Simple DCC Block Occupancy Detector

[kbriffa]

In Ames, Friberg and Loizeaux excellent book 'Digital Command Control' is this circuit diagram



Unfortunately the book fails to identify what the components of the circuit are. I seem to inderstand that the detector is made up of a bridge rectifier (D1) and an integrated circuit (IC1). However apart from that I haven't got a clue what type of IC this is or what is the current load of the bridge rectifier.

Any help identifying the components would be greatly appreciated.

Thanks
Kenneth

[poliss]

Circuit Diagram Symbols http://www.kpsec.freeuk.com/symbol.htm

[kbriffa]

Thanks for the link. Might be useful in the future but my understanding of electronics is very basic and I cannot really understand many of the terms used in that website so ideally could someone please tell me which particular IC is the one in the diagram (I can also tell from the symbols used that it contains an optoisolator).

Thanks in advance
Kenneth

[m.levin]

A quick look at the circuit and looks like a Bridge rectifier and a halls current sensor.
You can get a Current Sensor for AC currents, so the rectifier may not be needed.

[kbriffa]

Thanks.

Is this what you are referring to?

http://www.sypris.com/test-and-measurem ... ct-sensors
http://www.sypris.com/FileManager/Library/SHseries.pdf

Kenneth

[D0260]

halls current sensor

Don't think so. Its just an OptoIsolator with a darlington output.

An optoisolator is simply a Light Emitting Diode pointed at an opto(sensitive) transistor- current thru LED makes the transistor ( in case a pair of) conduct. No current thru LED, no light, opencircuit output. Cost about 40 or 50 p except if its in a nice plastic hang-up box when they are £1.99 or so. Usually in an 6 or 8pin DIP package with some pins not connected .

Need a data sheet try http://datasheets.name/datasheets/170713/4N33.html for a 4N33 device.
If you have some others without the darlington ( two transistor) output stage , just connect a 5p external transistor.
The differences between the various devices are the speed of operation and the isolation voltage etc - neither of which are critical in this application , so buy the cheapest you can find .

The bridge rectifier is just a simple 50 Volt bridge - again a 4 or 6 pin DIP is all that is required - 1Amp or maybe 1.5A capabilty if you double head or use coaches with lots of lights etc. Again the cheapest will do the job.

A suitable Hall Effect device COULD be used in this application , but in my experience is neither rugged or reliable enough for this sort of application , apart from costing more initially. This is NOT a Hall Effect device circuit diagram though!.

[m.levin]

halls current sensor

Don't think so. Its just an OptoIsolator with a darlington output.

An optoisolator is simply a Light Emitting Diode pointed at an opto(sensitive) transistor- current thru LED makes the transistor ( in case a pair of) conduct. No current thru LED, no light, opencircuit output. Cost about 40 or 50 p except if its in a nice plastic hang-up box when they are £1.99 or so. Usually in an 6 or 8pin DIP package with some pins not connected .

Need a data sheet try http://datasheets.name/datasheets/170713/4N33.html for a 4N33 device.
If you have some others without the darlington ( two transistor) output stage , just connect a 5p external transistor.

the Opto will switch, when it has voltage to it, Dcc track is always live. so useless in this applecation.
A hall Current sensor will detect the change in current, from the locomotive or resistance wheeled stock.

[D0260]

the Opto will switch, when it has voltage to it, Dcc track is always live. so useless in this applecation.
A hall Current sensor will detect the change in current, from the locomotive or resistance wheeled stock.

With no train on a track, no current flows, so NO lght from LED , Outout transistor is open circuit.
Train on the track - even stationary , takes enough current for the LED to start to emit and the output transistor conducts.
EXACTLY what is needed for an Occupancy detector....

The circuit shown is connected in series with either rail and its DCC supply NOT across the DCC suply.

[m.levin]

the Opto will switch, when it has voltage to it, Dcc track is always live. so useless in this applecation.
A hall Current sensor will detect the change in current, from the locomotive or resistance wheeled stock.

With no train on a track, no current flows, so NO lght from LED , Outout transistor is open circuit.
Train on the track - even stationary , takes enough current for the LED to start to emit and the output transistor conducts.
EXACTLY what is needed for an Occupancy detector....

Current doesn't flow. Current is drawn.
Voltage flows.
The LED will draw current so it will always output a signal from the Opto, until the track power is turned off.
So no its not EXACTLY what is needed for an Occupancy detector!

[D0260]

Think you need some electronics education.

I have tried to use use simple terms like 'current flows' , if you want me to explain it in technical terms that will leave you spluttering, just ask..

Read AGAIN . THIS CIRCUIT IS IN SERIES WITH THE DCC SUPPLY NOT IN PARALLEL WITH IT.

[D0260]

Take a look here

http://www.electronicsinfoline.com/Proj ... tors.shtml

for a slightly more sophisticated circuit diagram, in essence the same, except using a single transistor Optoisolator with external Op-Amp for the extra gain. ( and more controllable in characteristics.)

Ohhh and by the way , current flows , and Voltage is the potential difference and doesn't 'flow' anywhere.. I see, m.levin, you offer a DCC fitting service .... Uhhhh errrrr- you sure?

[NobbyClarke]

This is a pretty simple circuit for block detection, but has the disadvantage that it reduces the track voltage by something around 1.2v - the voltage drop across the diodes in the rectifier.

As this circuit is place in series with the track feed, rather than across it, when there is no load (train) within the blocks, the circuit is open series, so no current is flowing. When a load comes into the block, the circuit will be made, 1.2v will be dropped across the diodes in the rectifier - this voltage will provide enough forward voltage to activate the LED in the optocoupler and the output will go active.

If you do use this method, ensure the bridge rectifier has a suitable current rating, it will be handling all the track current for the block in question.

regards,

Nobby

[m.levin]

Thanks for your Education
Let me get this right
This is how I see that circuit.

Rail A -------- Always Live as its Dcc
Rail B -------- Always Live as its Dcc
Both Live at ~16v approx Dcc control
Then goes to the Rectifier Which converts that Ac voltage to a Dc output, about 19v Dc, as the tracks are live being Dcc you always get the a Dc voltage output, from the Rectifier.
Then goes to the Opto, which still has that voltage, from the Dcc ALWAYS LIVE track, to the input. the LED within draws current and then Always outputs a signal weather anything is or isn't on the track.

I simply Missed where you said about being in series and not parallel and you went straight is to arsehole mode.

[D0260]

That is neither how it is connected , nor how it works...

[Roger (RJ)]

As I understand it, the bridge rectifier is positioned so that it feeds power to the block . If there is nothing in the block, no current flows, therefore no pd across bridge and no output from opto.