Powering 12 VDC OO Gauge 2-Aspect Colour Light Signal Via DCC accessory bus

[CDRC]

Hi,
I'm at the stage of adding a number of 3 wire (black, red and green), 2 aspect colour light signals to my DCC layout. I shall be connecting them to DCC Concepts IP Digital turnout motors which in turn are powered by a stand alone accessory bus. I could power the signals via a separate 12vdc power pack or spare Hornby DC controller. This would however, add more wiring to the underneath of the layout somethimg I would like to avoid If possible. I would, if it could be done, like to add them to the accessory bus, thus reducing the wiring. I have research the issue with litte succes but a lot of confusion. I have established it can be accomplished by placing 2 diodes; wired in inverse-parallel.....,!!!!!!

[Chrissaf]

Now it really depends upon whether the LED signals you have purchased have been purchased as a kit that you build yourself or whether they have been purchased pre-built 'ready to run'.

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If they have been purchased as a kit, then you will have the opportunity to modify the circuit design, change components and add new ones. If they were purchased ready made, then any modifications are probably going to be too onerous to contemplate doing, if they are built with an integrated (rather than external) current limiting resistor.

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Firstly, you need to understand the additional electrical considerations needed to run the signals off of the DCC signal voltage, whether that be DCC track voltage or the DCC Accessory Bus voltage. In this context 'Accessory Bus' does not mean a general purpose power supply for accessories.

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DCC is Bi-polar which means that it has a square-wave voltage that swings both positive and negative relative to a zero volt line. It is not really accurate to call it AC, but it is a form of AC. Pure AC is usually considered to be sinusoidal rather than Bi-polar.

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Recommended modifications are needed to power LED signals using DCC power if applying 'best engineering practices'.

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The signals would most likely have been designed to work on 12 volts DC and not 28.8 volt Bi-polar. Thus if  'best engineering practices' are applied, then some modification needs to be performed to the original signal electrical design.

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LEDs can be damaged by excess reverse bias voltages, thus it is good engineering practice to include the 'inverse diode' LED protection that you have mentioned. Credit to you for taking your research that far and finding this reference for yourself. Let me explain this concept further.

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A silicon diode has a forward bias of typically 0.6 volts. A silicon diode has a tolerance to 'reverse bias' that greatly exceeds that of an LED which is also technically a diode. Thus if you place a silicon diode in reverse in parallel with the LED, then the reverse bias across the LED can never exceed 0.6 volts.

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This drawing will explain this better. Only the Red half of the 2 Aspect signal is detailed in the explanation text. It is assumed that your signals are made 'common cathode' rather than 'common anode'. Common Cathode is more prevalent in the UK market.

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/media/tinymce_upload/bc129ee209cd6615ef5165228521b3c4.jpg

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In the top drawing, the polarity of the voltage supply is the right way round and the Red LED is lit. This will create a stable voltage across the Red LED that is equal to its 'forward bias voltage' which is typically about 2 volts as indicated. Thus the 'reverse bias' of the silicon protection diode can never exceed this 2 volt value.

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Now because the supply voltage is DCC Bi-polar, the supply will also swing to reverse the supply voltage polarity as shown in the lower drawing. In this drawing, the silicon protection diode is now biased in the forward direction and a stable voltage of typically 0.6 volts is created across it. Thus the reverse bias voltage across the LED can never exceed this 0.6 volt value.

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If the silicon protection diode was not present, then the reverse bias voltage across the LED could potentially reach the DCC peak to peak voltage of 28.8 volts. This is in excess of LED recommended maximum values. The LED may not be instantly damaged, but it will be stressed and could have premature failure. The only saving grace, is that because the DCC Bi-polar signal has a nominal base frequency of about 7 to 9 Khz, these excessive LED reverse bias voltages are but brief pulses. But even so, it is best engineering practice to include the inverse protection diodes.

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These protection diodes MUST be connected on the LED side of the current protection resistor. This can be achieved if the resistor is fitted externally to the signal. However, if the signal is made with the resistor contained directly inside the signal hardware, then access to the LEDs for adding the protection diodes is not practical. It is also not practical to install the diode inside the signal either due to the space they take up.

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Just as an aside and for information. The latest Hornby LED based R406 signal includes these inverse protection diodes inside the signal at the factory because Hornby market the R406 for both DC and AC supplies.

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As well as including the protection diodes, I personally would increase the value of the current limiting resistor [R in the drawings] as the DCC Bi-polar voltage is higher than a pure 12 volts DC supply.

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Although I have given guidance on what needs to be done. Personally, if it was me, I would run the signals on a dedicated suitable 12 volts supply.

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IMPORTANT:

This is an extremely long reply and includes images. Please do not try using the 'Blue Button with White Arrow' to reply. Scroll down to the 'Reply Text Box' and place your reply text in that box and click the Green reply button.

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[96RAF]

I have made signal lights using 3-legged common cathode dual colour leds. These were designed to work from a separate DC power supply but they will work on DCC direct by connecting the cathode to one rail and by way of a change over switch connecting each anode to the other rail.

 

I originally used a separate DC power supply and Peco PL-13 switches mounted to my underboard point motors, but due to use of surface mounted motors I have powered them direct from the rails using a microswitch operated by the points tie bar to switch colours.

As they are dual leds, effectively back to back diodes, they self protect each other. I have not fitted additional diodes per standard recommendation (see Chris’s diagrams) and my test articles have survived hundreds of hours of use on DCC, no doubt due to the fact I have heavy current limiting resistors installed to control the brightness.

 

In your case you can use the point motor built in switch.

[Chrissaf]

The key sentence in Rob's reply is:

"As they are dual LEDs, effectively back to back diodes, they self protect each other."

thus additional protection diodes are superfluous.

[Flashbang]

Hi

If I read your question correctly, you want to feed and switch aspects of a colour light signal from DCC Concepts Digital Cobalt ip point motor.  

I assume your signals are actually LED lit? Are they Hornby or other make?  Older Hornby were filament lamp lit while the latest ones are LED lit.  

You can use the accessory DCC bus to feed LEDs as well as powering and operating the Digital ip motors.  

Connect one of the Accessory bus wires to Cobalt terminal 6 (S2-C). Connect signals aspect feed wires to Cobalt 4 & 5 terminals (S2-L) and (S2-R)    

Now, depending on the signal make and whether it has a series resistor is the next factor. Assuming it has an external series resistor then connect that to the signals return wire. Doesn't matter which way around the resistor is fitted. Connect the other end of the resistor to the other Accessory bus wire. 

Your signal will light and the aspect displayed will depend on which direction the Cobalt is set towards.  If red aspect shows when green is wanted swap around the two wires on 4 & 5 terminals

What you haven't now got is any added protection for the signals LEDs from the AC produced on the DCC Accessory Bus   So here I would simply add a series diode in each aspects feed wire with the diode inserted so as its banded end is towards the signal. However, this is assuming the LEDs are wired in the signal as Common Cathode.  That is why I asked what CL signal make you have at the beginning!  

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[Flashbang]

Sorry, I tried to add some extra words to my post above, but at the time the forum website locked up!  If the LED signal is Common Anode then simply reverse the two diodes in the drawing, so as their banded end is towards the Cobalt ip.

[CDRC]

Hi guys thanks very much for your excellent advice. Unfortunately, half of my post went missing I think the site locked up and I didn't discover this until checking the other day. In the last part of my post I referred to a piece of kit called a 'Full Bridge/Wave Rectifier'  which I understand converts AC to DC. Would this work? Using diodes would be less expensive so that's the route I think I shall follow. However, out of interest what's the opinion on the Rectifer option.

thanks again for the advice

Charles

[Flashbang]

No need for a bridge rectifier.  Just two simple diodes at around £0.6p each or less is all that's needed.  😎

[Chrissaf]

However, out of interest what's the opinion on the Rectifier option.

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Six of one, half dozen of the other.

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My solution just protects the LEDs from being damaged by 'reverse bias voltages' due to the Bi-Polar nature of the DCC signal.

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FlashBang's solution is to use single diodes to perform 'half wave rectification' prior to powering the LEDs. The LEDs are in essence powered by rapid pulses of DC (7 to 9 thousand times a second).

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Your 'bridge rectifier' option is similar to 'Flashbang's' except it provides 'full wave rectification', which in essence fills in the gaps to eliminate the pulses.

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These simplistic wave forms explain the difference:

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The top waveform demonstrates the initial DCC Bi-Polar wave form before any diodes are implemented.

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For operating an LED, with such a low current draw, there is little benefit in going 'full wave'. As you can see in the waveforms above. Full wave rectification fills in the gaps between the 'half wave' pulses. Great if you want to generate more overall power, but LEDs don't need a lot of power.

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So single diode 'half wave' rectification is more than adequate for powering LEDs.

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[CDRC]

Thanks guys. I understand the difference. 

Charles

[CDRC]

Hi guys,

Thought I'd let you know the lights have been connected as per Flashbang's diagram and they're working perfectly. Thanks for the assist. Can I ask a supplementary question? I have recently acquired a set of 3 Aspect lights which show three wires, one to each LED. I understand where the red and green would go ( as per your diagram) but to where would I connect the third wire, the amber LED?thanks again

Charles

[Flashbang]

Unfortunately you can't use the Cobalt for three aspect signalling easily. The three aspect needs know what the next signal ahead is displaying to get the correct aspect sequence.

Using the one Cobalt you can only show two aspects.

 

Have a look here where aspect sequencing and the switching is shown.  https://www.brian-lambert.co.uk/Electrical_Page_3.html#Bookmark13    Not that simple!