted by gunrunnerjohn on Tuesday, December 21, 2010 12:02 PM

No offense, but sticking a diode in seems harder than replacing the lamp with the 18V part. Wink

I'm not sure what bulb is in that signal, but a lower wattage one is certainly an option if it's available.

 

  • Member since
    December 2001
  • From: Austin, TX
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Posted by lionelsoni on Tuesday, December 21, 2010 12:28 PM

My point is that, while it's easy to swap lamps, the most popular 18-volt lamp will only make things worse, compared to the stock lamp.

Bob Nelson

  • Member since
    November 2010
  • 993 posts
Posted by gunrunnerjohn on Tuesday, December 21, 2010 4:07 PM

I'm at a loss as how you figure that.  The 18V bulb will be running below it's rated voltage and consuming far less power, hence less heat.  You can measure the current yourself and compute the difference in power dissipation.

  • Member since
    December 2001
  • From: Austin, TX
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Posted by lionelsoni on Tuesday, December 21, 2010 4:31 PM

The 18-volt lamp will be running below its rated voltage and consuming less power than it would at its rated voltage.  That says nothing about the power that it consumes compared to any other particular lamp, such as the number 53.

The number 53 runs cooler than the 1445 at any voltage.  Lamp ratings are a tradeoff among voltage, current, efficacy, and lifetime.  The ratings of the two lamps are

  #53  14.4 V  120 mA  1728 mW  1    mscp  1000 hr
  #53  18   V  136 mA  2442 mW  2.18 mscp    69 hr
#1445  14.4 V  135 mA  1944 mW   .7  mscp  2000 hr
#1445  18   V  150 mA  2700 mW  1.53 mscp   137 hr

The numbers in italics are values that I calculated using the rules that for incandescent lamps current varies as the .55 power of voltage, light output as the 3.5 power, and lifetime as the -12 power.  The 1445 is actually rated at 14.4 volts and 18 volts; but that doesn't give it any particular advantage over the 53 in terms of power dissipated at either voltage, as you can see.

For screw-based lamps, the number 52 is cooler than the 1447:

  #52  14.4 V  100 mA  1440 mW   .75 mscp  1000 hr
  #52  18   V  113 mA  2035 mW  1.64 mscp    69 hr
#1447  14.4 V  135 mA  1944 mW   .7  mscp  2000 hr
#1447  18   V  150 mA  2700 mW  1.53 mscp   137 hr

Bob Nelson

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