Block Occupation - Vehicle in Section or Train in Track

I like many others have been attempting to install a Block Occupation system
within my model railway. But as yet I have not been able to form a satisfactory
method that does not have one short coming or the other.
I have found that the most satisfactory method is to employ Optocouplers in the
feed to track circuit from my (Gaugemaster 100 series) controller. This provides
for any interference with the power circuit to be minimal but subject to a power
loss of some 1.3 vF consumed by the Opto's. It works fantastically well, but
wait for it, there is one downside to it and that is: The Opto's do not commence
working until they receive some 1.3v's, which means with the controller turned
off there is no output at all from the Block Occupation circuit. I can always
leave the controler set at the 10% position but then I can't always hope that I
will remember to do so and as for visitors using the system then it is very
likely that they would fall if not at the first fence then almost definately at
the second fence.
So you guys out there with super intelligent electronics know how do I induce
enough power to switch on the optocouplers when the controller (or controllers
are) is at "OFF" when the controller leakage current equals just
+195mV's.
Controller: Gaugemaster 100 series
Controller output: 0.195v to 14.5v-DC.
Controller auxiliary output 16v~AC
Auxiliary circuit used on the Transistor side of the Optocouplers is +13.75v-DC
Optocouplers are: 4N35's
Track feed utilised is: controller positive with the optocouplers wired in
series with the return wire being the continuation of +v controller back to the
+ rail. Twin optocouplers are utilsed to determine the direction of track
current.
Any help would very gratefully received.
Cheers,
H Chris Spreckley
Reply to
Totnado
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Take a 9V battery and make a 1-2mA FET-based constant current source[1]. Connect it via an ordinary diode to your track.
When the controller is off it will hopefully be disconnected from the track rather than shorting it, in which case your constant current source will provide just enough power to trigger the optocoupler. When the controller is on in the opposite direction to the battery you still drain the same current; when it's on in the same direction eventually the diode will stop the reverse current.
[1]
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If your controller shorts the track when it's in the "off" position then I don't know the answer.
Reply to
Ian Jackson
Connecting the track feed through the LED in the opto-coupler like that is not a good idea.
Have a root around here
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You'll see optos are always used to detect the voltage across diodes or resistors, never to conduct the track current directly.
Ordinary diodes (Vf about 0.7V) with a comparator would be better. You can still use an opto-isolator at the output.
Another way to do block detection on DC is to add a high frequency AC signal to the DC, that is on all the time. You can then detect this even when the locos are stationary. I think MERG have a design in one of their technical bulletins but you would need to be, or know, a member to get hold of it.
MBQ
Reply to
manatbandq
Thanks for the input. However been there done that seen that, you know what I mean. I did something similar but never quite got it right and blew up three opto's in the process.
Cheers Totnado
Reply to
Totnado
Hi I know of Rogers circuits, they use though the bridge rectifier to sort out the vehicle direction whereas using the two optocouplers to do the same job uses only 0.8vF as against master Paisley's 3vF and given all that he still ends up without section occupied when the controller of a DC., controlled circuit is at rest and in my instance providing just +0.195v leakage current.
So may I ask just once more does anybody know how to provide "Train in Section" when given only 200mV's. I am currently toying with using an OPamp., immediately after entering the Block Occ., circuitry and before t he Opto's but don't at this moment in time know whether sufficient voltage can be raised to actuate the LED's in the opto or not.
Any ideas?
Reply to
Totnado
Putting the track supply straight through the OC does seem quite bold.
I use a variant of this circuit:
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I don't have an R1 in the track side of the circuit. It doesn't seem necessary and it's not clear why it's there.
And I don't use an opamp; instead I simply take the common of the OC output phototransistor and the pullup resistor (R3 in that diagram) and connect it directly to the digital IO pin on a PIC microcontroller. I'm using DCC so I only need to detect one direction of current flow; there will be flow in the other direction in a moment anyway.
My system is sufficiently sensitive that it can detect the current draw of a DCC decoder with all its outputs turned off; conversely it doesn't ever seem to produce false detections. It will generally detect fingers if they are bridging the track, though :-).
If you're interested I can look up the actual exact component values and/or post a photo of my circuit diagram.
If you don't have a microcontroller a 74LS- or 74HC- series Schottky buffer chip, or an LED driver chip, ought to do fine.
Reply to
Ian Jackson
As posted, inject a HF signal into the track circuit and detect that.
Did you listen to the bit about not feeding the layout through the opto coupler?
The best you could hope is one diode schottky diode drop which will be less than a normnal diode and a lot less than an LED. Then some form of detector/amplifier with the isolator on the putput of that.
MBQ
Reply to
manatbandq
It's easy with DCC :-)
MBQ
Reply to
manatbandq
I don't think you realy understand the circuits. An opto-coupler uses an LED which works with 10 - 20mA current, maximum. Connecting the layout through an opto is a really bad idea, when you consider even a modern loco will be drawing 100mA or more.
The LED in the opto-coupler will drop almost 2V. The bridge rectifier in Paisley's circuits does sort sort out the bidirectional detection, but it's also there to cause a large enough voltage drop to light the LED in the opto-coupler.
What do you mean by "+0.195v leakage current"? Current is not measured in volts. Are you saying that is the voltage from the controller when it is nominally off?
MBQ
Reply to
manatbandq
I must learn to think before hitting "send".
AND he uses a bidirectional opto-coupler.
MBQ
Reply to
manatbandq
Hi I've been trying to do this but unsuccessfully for some time, I seem to be getting either dead shorts or standing the IC's too close to the sun.
You said "Take a 9V battery and make a 1-2mA FET-based constant current source[1]. Connect it via an ordinary diode to your track."
You wouldn't care to explain this in a simple language or provide a drawing of just how you would do it, Can it be done off the +13.75Vcc circuit on the other side of the opto's perhaps by using a 7805 or a variable voltage regulator? I have a VR., that I can successfully tame to +2.5v. I've tried to 'joint' it in off the +13.75Vcc but unsuccessfully.
Hope you can get back to me if the drawing was too difficult to send on this forum perhaps you could send it direct to: hcs at chrisspreckley dot com could you.
Cheers
Reply to
Totnado
I've got a feeling this could go on and on I understand where I am up to conclusively. Now I'm not sure just what you mean by not feeding the track through the Opto's as Opto's use two separate circuits one of them being a sensory one off the track the earth must returns to the track, do you understand the circuit? The transistor/Vcc side of things is to run whatever, but track side only.
Master Paisley uses the Opto after the BR., I chose to dump the BR., and use two Opto's in order to loose less of the voltage and of course current.
Leakage current is the current drifting out of the controller when the controller itself is to all intents shut down.
Any more ambiguity?
Reply to
Totnado
Hi Ian, Many thanks for the reply. It is so very easy when considering DCC., however almost all large track layouts are still using DC., and my club the MMRS., is no exception so where I would like to design (he says rather humbly) curuits for DCC., the reality is that these circuits take years to construct and some are inventions upon inventions and a change to all the electrics is just not on.
But if you would like to send me details of your block occupation circuit then I'd be delighted to see it. If you have a mond to do so then please use Chris at ChrisSpreckley dot com. Cheers and thanks, Totnado
Reply to
Totnado
Indeed.
I'm still not do sure...
The sensing side of the opto is an LED. From your description you seem to have connected the feed to the track through the LED. Good luck with that.
Which just confirms what I said about understanding the circuit. The bridge rectifier or diodes are there to carry the track current. A small current flows through the opto due to the voltage across the diodes, limmited by the resistor in series with the opto.
But you quoted a *voltage*. Where are you measuring that voltage?
MBQ
Reply to
manatbandq
May be I am misunderstanding the circuit that you are actually using, but in the ones that are on the web, the current to the track goes through the bridge rectifier (in order to create a voltage drop). If you have replaced the bridge rectifier with opto isolators then is the current to the track not passing directly through the optos? If that is the case then that current will be well in excess of the maximum for the led's in the optos and most likely will destroy them.
Jeff
Reply to
Jef
What happens when you short the track?
I already have.
Try reading all the messages in the thread.
MBQ
Reply to
manatbandq
The passengers have to get out and walk to the station.
Reply to
Chris Wilson

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