Trains.com

Subscriber & Member Login

Login, or register today to interact in our online community, comment on articles, receive our newsletter, manage your account online and more!

Controlling Staging Tracks

1624 views
7 replies
1 rating 2 rating 3 rating 4 rating 5 rating
  • Member since
    November 2007
  • 8 posts
Controlling Staging Tracks
Posted by befudld on Saturday, July 4, 2009 11:23 AM

On page 41 of The DCC Guide by Don Fiehmann there is a circuit diagram to use a push button for powering staging tracks.  I made one of these and it works like a champ powering the track but the LED doesn't work.

I used a 2.5V LED, should it be a 12V?  I hooked the "Rail A" side to the circuit, is the schematic incorrect and both sides need to be wired to the circuit?  If so, how?

The only thing I have tried thus far is reversing the polarity of the LED.

Any help would be appreciated.

Aksel

Tags: DCC , LED
  • Member since
    February 2002
  • From: Reading, PA
  • 30,002 posts
Posted by rrinker on Saturday, July 4, 2009 4:52 PM

 Not having the book, I can only guess at what the circuit is, but if the LED gets full DCC track voltage across it, then the problem is you have fried it. An LED needs a dropping resistor 99.9% of the time (note to the other electronic types - if the circuit you hook the LED to already has a resistor, then it has a dropping resistor, doesn't it?). LEDs are current devices, not voltage. Yes they need at least a certain voltage, that's the rating given for them, but it's the current that controls the brightness. The relationship between voltage and current is probably he most critical thing to learn in electronics. Once you have that figured out it all makes a heck of a lot more sense. If you are suffering from insomnia some night, google "kirchoff's laws" and read one of the many explanatory web sites. This is the background on why dropping resistors are needed and how they work,.

 At any rate, to use a typical LED with HO/N DCC voltages, you need about a 1K resistor in series with the LED to keep the current levels to a point where the LED will glow nicely but not burn out. LEDs also do not like large amounts of voltage in the reverse direction, which happens about half the time with DCC - I would recommend a type of LED called a bicolor LED - the version with 2 terminals looks just like an ordinary red or green LED but inside it is two LEDs wired back to back - usually one red and one green, so on DCC it usually looks somewhat orange. Also the polarity of these or even an ordinary LED used with DCC does not matter. The polarity of an LED used for DC or as lights on a DCC decoder DOES matter because an LED will only pass current in one direction, just like an ordinary rectifier diode.

                                       --Randy


Modeling the Reading Railroad in the 1950's

 

Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.

  • Member since
    November 2007
  • 8 posts
Posted by befudld on Sunday, July 5, 2009 2:29 AM

Randy,

Thanks for the response.

 There are two resistors in the circuit, both 1K ohm.  One in series the other in parallel to the LED and diode.  The original LED was .23mA.  These should have been enough.  Have not tried a bi-polar.

Aksel

  • Member since
    February 2002
  • From: Reading, PA
  • 30,002 posts
Posted by rrinker on Sunday, July 5, 2009 11:55 AM

 Not sure why there would be a resistor in parallel with the LED - that isn't goign to work for any sort of basic indication that I can think of, however since I haven't seen the circuit I have no idea exactly what is trying to be accomplished. Is the idea for the LED to be on when you push the button to apply power, or is there some other function?

                     --Randy

 


Modeling the Reading Railroad in the 1950's

 

Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.

  • Member since
    November 2007
  • 8 posts
Posted by befudld on Monday, July 6, 2009 1:35 PM

The circuit is meant to operate like a wall switch that glows when the room light is off.  When the pushbutton is pressed power is restored to the track and the LED goes out.  Presumably the LED should light to indicate a locamotive has reached the isolated track of a switch lead.

I drew the circuit in Word but something wont let me paste it here.Banged Head

Aksel

  • Member since
    November 2007
  • 8 posts
Posted by befudld on Monday, July 6, 2009 2:53 PM

Here is a feeble atempt to draw the circuit.  Hope it makes sense.  The resistors are rated at 1/2 Watt.  The arrows show diode polarity.  Due to size constraints the isolated track should be on the other side of the pushbutton.

 

DCC SYSTEM--------------I------------------- NORMALLY OFF  ----------------------I------------ISOLATED TRACK

                                   I                     PUSHBUTTON                           I

                             1K OHM                                                                I

                            RESISTOR                                                             I

                                   I---------I-------- 1K OHM RESISTOR -----------I        I

                                             I--------------- LED I>I------------------I-----I--------I

                                             I--------I<I 1N4001 --------------------I

                                                           DIODE

  • Member since
    February 2002
  • From: Reading, PA
  • 30,002 posts
Posted by rrinker on Monday, July 6, 2009 9:15 PM

 OK, bi-color LED not needed, that other diode handles the reverse current and protects the LED.

 Powered loco on tracks, button NOT pushed, loco gets DCC voltage dropped by 2K of resistence. LED sees voltage across the second 1K resistor - I think this will probably result in not enough voltage and/or not enough current for the LED. Without the resistor that is parallel to the diodes though, the loco will probably get enough power to move - although the signal will probably be ragged because an LED and a 1N4001 generally do not switch on and off fast enough for the DCC signal. Hmmm. I can see what is trying to be done but I'm not sure it will work exactly as designed.

                                     --Randy

 


Modeling the Reading Railroad in the 1950's

 

Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.

  • Member since
    November 2007
  • 8 posts
Posted by befudld on Sunday, July 12, 2009 3:23 PM

I rebuilt the circuit, must've done something wrong.  Circuit works as advertised.  Thanks for the input anyway.

Aksel

Subscriber & Member Login

Login, or register today to interact in our online community, comment on articles, receive our newsletter, manage your account online and more!

Users Online

There are no community member online

Search the Community

ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT
Model Railroader Newsletter See all
Sign up for our FREE e-newsletter and get model railroad news in your inbox!