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DCC occupancy detections false positives

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Old.Professor - 17 May 2008 21:27 GMT
I just installed DCC block occupancy detectors and am getting false
positives only when the layout is in an operating session with lots of
people and trains. When I'm by myself there are no false indications.
I'm at a loss what to do, or how to test if the changes have been
effective.

I'm using a variant on the VT-5 inductive (transformer-type) occupancy
detector (see http://home.cogeco.ca/~rpaisley4/DccBODvt5.html).
Specifically, the variant by Richard Napper is described on page 18 in
http://www.mcor-nmra.org/Publications/Kibitzer/Volume56-2.pdf .

During the operating session I saw the occupancy indication change
from unoccupied to occupied and then back to unoccupied in about 30
seconds.  Most of the time it showed occupied when there was nothing
on the track.  Now that I'm by myself, it is steadily showing
unoccupied.

I had a thought that there might be stray fields.  I was going to try
using aluminum foil to shield the detector, but then I got to
wondering if that would be effective if the fields were magnetic.

Suggestions would be very much appreciated.

I also posed this request to mrrelectronics@yahoogroups.com.
Charles Davis - 17 May 2008 21:48 GMT
> I just installed DCC block occupancy detectors and am getting false
> positives only when the layout is in an operating session with lots of
[quoted text clipped - 12 lines]
> on the track.  Now that I'm by myself, it is steadily showing
> unoccupied.

You need to check the 'sensitivity'

I.E. a DRY finger across the rails shouldn't trip things, a WET finger
should.

> I had a thought that there might be stray fields.  I was going to try
> using aluminum foil to shield the detector, but then I got to
> wondering if that would be effective if the fields were magnetic.

There isn't any 'magnetic field' to shield.

Chuck D.

> Suggestions would be very much appreciated.
>
> I also posed this request to mrrelectronics@yahoogroups.com.
Charles Davis - 17 May 2008 22:16 GMT
>> I just installed DCC block occupancy detectors and am getting false
>> positives only when the layout is in an operating session with lots of
[quoted text clipped - 23 lines]
>
> There isn't any 'magnetic field' to shield.

To expand on this a bit.
The only 'magnetic fields' involved, are the magnetic coupling of the
'track feeder' passing through the 'sensor coil'. And those, you need,
and they seems to be working. (Too well at times.)

The DCC signal is 'loaded' (current drawn) to indicate occupancy.

Something is loading things, the problem is WHAT?

The available DCC signal, can be loaded (some current drawn) by either
capacitance or resistance.
Capacitance can be as simple as the two wires paralleling each other
over a long distance.
Resistance can be as subtle as DAMP roadbed.

We need to determine what the cause of the problem is.

Chuck

> Chuck D.
>
>> Suggestions would be very much appreciated.
>>
>> I also posed this request to mrrelectronics@yahoogroups.com.
BDK - 17 May 2008 23:27 GMT
> >> I just installed DCC block occupancy detectors and am getting false
> >> positives only when the layout is in an operating session with lots of
[quoted text clipped - 48 lines]
> >>
> >> I also posed this request to mrrelectronics@yahoogroups.com.

I'm wondering if maybe he's got some sort of slightly conducting glue or
paint on the tracks, or the ballast? I saw a layout where the ballast
was made by some guy and then sold at a train show. You could get a
solid ohmmeter reading just by touching it to the ballast and either
rail. It got warm in some places when you put 12V to it for a while.
This was back in the late 70's.

The guy tore it all up and redid the track and ballast. He was just
about to start putting the buildings up.

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David Nebenzahl - 17 May 2008 23:27 GMT
On 5/17/2008 2:16 PM Charles Davis spake thus:

> The available DCC signal, can be loaded (some current drawn) by either
> capacitance or resistance.
> Capacitance can be as simple as the two wires paralleling each other
> over a long distance.

This is so unlikely (capacitive loading from just the 2 rails) as to be
outside the realm of possibility.

If you had a pair of insulated wires tightly twisted together for a long
run, you *might* start to see some effects due to capacitive loading,
but even this is unlikely.

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Charles Davis - 18 May 2008 00:46 GMT
> On 5/17/2008 2:16 PM Charles Davis spake thus:
>
[quoted text clipped - 9 lines]
> run, you *might* start to see some effects due to capacitive loading,
> but even this is unlikely.

Not that UNLIKELY at the frequencies (& harmonics) generated by a DCC
signal. Remember, they TRY for a 'Square Wave' with DCC. You can't look
at things as if you were working with Analog Sine-Wave RF, and draw
conclusions based on that universe.

When I typed the above ----
">> Capacitance can be as simple as the two wires paralleling each other
>> over a long distance."
I wasn't thinking of a 10' run to a close block. more on the order of
exceeding 75'.
Chuck D.
David Nebenzahl - 18 May 2008 02:25 GMT
On 5/17/2008 4:46 PM Charles Davis spake thus:

>> On 5/17/2008 2:16 PM Charles Davis spake thus:
>>
[quoted text clipped - 20 lines]
> I wasn't thinking of a 10' run to a close block. more on the order of
> exceeding 75'.

Sorry, TYPING IN ALL CAPS still doesn't make it any more correct. Even
taking harmonics into account, a DCC signal is still well down in the
kilohertz region, nowhere near radio frequencies. Capacitance effects,
if any, are negligible.

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Charles Davis - 18 May 2008 04:52 GMT
> On 5/17/2008 4:46 PM Charles Davis spake thus:
>
[quoted text clipped - 29 lines]
> kilohertz region, nowhere near radio frequencies. Capacitance effects,
> if any, are negligible.

Right, but the poster was talking about a 'bleeding edge' type of
sensitivity.
And no, the fact of the 'Square Wave' signal wave form generates all
sorts of higher frequency harmonics. In face, those harmonics are a not
insignificant portion of the power being pumped out by the DCC Power
Station/ Booster.

Chuck D.
David Nebenzahl - 18 May 2008 20:41 GMT
On 5/17/2008 8:52 PM Charles Davis spake thus:

>> On 5/17/2008 4:46 PM Charles Davis spake thus:
>>
[quoted text clipped - 36 lines]
> insignificant portion of the power being pumped out by the DCC Power
> Station/ Booster.

So show me where I disagreed with that; I said above, if you bothered to
read, "even taking harmonics into account".

So let's look at those harmonics, shall we? First of all we need to know
the fundamental frequency of a DCC signal. Referring to the DCC standard
itself (I'm reading from one of the PDFs on the NMRA site), the shortest
interval defined is for a "1" bit, which they define as being about 100
microseconds wide. (They actually allow more time than that, but let's
use the lower figure for the sake of discussion.)

Since freqquency is the reciprocal of bandwidth, that means the the
highest frequency in a DCC signal *should* be about 10 kHz. For the sake
of discussion, let's double that, to 20 kHz.

As we all know, square wave harmonics are the odd multiples of the
fundamental frequency. These harmonics diminish the higher they go; in
other words, the strongest harmonics are the lowest-order ones.

So we now have harmonics at the following frequencies:

  - 60 kHz  (3f)
  - 100 kHz  (5f)
  - 140 kHz  (7f)
  - 180 kHz  (9f)
  - 220 kHz  (11f)

and so on. So we can see that we have a diminishing set of harmonics
that is *well* below a megahertz until we get well up into the series.
Which says that any higher-frequency harmonics are going to be so weak
as to have negliglbie effect. Which means that the capacitive effect of
two conductors in the form of HO track can be disregarded.

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David Nebenzahl - 18 May 2008 20:46 GMT
On 5/18/2008 12:41 PM David Nebenzahl spake thus:

> Since freqquency is the reciprocal of bandwidth,

which should have read "Since frequency is the reciprocal of wavelength".

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Charles Davis - 19 May 2008 04:51 GMT
> On 5/17/2008 8:52 PM Charles Davis spake thus:
>
[quoted text clipped - 71 lines]
> as to have negliglbie effect. Which means that the capacitive effect of
> two conductors in the form of HO track can be disregarded.

Hi David;
Sorry you are so hung up on "theory'.
The 'Old Professor', was asking about a 'Real World' problem.
In the 'Real World', observed problems, 'Trump' theory every time.

Capacitive loading cannot be ignored. It may be only a minute part of
the detected load, but that 'minute' part may be the final 'smidgen'
that trips the detector.

Go argue with Greg for a while.

Chuck D.
Ian Jackson - 19 May 2008 12:45 GMT
>Which says that any higher-frequency harmonics are going to be so weak
>as to have negliglbie effect. Which means that the capacitive effect of
>two conductors in the form of HO track can be disregarded.

Well, I would more or less agree with this analysis _but_ the detector
is inductively coupled, which means it will be much more sensitive to
higher than to lower frequencies.

My own block occupancy detectors work off the voltage drop from two
diodes in series, which is used to feed an optocoupler[1].  I find
it's very reliable: no spurious detections even though it will detect
a firmly-pressed-down dry finger (across N-scale track) or an inactive
function decoder drawing a fraction of a milliamp.

I've got lots of big runs of wire in parallel and quite a bit of EMI
(I had to shield a different part of my circuit with paper-wrapped
tinfoil) and I see no problems.

[1] circuit, best I can do in ASCII art:

          from booster output
                  |
                  |                         5V
         +--------+-----------+          ------+---------
         |        |           |                |
         |        |           |               +-+
         |        |          +-+              | |
       -----    +---+        | |              | |
        / \      \ /         | |              +-+
       +---+    -----        +-+               |      detector output
         |        |           |             +--+--------------
         |        |           |          | /         to microcontroller
         |        |           |      __  |/
       -----    +---+       +---+ ,. /|  |\
        / \      \ /         \ / ' ~~    | V
       +---+    -----       -----           |
         |        |           |             |
         |        |           |             |
         |        |           |             |  GND
         +--------+-----------+          ---+--------------
                  |
                  |
               to track

This has a minimum number of parts (and solder joints) per block: one
bridge rectifier, 1/8th of an octal optocoupler, 1 ordinary resistor,
1/nth of a SIL resistor array for the pullup, and of course one
microcontroller pin.  I can fit 21 of these on one modest PCB along
with the microcontroller, circuitry to control a dozen point motors,
etc.

I built the booster myself and its output voltage is higher than the
12V nominal track voltage to compensate for the 0.7ish per diode
forward drop.

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David Nebenzahl - 18 May 2008 23:43 GMT
On 5/17/2008 8:52 PM Charles Davis spake thus:

> Right, but the poster was talking about a 'bleeding edge' type of
> sensitivity.
> And no, the fact of the 'Square Wave' signal wave form generates all
> sorts of higher frequency harmonics. In face, those harmonics are a not
> insignificant portion of the power being pumped out by the DCC Power
> Station/ Booster.

I'll let my last words on this subject be not my own, but those of the
DCC experts at Tried & True Trains (http://www.tttrains.com/dcc/dccfaq.htm):

  At 10 KHz, the DCC signal is essentially immune from the problems
  with reflections and standing-waves which higher frequency tone
  systems can experience. The DCC specification requires that the
  decoders be able to reject input above 100 KHz. All useful DCC signal
  information is below 100 KHz, and the behavior of wiring at
  frequencies above 100 KHz is irrelevant to DCC operation.

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Charles Davis - 19 May 2008 04:56 GMT
> On 5/17/2008 8:52 PM Charles Davis spake thus:
>
[quoted text clipped - 15 lines]
>   information is below 100 KHz, and the behavior of wiring at
>   frequencies above 100 KHz is irrelevant to DCC operation.

But the question wasn't about the DCC base to decoder information transfer.

Chuck D.
Klaus D. Mikkelsen - 18 May 2008 07:29 GMT
> Sorry, TYPING IN ALL CAPS still doesn't make it any more correct. Even
> taking harmonics into account, a DCC signal is still well down in the
> kilohertz region, nowhere near radio frequencies. Capacitance effects,
> if any, are negligible.

Hmmm, can you tell me the harmonic content of a 10kHz square wave?

In case you need help, then read:
http://en.wikipedia.org/wiki/Square_wave

Especially the lines:
"An ideal square wave requires that the signal changes from the high to
the low state cleanly and instantaneously. This is impossible to achieve
in real-world systems, as it would require infinite bandwidth."

Klaus
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David Nebenzahl - 18 May 2008 21:04 GMT
On 5/17/2008 11:29 PM Klaus D. Mikkelsen spake thus:

>> Sorry, TYPING IN ALL CAPS still doesn't make it any more correct. Even
>> taking harmonics into account, a DCC signal is still well down in the
>> kilohertz region, nowhere near radio frequencies. Capacitance effects,
>> if any, are negligible.
>
> Hmmm, can you tell me the harmonic content of a 10kHz square wave?

Yes, that's easy: harmonics at 30 kHz, 50 kHz, 70 kHz, 90 kHz, etc. See
my other post in this thread.

> In case you need help, then read:
> http://en.wikipedia.org/wiki/Square_wave

Sorry, I don't use Wikipedia ("the encyclopedia any junior-high-school
idiot can edit") for any serious information.

> Especially the lines:
> "An ideal square wave requires that the signal changes from the high to
> the low state cleanly and instantaneously. This is impossible to achieve
> in real-world systems, as it would require infinite bandwidth."

Misleading; you're forgetting that the amplitide of the harmonic series
is constantly decreasing, so that the higher-order harmonics can be
practically disregarded above some limit.

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Klaus D. Mikkelsen - 18 May 2008 21:49 GMT
> Yes, that's easy: harmonics at 30 kHz, 50 kHz, 70 kHz, 90 kHz, etc. See
> my other post in this thread.

Okay

> Misleading; you're forgetting that the amplitide of the harmonic series
> is constantly decreasing, so that the higher-order harmonics can be
> practically disregarded above some limit.

I know this very well.

But that's looking at at square wave. What is the frequency content of a
signal changing from 0 to 1 ?

Only the slew rate of the driver sets the limit.

Klaus
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David Nebenzahl - 18 May 2008 22:03 GMT
On 5/18/2008 1:49 PM Klaus D. Mikkelsen spake thus:

>> Yes, that's easy: harmonics at 30 kHz, 50 kHz, 70 kHz, 90 kHz, etc. See
>> my other post in this thread.
[quoted text clipped - 9 lines]
> But that's looking at at square wave. What is the frequency content of a
> signal changing from 0 to 1 ?

Well, that's basically half a square wave, right?

> Only the slew rate of the driver sets the limit.

Correct, which is why we have to approximate the frequency. See my other
post, where I derived a plausible upper frequency of 20 kHz from taking
the stated parameters for the shortest-interval signal (a "1" bit) and
doubling them. Maybe you could triple it for safety's sake.

Of course, this doesn't take into account the various "glitches" that
DCC controllers inevitably output ...

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Klaus D. Mikkelsen - 18 May 2008 22:06 GMT
> Well, that's basically half a square wave, right?

No, because the harmonics of your square is mainly to get the top of the
square "flat". It really doen't tell you anything about the rising edge.
 
> Correct, which is why we have to approximate the frequency.

No, the leading edge of the square don't care if the pulse is 1uS or 1mS
long, it still has to be very steep.

Klaus
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David Nebenzahl - 18 May 2008 22:39 GMT
On 5/18/2008 2:06 PM Klaus D. Mikkelsen spake thus:

>> Well, that's basically half a square wave, right?
>
[quoted text clipped - 5 lines]
> No, the leading edge of the square don't care if the pulse is 1uS or 1mS
> long, it still has to be very steep.

Well, yes. We're really splitting hairs now, so this might be my last
word on the subject. Basically, the steeper the rise time, the closer an
approximation to a square wave it is, with all the applicable stuff
about harmonic generation (odd-numbered harmonics). If it's less steep,
then it's closer to a sine wave. So everything I've said applies, to
some extent or other.

Note that in the real world, everything is only an approximation of
these idealized concepts, like square waves, which don't really exist
anywhere in reality. There's always stuff like overshoot, ringing,
sloped edges, etc. Look at a "square wave" on a 'scope sometime for
amusement.

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Klaus D. Mikkelsen - 19 May 2008 05:24 GMT
> Well, yes.

Thanks.

> Note that in the real world, everything is only an approximation of
> these idealized concepts, like square waves, which don't really exist
> anywhere in reality. There's always stuff like overshoot, ringing,
> sloped edges, etc. Look at a "square wave" on a 'scope sometime for
> amusement.

And in the ideal word you do not have problems with occupancy detectors
- real world is something else.

Klaus
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Bernhard Agthe - 20 May 2008 12:05 GMT
> Sorry, TYPING IN ALL CAPS still doesn't make it any more correct. Even
> taking harmonics into account, a DCC signal is still well down in the
> kilohertz region, nowhere near radio frequencies. Capacitance effects,
> if any, are negligible.

Just to amuse you, I once had a calculation of shorting-current in
German (real) catenary performed. You start off with 15 000 Volt at
16 2/3 Hz (stupid frequency, but anyway): short it at the end which is
far away from the supply point and you get a short blast of some 10 000
Amps - just from two "wires" running not even directly next to each
other (a few meter distance) for some few kilometers. You have a
non-neglectible capacity in there and a non-neglectible inductance.

Now, scale it down: the feeder wires run next to each other (a milimeter
or two of distance), for 50 or 100 ft. and you have frequencies in the
kiloHz range. You won't get the 10 000 Amps blast, but you will have
some "nice" side effects. Not much, but probably enough, especially if
you add some more stuff:

- the wires may be running parallel next to each other (and not be
drilled around each other) which *might* give you some effect of
radio-frequency interference,

- the wires might be soldered together for length and plugged into the
booster, they will be soldered to the rails. As we're talking about AC
signals, each change of AC resistance might produce some little
reflection which lowers signal quality,

- as described in the original post, this happens when there are lots of
people in the room which means higher air humidity. With some of the
afore mentioned effects this might just be enough to "trigger" the false
detection

- many people also means there might be a cell phone (or dect phone)
disturbing the signalling, or there might be lots of wireless
controllers, probably even somebody leaning against the benchwork
causing a minor change in geometry which allows some leak current...

My suggestion would be to (1) check the cables and tracks, especially if
this happens in certain places only, (2) try changing the ventilation of
the room for the next OP session, (3) move the detectors as close to the
track as possible in terms of cable length, (4) if you exchange cables
somewhere, try to use shielded (computer data) cables and ground the
shields. Surely, this is a very last resort as it is major work. There
were other suggestions in this thread which are easier to follow ;-)

Good luck and lots of fun...

PS: This post is based on the (educated) opinion of the author.
David Nebenzahl - 21 May 2008 05:40 GMT
On 5/20/2008 6:00 AM Bernhard Agthe spake thus:

>> Sorry, TYPING IN ALL CAPS still doesn't make it any more correct. Even
>> taking harmonics into account, a DCC signal is still well down in the
[quoted text clipped - 8 lines]
> other (a few meter distance) for some few kilometers. You have a
> non-neglectible capacity in there and a non-neglectible inductance.

Interesting: I'm a little confused by some of the details (i.e., what
exactly is being shorted to what) but get the general picture. Of
course, this spectacular "blast" only occurred on paper and pencil. I'm
curious just how you derived this calculation.

Specifically, and here I'd like to relate this to the original
discussion I was having, how do you calculate capacitance between
conductors in air? And can you tell us what the capacitance of typical
HO track would be, say per foot, in microfarads (or, more likely, in
picofarads)?

I'm still skeptical that any such capacitance could have any measurable
real-world effects at the relatively low frequencies found in any DCC
system (and even more skeptical of your claim at a nearly subsonic
frequency), but am open to being persuaded.

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Bernhard Agthe - 21 May 2008 09:15 GMT
Hi,

>> Just to amuse you, I once had a calculation of shorting-current in
>> German (real) catenary performed. You start off with 15 000 Volt at
[quoted text clipped - 8 lines]
> course, this spectacular "blast" only occurred on paper and pencil. I'm
> curious just how you derived this calculation.

They demonstrated a short between the overhead wire and the tracks at a
point far from the feeder. Similar things would happen in long-distance
power lines ;-) It all starts out with rather simple "school-physics",
but gets wierd (sorry, mathematics is my weak point ;-) at the moment of
the short because you have the following problem: there is a very sharp
voltage change. Because the system has a non-zero capacity and
inductivity, there is a compensating effect in the form of a current
overswing. As it's a few years back, I would have to look up my
university records which are back at my parents place...

(OT:)
Actually, in case of AC, there is a point where the voltage reaches zero
every so many microseconds, which is not the case in DC train systems.
While the voltage is lower, usually ~3000 Volt, the big problem there is
the engine's main switch. They do have severe arcs and sometimes the
switches burn "on" - which will mean the unlucky engine may have its
wheels soldered solidly on the rails (probably with the arc jumping
about happily inside the engine). So lineside "short detection" is a
real task in case of DC power ;-)
(/OT)

> Specifically, and here I'd like to relate this to the original
> discussion I was having, how do you calculate capacitance between
> conductors in air? And can you tell us what the capacitance of typical
> HO track would be, say per foot, in microfarads (or, more likely, in
> picofarads)?

Capacity is not a matter of enclosed conductors. It also works with air
as the isolator medium. Basically you can use the formula from your
school's physics classes (sorry if I kind of expect you to have the same
school background as I had, but my physics classes were really good,
just too short):

C = eA / d

with C=Capacity, e the dielectric factor (describing the isolator), A
the facing area of the plates (wires) and d the distance between the
wires. You can look up the general scheme at [1].

[1] <http://en.wikipedia.org/wiki/Capacitor>

In fact I do agree we're talking about very small capacities (and
inductances), but in a case where the signal is already of a bad
quality, the distortion of such a small capacity may be enough.

As I recall my Comm-Tech courses, there are a few more problems, namely
- different frequencies having different "running speeds" - which is a
problem in case of square waves, Gauss-Impulses were suggested as
better, even sine-waves.
- compare the wavelength to cable-size: 10kHz makes about 30km
wavelength, so we can sure discard this and assume DC calculation rules
(as to AC rules where you have to care for phase-shift),
- the receiver filter may be matched to the impulse form, but it doesn't
take into account the noise received and the transfer function of the
wires and tracks,
- Usually the cables on model railways are not shielded while computer
cables are usually shielded even if they're only a meter (3 ft.) long.
With small signals, there may be enough noise received...

That was why I suggest (and concur with several other suggestions here)
the OP should check the "simple" things first - sensitivity, room
ventilation, the occasional "cold" soldering, a hidden leak producing
some minor current flow, interference by cell phones... If those simple
things don't help, the use of shielded cables for long-distance routing
may or may not help - as the placement of the detectors may or may not
have an influence...

> I'm still skeptical that any such capacitance could have any measurable
> real-world effects at the relatively low frequencies found in any DCC
> system (and even more skeptical of your claim at a nearly subsonic
> frequency), but am open to being persuaded.

Probably not, but I would not rule this phenomenon to be completely
absurd. I have seen fuses go just by plugging a cable into a wall
outlet. The cable was too long and had been coiled in a place - the
resulting inductivity was enough - after I had re-cabled the whole
place, everything was fine. So eventually, the residual capacities may
grow large enough to actually matter - as I tried to demonstrate in the
Train-short-power-blast example above. Go figure what happened when they
encountered something like this for the first time 100 years ago ;-)

Have some fun...
PV - 19 May 2008 18:59 GMT
cad@gamewood.net writes:
>The DCC signal is 'loaded' (current drawn) to indicate occupancy.
>
>Something is loading things, the problem is WHAT?

Do you use current-based reversing loops? I had a couple problems with
false triggering early on, when the occupancy sensor was on the track feeder
hooked to the reverser. If a train entered the block from the opposite
end, the occupancy detector would trigger for a bit (long enough to cause
the computer to throw that switch thinking a train was coming), even though
the sensed block was seperate - throwing the relay in the AR-1 triggered
the detector.

The key thing with these detectors is that you MUST be certain that there
is nothing but train that can pull current from the sensed block. Breaking
the block up so that the sense block and the power block didn't start in
the same place seems to have solved it. *
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Klaus D. Mikkelsen - 18 May 2008 07:34 GMT
> I'm using a variant on the VT-5 inductive (transformer-type) occupancy
> detector (see http://home.cogeco.ca/~rpaisley4/DccBODvt5.html).
> Specifically, the variant by Richard Napper is described on page 18 in
> http://www.mcor-nmra.org/Publications/Kibitzer/Volume56-2.pdf .

Alone that you are trying to send sqare waves through a transformer can
be enough to draw current and induct voltage to the secondary side of
the transformer.

My recommendations is to use 4 diodes and an optocoupler for track
occupancy detection.

Klaus
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Old.Professor - 23 May 2008 16:58 GMT
I want to thank everybody who wrote suggestions and discussion. I'll
try to answer all the questions and share my proposed line of attack
for your comments. Whatever I do, I really would like to find some
test that I can apply by myself. Having to wait for the next
operating session at my house doesn't appeal. Nor does trying to
debug the detectors while the guys are trying to run trains.

My layout is in a 20' x 24' room. I have a dehumidifier, but haven't
turned it on yet for the season. In the winter I use a humidifier,
but in spring and fall I don't need either.

The block being detected is 14' of straight track followed by a rising
curve about 15' long that passes over the straight track. Both ends
are gaped. Power from the booster is distributed using 12 gauge wire
power bus running roughly parallel to the track. When I decided where
the detection block was, I disconnected all the feeders on one rail
from the bus and connected them to a sub-bus that is about 18' long.
The detector circuit board is near one end of the sub-bus. The wire
from the detector to the bus is about 1' long.

I measured resistance between the rails using an auto-ranging ohm
meter; it measured as an open circuit. I realize that's a DC
measurement, but I don't have any way of measuring at DCC frequencies.

It seems that some people refer to one pass of the track lead through
the coil as one-half turn while others call that one turn; I'll use
the latter. If there's a difference in configuration between one-half
turn and one turn it escapes me.

I am planning to have 5 detection blocks. So far I have two
configured. Only one of the detectors is showing the false detection.
That presents several ideas. (1) I could switch the detectors and see
what happens. (2) I could replace the detector that is giving the
false positive with one that I haven't used yet. (3) I could try
putting both detectors on the same block in series.

I adjusted the sensitivity to detect the caboose by experimenting with
number of turns through the inductor coil and the resistance across
the track. With one turn it could detect 2K ohm. With two turns it
detected 5K. I decided that was sensitive enough and wired 9.6K
resistors between the wheels of two axils on the cabooses. In
general, I'm still using plastic wheels, so I purchased new metal
wheel sets. I confirmed detection of cabooses.

I tried the dry finger test. The false positive detector switched
states and the other did not, confirming that the first detector is
more sensitive.

That's all I can think of now. I'd appreciate comments on my ideas of
things to try or other ideas.
Rick Jones - 24 May 2008 00:24 GMT
> I just installed DCC block occupancy detectors and am getting false
> positives only when the layout is in an operating session with lots of
> people and trains. When I'm by myself there are no false indications.
> I'm at a loss what to do, or how to test if the changes have been
> effective.

   I'd like to take the thread off on a new tangent. I notice that the
commercial occupancy detectors are pretty expensive on a per block
basis, around $5 or more. What is the cheapest detector circuit that
could be built? I've been checking out the various current sensors
available and I still haven't found anything that could get the cost
down to $1 per block or less.

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Ian Jackson - 27 May 2008 15:35 GMT
>    I'd like to take the thread off on a new tangent. I notice that the
>commercial occupancy detectors are pretty expensive on a per block
>basis, around $5 or more. What is the cheapest detector circuit that
>could be built? I've been checking out the various current sensors
>available and I still haven't found anything that could get the cost
>down to $1 per block or less.

Well, not in dollars, but my circuit (see my previous posting) can
be made as cheaply as this:

1/4  quad optocoupler eg TLP521-4            L1.08    27p
1x   bridge rectifier (4 diodes) eg Fairchild DF08M        15.2p
2x   resistors eg 2% 0.1W                L0.001    0.2p
                                                               -----
                                42.4p

which is a little under $1.  I'm assuming that you're having lots and
lots of these - at least a hundred - so I used Farnell, a serious UK
electronics supplier, and assumed (for example) the 50-off price for
resistors and the 25-off price for quad optocouplers.  You need to
budget for some board to mount it on too which will at a guess will
add 25-50p to the cost per block.  Also these prices don't include VAT
(UK sales tax).

Resistor values I use are 56R on the optocoupler input, but this will
depend on the desired sensitivity and the voltage drop provided by
your actual bridge rectifier.  I use a 100K pull-up on the output.

In my setup I then feed the outputs to PIC18F458 microcontrollers for
scanning and multiplexing; the PIC has quite good properties on its
inputs which means I can feed the output from my circuit straight into
its general IO pins as they have an excellent high input impedance of
1MR or so, don't seem to mind wavering levels, and which I can program
to debounce the incoming signal before reporting detections.

In my setup, if you count the cost of the microcontroller too, and the
custom PCB to mount it on, and all of the gubbins for reprogramming
the microcontrollers, talking to them from the computer, etc. then it
works out as a lot more than $1 per block :-).

But I don't know what you want to do with your output.  If you need an
output which remains constant rather than flickering between `present'
and `not present' at 10-20kHz (due to the NMRA DCC squarewave, with
the detection circuit only working during one half of each cycle) you
need to at least add a capacitor (say, 10nF at 5.3p) and you'll
probably need to use something to provide a hysteresis, eg a schmitt
trigger 74HCT14 (10.5p for 10-off but you get six to a chip) - that
will give you a TTL-level output for another 7.05p per block if you
don't bother to include a decoupling capacitor.  Total: 49.9p which is
pretty close to US$1 :-).

Of course all of this more or less assumes you have a suitable 5V
power supply and want an output you can feed to something resembling a
TTL input.  And as I say I haven't included the cost of some matrix
board for mounting, which will be a significant fraction of the cost.

I think the real cost of anything along these lines depends on what
you're going to do with the output from the detectors.

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Rick Jones - 27 May 2008 23:19 GMT
>>    I'd like to take the thread off on a new tangent. I notice that the
>> commercial occupancy detectors are pretty expensive on a per block
[quoted text clipped - 13 lines]
>
> which is a little under $1.

   I'm afraid I was unable to make heads or tails of the attempt at an
ASCII schematic. Could you perhaps draw up a version in Paint that could
be posted somewhere?

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Ian Jackson - 28 May 2008 11:16 GMT
>> Well, not in dollars, but my circuit (see my previous posting) can
>> be made as cheaply as this:
...
>> which is a little under $1.
>
>    I'm afraid I was unable to make heads or tails of the attempt at an
>ASCII schematic. Could you perhaps draw up a version in Paint that could
>be posted somewhere?

Did you look at the schematic in a fixed width font ?
I can't be bothered to draw another diagram but here's a screenshot of
what it looked like for me:
http://www.chiark.greenend.org.uk/~ijackson/2008/block-detection.png

The four ordinary diodes on the left are most easily found together in
a bridge rectifier: short out the rectifier's `+' and `-' terminals
and connect the two `~' terminals to booster output and track.  (This
ties the two inter-diode midpoints together, which is not a problem.)

The thing with a diode, a little arrow, and part of a transistor is
the optocoupler.  The rectangular boxes are the resistors.

I do have a more useful schematic at home that I could take a digital
photo of and put online if it would help.

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Rick Jones - 29 May 2008 01:04 GMT
>>> Well, not in dollars, but my circuit (see my previous posting) can
>>> be made as cheaply as this:
[quoted text clipped - 19 lines]
> I do have a more useful schematic at home that I could take a digital
> photo of and put online if it would help.

   OK, the diagram at that link looks better. I can make it out now.
Thanks.

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Rick Jones - 06 Jun 2008 00:51 GMT
>  1x   bridge rectifier (4 diodes) eg Fairchild DF08M        15.2p

   Those are only rated at 1.5A. I doubt that I'd trust anything rated
less than 5 amps, and preferably more just to be safe. My old brass
locos could easily pull 2 amps when heavily loaded. Double head a couple
and... well, you get the idea.
   Also I sort of wonder about the reliability of using standard
rectifier diodes or diode bridges on the higher frequency DCC power. I
suspect some of those bridges designed for use on 50 or 60 Hz power may
not have a fast enough shutoff time when the signal reverses.

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Robert Heller - 06 Jun 2008 01:38 GMT
> >  1x   bridge rectifier (4 diodes) eg Fairchild DF08M        15.2p
>
[quoted text clipped - 6 lines]
> suspect some of those bridges designed for use on 50 or 60 Hz power may
> not have a fast enough shutoff time when the signal reverses.

I suspect that silicon rectifiers diodes less than 10 amps switch at
speeds well above 'higher frequency DCC'.  DCC only switches in the
kilohertz range.  

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Greg Procter - 06 Jun 2008 02:19 GMT
> > >  1x   bridge rectifier (4 diodes) eg Fairchild DF08M                15.2p
> >
[quoted text clipped - 10 lines]
> speeds well above 'higher frequency DCC'.  DCC only switches in the
> kilohertz range.

4.5/9kHz.
Ian Jackson - 06 Jun 2008 12:22 GMT
>>  1x   bridge rectifier (4 diodes) eg Fairchild DF08M        15.2p
>
>    Those are only rated at 1.5A. I doubt that I'd trust anything rated
>less than 5 amps, and preferably more just to be safe. My old brass
>locos could easily pull 2 amps when heavily loaded. Double head a couple
>and... well, you get the idea.

The OP didn't specify their scale IIRC.  Mine's N so 1.5A is plenty.

The best thing about working with the proper electronics industry is
that you can just look up this kind of question.  Farnell, for
example, provide a link to their copy of the datasheet PDF alongside
each part listing in their web search system.

The 1.5A is average rectified forward current, so since you've got two
pairs of diodes each of which only needs to carry the load half the
time, you're up to 3A average.  That'll be primarily thermally limited
so if you want more you need to buy a more expensive part in a bigger
package.  The occasional surge it will allegedly cope with is 50A
which I hope is fine even for a large scale setup.

>    Also I sort of wonder about the reliability of using standard
>rectifier diodes or diode bridges on the higher frequency DCC power. I
>suspect some of those bridges designed for use on 50 or 60 Hz power may
>not have a fast enough shutoff time when the signal reverses.

My actual layout uses a higher-rated bridge rectifier, SB601 (Farnell
7278276) and similar, rated for 6A average.  I chose such a chunky
part because I wanted to keep the forward drop relatively independent
of load so wanted big junctions with good thermal properties.  The OP
was trying to optimise for cost in which case those parts at over $2
each aren't the ideal answer.

But even so I've not had any trouble with shutoff times.  I've
conducted tests running 3A through a single block, and everything has
still worked just fine (my tests included looking at the on-track
waveform shape as seen on a scope, under heavy load).  My test load -
a nominally 2W rated 5R resistor - did get very hot as was to be
expected from pumping 40W into it!

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Ian Jackson - 06 Jun 2008 14:23 GMT
>>    Those are only rated at 1.5A. I doubt that I'd trust anything rated
>>less than 5 amps, and preferably more just to be safe. [...]
>
>The OP didn't specify their scale IIRC.  Mine's N so 1.5A is plenty.
...
>The 1.5A is average rectified forward current, so since you've got two
>pairs of diodes each of which only needs to carry the load half the
>time, you're up to 3A average.   ...

Actually on reflection I think this isn't true and the 1.5A figure is
supposed to be for the whole part, rather than per diode.

So if you have locos which pull 1-2A probably you'd be better off with
a bigger part, perhaps something like KBU4D at 57p which entirely
blows the US$1 per block budget.  Instead you can just use four
ordinary diodes eg 1N5400G at 5.9p each, which is twice as much
soldering but much cheaper.

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Rick Jones - 07 Jun 2008 03:39 GMT
>>>    Those are only rated at 1.5A. I doubt that I'd trust anything rated
>>> less than 5 amps, and preferably more just to be safe. [...]
[quoted text clipped - 3 lines]
>> pairs of diodes each of which only needs to carry the load half the
>> time, you're up to 3A average.   ...

   The data sheet I read referred to the rating as "continuous
current", or the maximum amount of current you could put through it on a
constant basis without stressing it abnormally. Not to be confused with
surge current which is a higher amount a device can stand for a very
short, non-repetitive spike.
   At my job, which is repairing industrial controls, we normally order
parts from Mouser, Allied, Newark (the U.S. counterpart of Farnell), and
Digi-Key.

> Actually on reflection I think this isn't true and the 1.5A figure is
> supposed to be for the whole part, rather than per diode.

   In the case of a bridge of any sort, the continuous current is the
amount you have to expect any individual device in the bridge to be able
to handle. So yes, the 1.5A rating would be per diode. I think a normal
rule of thumb is to use a device capable of 150% of the maximum current
you expect to be drawn, and 200% of the voltage. This correlates with
the parts used in the equipment that I repair on a daily basis.

> So if you have locos which pull 1-2A probably you'd be better off with
> a bigger part, perhaps something like KBU4D at 57p which entirely
> blows the US$1 per block budget.  Instead you can just use four
> ordinary diodes eg 1N5400G at 5.9p each, which is twice as much
> soldering but much cheaper.

   Checking Mouser's web site I see they have some discontinued STBR406
600V/5A bridges they're unloading for $.99 ea. Those would work, if I
had the spare cash to spend ATM.

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Bernhard Agthe - 06 Jun 2008 11:33 GMT
Hi,

this is slightly OT, but I'll try anyway.

Yesterday I finished my control pult. It connects to the layout via two
36-contact Centronics plugs. Most of the contacts are used already, so
any occupancy detection I want to add has to be in the control pult only
;-) Since I use a DC system with two throttles and a common return, this
is an interesting challenge ;-)

Anyway my idea is to use a high frequency (some KHz) which I apply to
the common return via some capacity. Detect would be on the cables for
block power supply, again coupled via capacitors. The detectors would be
used to back-light the switches for each block, indicating a train in
that block.

The interesting part is, that the signals would have to travel for up to
4 meters in quasi-parallel wires, through track and train and still be
detectable without interference between neighbour lines... So to say,
the wires themselves will provide some capacity - plus the coupling into
and out of the track power system by capacitors leaves very low
tolerances anyway...

The "sender" would basically be an oscillator circuit and the receiver
would sense any HF voltage parts and use them to power a LED.

Does anyone have any experience with such a system - I'm a hobby
electrician, but have little to no experience with HF stuff...

Actually I'm pretty sure, the principle will work in other than my
special conditions, even with DCC, if the parameters are right ;-)

Thanks in advance for any input on the matter..

Ciao...
Wolf Kirchmeir - 21 Jun 2008 16:18 GMT
> Hi,
>
[quoted text clipped - 11 lines]
> used to back-light the switches for each block, indicating a train in
> that block.

There's actually a US patent on this idea/design. the abstract is vague
enough that it would cover your design, too, I think. I didn't look at
the details.

http://www.freepatentsonline.com/5752677.html

> The interesting part is, that the signals would have to travel for up to
> 4 meters in quasi-parallel wires, through track and train and still be
[quoted text clipped - 15 lines]
>
> Ciao...

For a survey of occupancy detection ideas, look at
http://homepages.paradise.net.nz/~rdmurg/clinic/detectit.htm

HTH

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Bernhard Agthe - 23 Jun 2008 11:52 GMT
Hi,

> For a survey of occupancy detection ideas, look at
> http://homepages.paradise.net.nz/~rdmurg/clinic/detectit.htm

Thanks for your input on the topic, I do appreciate it.

Probably I will go for the approach of inserting a high-freq. voltage
and trying to detect it, but I'm still not sure whether to use
transformers or (probably) capacitors - I will report here if I find any
interesting things ;-)

Ciao...
Klaus D. Mikkelsen - 23 Jun 2008 22:04 GMT
> Probably I will go for the approach of inserting a high-freq. voltage
> and trying to detect it, but I'm still not sure whether to use
> transformers or (probably) capacitors - I will report here if I find any
> interesting things ;-)

I would defenitly go for an opto coupled system.

Much more easy to work with.

Klaus
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Bernhard Agthe - 25 Jun 2008 11:27 GMT
Hi,

>> Probably I will go for the approach of inserting a high-freq. voltage
>> and trying to detect it, but I'm still not sure whether to use
[quoted text clipped - 4 lines]
>
> Much more easy to work with.

In a way you're right. Opto's are nice. But I'm not sure they'd work in
my case. Just assume, I generate some tickle-voltage oscillating with
say 1kHz at a few Volt. How can I insert it into a non-powered track
with an Opto? With a transformer this has been done often (light power
on DC?), I'm not sure whether a capacitor will work (it should by theory ;-)

The detection is the other side, the tickle voltage is sent through say
a non-lighted car with a 1K-Ohm resistor accross the axle. The inserted
voltage will result in a current of a few milli-amps - hardly enough to
"enlighten" an opto... I'm not sure if there's a way to uncouple that
current from track power reliably (again with capacitor or transformer).

So in principle I'm sure my system should work, but I'm also sure,
practical conditions will leave very little tolerance, if this works at
all...

Actually I did have another idea last night. There are capacitance based
sensors to detect whether a room is filled (by a person) or not. These
systems work with the air of that room as dielectricum and sense a
change in capacity. Although I'm sure this would work in a model
railroad scenario (provided the train occupies a significant part of
that section of track) there is lots of doubt if this will work with
train power supply on the same wire...

Hmmm... I'm almost through track-laying, next step is wiring the layout
and then some scenery, this leaves enough time to think about occupancy
- until then the "optical" solution must suffice: just look if the track
is free ;-)

Ciao...
Ian Jackson - 25 Jun 2008 15:13 GMT
>The detection is the other side, the tickle voltage is sent through say
>a non-lighted car with a 1K-Ohm resistor accross the axle. The inserted
>voltage will result in a current of a few milli-amps - hardly enough to
>"enlighten" an opto... I'm not sure if there's a way to uncouple that
>current from track power reliably (again with capacitor or transformer).

Optocouplers are a lot more sensitive than you're giving them credit
for.  My system is able to detect the leakage current when I put my
dry finger across the DCC-energised track, which I haven't measured
but will be perhaps a few tens of microamps.  The real question is
what the impedance is of the circuit at the optocoupler output.  My
circuit is built around microcontrollers whose input pins are at least
1MR.

Look at the datasheet for the optocoupler you're using.  The figure
you're looking for is `current transfer ratio'.  I use TLP521-A whose
datasheet quotes 50% minimum.  The impedance presented by the circuit
on the output in my system is dominated by a 100kR pullup resistor;
that output (ie, the microcontroller input pin) will go from 5V to 0V
if the optocoupler's phototransistor draws 50uA.

The actual trip point of the microcontroller input pin is somewhere
around 1V but we can see that given the quoted current transfer ratio
of 50% to 600%, my circuit will be trip at detection currents of
approximately 8uA to 100uA.

Of course you can't run a track occupancy indicator on a layout
diagram - even an LED - directly off this circuit.  The current
transfer ratio of 50% is hopelessly low.  But this is easily fixed by
(say) using a FET as an amplifier.  FETs have gate impedances of 1MR
or more.

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Bernhard Agthe - 26 Jun 2008 14:42 GMT
Hi,

> Optocouplers are a lot more sensitive than you're giving them credit
> for.  My system is able to detect the leakage current when I put my
[quoted text clipped - 3 lines]
> circuit is built around microcontrollers whose input pins are at least
> 1MR.
[...]

Thanks for your input, I will add this to my list of things to try ;-) I
will have to use a shunt to take the bulk of the train current and
generate a voltage drop, if I understand this right?

Ciao...
Klaus D. Mikkelsen - 25 Jun 2008 15:54 GMT
> In a way you're right. Opto's are nice. But I'm not sure they'd work in
> my case. Just assume, I generate some tickle-voltage oscillating with
> say 1kHz at a few Volt. How can I insert it into a non-powered track
> with an Opto? With a transformer this has been done often (light power
> on DC?), I'm not sure whether a capacitor will work (it should by theory ;-)

Why use the oscillator and all that - it is way to advanced and
difficult.

The optocoupler works, if you feed the track voltage to the tracks via a
10Kohms resistor and nothing else.

Klaus
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Dale Gloer - 26 Jun 2008 16:17 GMT
Exactly, an oscillator is overkill.  Just use an isolated low voltage AC
supply, say 12 to 15 volts and a current limiting resistor to each
detection section.  I used Bruce Chubb Optimized Detectors, a 15 volt AC
supply and 1K0 current limiting resistors.  My rolling stock has 10k0
resistors, 1 axle per car.  This worked just fine and did not cause any
problems even after I converted to DCC but hadn't got around to updating
all the wiring.

Dale Gloer

>> In a way you're right. Opto's are nice. But I'm not sure they'd work in
>> my case. Just assume, I generate some tickle-voltage oscillating with
[quoted text clipped - 9 lines]
>
> Klaus
Bernhard Agthe - 27 Jun 2008 10:01 GMT
Hi,

> Exactly, an oscillator is overkill.  Just use an isolated low voltage AC
> supply, say 12 to 15 volts and a current limiting resistor to each
[quoted text clipped - 3 lines]
> problems even after I converted to DCC but hadn't got around to updating
> all the wiring.

Thank you all for your input, this looks much more simple than I had
imagined ;-)

Ciao...
 
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