Kato turnout controller using capacitors

[peteski]

Vishay still has a few legacy ex-Sprague parts among their product offerings with 15V ratings.

I was waiting for someone to prove me wrong.    Yes, I also remember some old Japanese radios using 15V capacitors, but the key is "vintage".  And as I mentioned, I would never recommend using electrolytic caps at their rated voltage.  They are more forgiving then tantalums, but still  . . .
Steve, I'm not sure how you will operate 2 separate switches. You will first select the route, then press the pushbutton switch?
Also, adding series resistors for chargin the cap(s) will reduce the maximum current, and also slow down the time to charge. If you add more capacitors, you woudl have to use smaller value resistor to keep the charging time reasonably short.

Think of each cap as an empty bucket and the resistor as a hose which fills them.  To fill more buckets at once you will need a larger diameter hose (less resistance, more "water" or current flow)

Now that I see the Stillwell circuit, I really like it.  It is quite clever, and it's somewhat like Tortoise wiring where the constant power (through a 1k resistors) powers the status LEDs.  Power it from a filtered DC supply and forget about ripple current.  Why complicate things?

[mmagliaro]

Max, If I add more caps to power more coils with the same switch, I guess I should be, in effect, adding resistors in series to keep the peak current to about the same level, right?  So, for a double crossover with 4 coils, the capacitor would be something like 4000 µF and the resistor would need to be more like 40 ohms and a higher watt rating, too, right?  I am thinking just size the resistor for the 4 capacitors, whatever they turn out to be, and use that for any lesser number of capacitors, too.  I am thinking that the capacitor(s) would still recharge plenty fast enough, especially if they are a bit oversized for the coil current requirement.

Additional thoughts?

No.  If you increase the resistor to 40 ohms for a 4000 uF cap (instead of 10 for a 1000 uF cap), you will reduce the current draw by a factor of 4.  The capacitor size has nothing to do with the current peak once you put that resistor in there.  Think of the cap as a dead short at the instant you turn it on.  The only thing limiting the current is the resistor.  If you keep the R at 10 ohms and put 4000 uF on it, you still get (at 15v) 15/10 = 1.5 A.  The only thing that changes is the time it takes for the caps to recharge.

[Carolina Northern]

I don't post much anymore, but when Ray's circuit comes up.....(he always went by Ray, George was his father).

Ray's circuit has been given kudos and brickbats since day one. All I have is pluses - it just works.

What people forget is Ray was an IBM engineer from the old days. He's the guy who came up with how to send data over fiber optic cable. His name is on the patent for it.

Going back to 1958, he's the first person to come up with diode drop constant headlights.

Lynn Westcott credited Ray with why he wanted to learn electronics and how apply it to model railroading.

He tweaked that circuit over the years, but the principle is sound. 
I would believe him when he says how to do it.

My experience is 4700 mf is the right cap for the four way.

Don

[Maletrain]

Max, Yes, I see your point (brain fart on my part). I will stick with the 10 ohms in the main feed from the power source.  I will probably be feeding several individual push button controllers for different turnouts or groups of turnouts, with different amounts of capacitance attached to the individual buttons.  Only one group of capacitors attached to one push button will be fired at any given time, and the recharge time will differ between them depending on how many 1000 µF (?) capacitors are in parallel connected to a given button.  Do you have an estimate for the recharge time to, say 95% or 99% of 15 volts, for 4000 to 5000 µF though 10 ohms of resistance?  (My idea of "research" is to first ask an expert, rather than start by going to Google or trying to find my college text books.)

Peteski, We are getting off on a misunderstanding tangent.  I do not have any wish to use caps rated at 15 volts.  I was just initially confused by the apparent confluence of some different terminology used in the 2 separate papers by Stillwell and Bronson.  Stillwell used the calculation "22000/22" where the 22 is what he calls the "resistance" in ohms for the single solenoid coil in a Kato turnout.  I do not physically understand where that 22000 comes from, nor the basis in physics for the calculation.  I am only assuming that it comes from some empirical relationship developed for the discharge of a capacitor through the impedance of a solenoid coil that is sufficient to move the solenoid slug.  The confusion in my mind occurred when I saw Bronson's paper showing "2200/35" by the caps and mistook that for some sort of calculation instead of a specification for capacitance and voltage rating.  It was not making sense to me until I realized they were two different things.  So, if you want to explain something that I am totally missing, that would be where the "22000/22 = 1000 µF" comes from in Stillwells paper (on page 1).  [Also note that Stillwell was using "mfd" for µF.]

To answer the question about how I intend to use the controller I sketched, yes, I would first set the DPDT toggle switch to the position that I want the turnout to be in, then push the momentary SPDT button to make the turnout actually move to that position.  That is the same way that a very popular control system works (for instance see

), except that system requires short presses of the momentary push button to avoid killing the turnout solenoid coil.  Mike Fifer does essentially the same thing with just a momentary, center off DPDT toggle, but that has the same potential solenoid burnout problem as well as no indication of the turnout position. (See https://www.fiferhobby.com/how-to-make-kato-turnout-control-switches/ )  Similar systems and others add complexities to provide LED indicators of turnout position.  But, with Kato turnouts, it is not so easy to show actual physical position, so most of these show only the last position created by electrically throwing the turnout, and are not able to accommodate somebody sliding the manual control slider on the turnout itself.

So, what I am aiming for is a very simple system that shows position (so far as it can be done with a toggle position) and avoids the coil burnout potential.  If I can't see the turnout from where I am at a panel, I can just push the button with the toggle in the position I want, and that should assure that any previous manual manipulations are countered.

When I started this thread, I was hoping somebody had the empirical experience to tell me what number of µF at a specific feed voltage is needed to reliably throw a single Kato turnout solenoid.  There are some indications of issues with the papers I am using.  Stillwell writes about using "12 volt" power supplies, while Bronson says that he needed at least 15 volts to get reliable throws.  And some designs add capacitors across the power feed to supplement the peak current available from the power supply when a turnout is thrown. 

What I would like is a design that provides reliability for a successful turnout throw that is independent of the peak current available from the power source.  I understand that the actual open circuit voltage produced by the power source will be important to that.  Unfortunately, the specs on wall warts and other power supply devices often do not give any hint of the open circuit voltage that they will eventually charge a capacitor to.  For example, the wall wart I am using now is rated at 14.5 VDC and 0.8 A, but puts out 18.67 VDC with about 2 VAC ripple on an open circuit.  That would definitely make a Kato solenoid buzz if connected with the simple Stillwell circuit.  But, for now, I am using it with the Kato mechanical devices, so the power is only supplied to a solenoid for a fraction of a second with those momentary switches.  I worked my way up in power supply voltages until I had one powerful enough to throw a Kato double crossover.  But, that is about 4 times the power I need to throw a single Kato turnout.  The design I am trying to develop would depend only on the capacitance I have provided in each individual actuating circuit (and the open circuit voltage of the power supply, of course).

[mmagliaro]

A few things...

While Stilwell's document says "1000 uF per coil", that number isn't just for any coil.  He based that on a 22 ohm Kato switch machine coil.  He calculates it based on the formula  C = 22000/R
If you look at a capacitor discharge curve, that gives you a discharge where the charge voltage on the cap drops to 1/3 of its
value in about 22 mSec.  And lets further assume he was shooting for 20 mSec to be the time where the cap was 2/3 discharged.
So, as to why 20 mSec is a good number, I don't know.  He may have just experimented with a bunch of turnout coils
and decided 20 mSec was a long enough time that machines would always reliably throw.

As for how long it will take your caps to recharge to 95% with 4700 uF and a 10 ohm resistor in there....  the same capacitive charging principle applies.   Here's a handy page that shows you the charge equation for a capacitor, but also lets you put in values and plot a handy dandy graph of the charge voltage.  For 12v, 4700uF, 10 ohm, it will take about 150 mSec to charge back up to 11.5 volts (from a 12v supply).
https://www.redcrab-software.com/en/Calculator/Electrics/C-Charge-State

I doubt that 150 mSec will be onerous to wait for if you need to throw the switches and then throw them again soon after.

Also... without the 10 ohm limiting resistor in there, that 4700 uF cap will draw a 35 A peak to recharge, and the hump of that peak is over 30 A for about 1 mSec.   It may only be 1 mSec, but I'd still put that resistor in there.

[Maletrain]

Thanks Max!  I'll use that calculator a lot.

Recharge times look inconsequential at that level.

[mmagliaro]

Thanks Max!  I'll use that calculator a lot.

Recharge times look inconsequential at that level.

I agree.  But if you do want to get the recharge time down to "near instantaneous", but still avoid the current surge, there are ways,
but the circuit gets more complicated.  Considering that the bulk of the current surge only lasts about 3 mS (that's about how long it takes for the recharge current to drop under 5 A), the circuit would need to limit the current for a really brief time, say, 10 mSec, and then let it charge without the resistor after that.  That would get you something like a 10-15 mSec recharge instead of 150.

[peteski]

Does Kato provide and specs or ratings for what is required for powering their switch machines?  Kato must sell some sort to power supply for this.  Something like the accessory output from their power-pack.throttle? No, I don't own one.

[Maletrain]

Max, I am happy with simple and recharge times under a second - very happy.

Peteski, Kato sells a little rectifier that snaps onto the side of their mechanical turnout thrower, and that has "AC 17 V" cast into the plastic.  The power from this gets daisy chained through to other mechanical throwers that snap together with what look like 9 volt battery connectors but with the individual polarity snaps spaced much farther apart than on the batteries.  The individual throwers say "DC 12 V" and have the connection snaps marked as "+" and "-".  When I power the little rectifier from my nominally 14.5 VDC wall wart that actually outputs 18.67 VDC 2VAC at open circuit, the Kato inverter reads 18.15 VDC and still shows the 2 VAC on top of that (?). 

As I posted before, I tried wall warts with lower nominal voltage ratings, and they would not throw all 4 of the solenoids on the Kato double cross over.  Another somewhat odd observation was that sequential throwing of the turnout when it was getting marginal amounts of power showed that it was getting less effective with additional throws.  That is, it might throw 3 of the 4 all of the way and not get the 4th point all the way across on the first attempt, and on subsequent attempts, it would get to the situation where the 4th point didn't move and the 3rd didn't get all the way across.  I am suspecting the solenoids got warm and had higher electrical resistance, but it might also have been an issue with the various power supplies I was trying.  But, this nominally 14.5 VDC 800 mA wall wart reliably throws all the points of the double crossover back and forth a few times at least without issue. 

I don't own a Kato OEM power pack to see what it puts out.  I originally powered the Kato turnouts with a tap into the power output of a Bachmann wall wart that powered its Bachmann short-address-only little DCC system, and that actually measured 18 volts output.  I have forgotten if it was AC or DC, but I am thinking it was DC.

[peteski]

As I posted before, I tried wall warts with lower nominal voltage ratings, and they would not throw all 4 of the solenoids on the Kato double cross over.  Another somewhat odd observation was that sequential throwing of the turnout when it was getting marginal amounts of power showed that it was getting less effective with additional throws.  That is, it might throw 3 of the 4 all of the way and not get the 4th point all the way across on the first attempt, and on subsequent attempts, it would get to the situation where the 4th point didn't move and the 3rd didn't get all the way across.  I am suspecting the solenoids got warm and had higher electrical resistance, but it might also have been an issue with the various power supplies I was trying.  But, this nominally 14.5 VDC 800 mA wall wart reliably throws all the points of the double crossover back and forth a few times at least without issue. 

So the specs are kind of vague.  Going back to the DC throttle days, turnouts were powered from the accessory terminals output, which was usually rated 16V AC (basically it was the direct output of the transformer before the rectifier).

Going back to your example, if   some wall-wart can throw single turnout reliably, but not multiple, the voltage is not your problem - the amount of current available is the problem. But in the cheap power supplies (where transformer output is feed to a rectifier with no regulation) the voltage is related to the amount of consumed current. If the current consumption is low enough (single coil being operated) the voltage will stay high enough. When multiple coils are connected that will increase the total current draw, so the voltage supplied from the power pack drops, causing the turnouts not to throw. 

If the coils have 22 ohm resistance then your 14.5VDC pack should supply at least 14.5/22=0.65A.  Since it is rated at 0.8A, it provides full  voltage at 0.65A.  But with 2 coils together, that creates 11 ohm load and now it will consume  1.32A@14.5V, which is overloading the power pack, dropping the voltage down from the specified 14.5V. With 4 coils you have 5.5 ohm load and see where I'm going with this.

But with the same pack and a capacitive discharge unit, large enough capacitor charged by that same power pack will produce an ample short pulse of current to throw all 4  turnouts.  I'm sure our Gary Hinshaw could come up with mathematical solution for the size of the cap to supply high enough voltage for long enough period to throw all 4 turnouts, but I would just experiment. There are too many variables involved. 

BTW, I'm not sure what that "18.67 VDC 2VAC" is. Are you measuring the open circut voltage using both, DC and AC range?  If yes, then that 2VAC is likely the AC ripple from rectified and filtered transformer output (with no voltage regulator).

[Maletrain]

Peteski, I agree with all you just posted. 

Regarding the measurements of open circuit output from the nominal 14.5 VDC wall wart, yes, I am just using a multimeter on DC and then AC settings to see 18.67 volts and 2.05 volts, respectively.

Since we are talking about the way that wall wart works, I will add that the AC reading has a repetitive set of reading jumps that I don't understand.  It mostly  reads a steady 2.05 volts, but suddenly jumps to 19.something and quickly goes back to 2.05 volts, then goes to 100.0 volts with the "1" flashing and a beeping sound, then quickly back to 2.05 volts again.  This repeats with what seems to be a steady frequency.  The only load when doing this is the little Radio Shack digital multimeter.  ¯\_(ツ)_/¯  I don't think I really need to understand that, but I'm throwing it out there to see if it rings any bells on your end.

[Maletrain]

Don, What power supply are you using with Stillwells circuit and 4700 µF to throw the Kato double crossover?  Do you know the open circuit voltage?

[peteski]

Since we are talking about the way that wall wart works, I will add that the AC reading has a repetitive set of reading jumps that I don't understand.  It mostly  reads a steady 2.05 volts, but suddenly jumps to 19.something and quickly goes back to 2.05 volts, then goes to 100.0 volts with the "1" flashing and a beeping sound, then quickly back to 2.05 volts again.  This repeats with what seems to be a steady frequency.  The only load when doing this is the little Radio Shack digital multimeter.  ¯\_(ツ)_/¯  I don't think I really need to understand that, but I'm throwing it out there to see if it rings any bells on your end.

I love my little clam-shell Radio Shack multimeter. I use it all the time. For a small cheap meter it actually has higher measured range than most typical 3¹/₂ digit meters.  unlike those which overflow at 1999, this one keeps on going up to 3999.  As far as the weirdness goes, it is probably related to auto-ranging logic in the meter. Probably something in the voltage being measured confuses it.  Yeah, I wouldn't worry about it.

[mmagliaro]

Okay, I know you said 150 mSec was plenty fast enough for the recharge time, but I couldn't help myself... so I did some poking in
my circuit simulator and came up with this...

S1 is your pushbutton.  When in the position shown, it recharges the cap (the normally closed position).
When you push the button, S1 flips the other way, and discharges the cap into the switch machine (or machines)

Some things to note here:
The circuit actually lowers the negative terminal of the cap (the one connected to the transistor) as it charges, as opposed
to the way you might always think of charging as increasing the positive terminal.  But it accomplishes the same end - creating
a 12v potential across the cap.   Note the charging time in the voltage graph (upper right): only about 16 mSec!  And note the max current peak - only 4A that only lasts 16 mSec (the second graph at lower left).   Another interesting side effect of this, if you look at the voltage graph: it charges LINEARLY as opposed to the usual exponential curve charge that a cap has.   Doesn't really matter here, but could be useful for something else.

The transistor is a beefy one: 90W 15A, and you'll need it (max power when charging cap is a spike of about 50W).  It's still a pretty small, cheap device (about $1.50 when you are buying 10 of them at a time).  R1 shoulld be a 1 watt resistor.  The base power is about 1/2 watt.   Diode D1 is in there just in case the switch machine coil puts a nasty reverse spike back into the circuit which would destroy the transistor or the switch contacts.

[peteski]

Max, that reverse voltage spike generated by the coil only happens when a fully energized coil is disconnected form its voltage source.  The electromagnetic field collapses, creating reverse voltage across the coil terminals.  In your circuit the coil is only powered by the capacitor which discharges very quickly (even before someone takes their finger off the switch), so since the coil is not energized when the switch is released, there is no voltage spike. I guess a diode would not hurt though.