Hi

I know this is a model railroad turnout, but does it have contacts that indicate the position of the switch?
A real electric switch machine has position indication contacts that are used to provide feedback to the control circuit logic that the switch has actually moved to the requested position. When power is applied to the switch motor, power is maintained to the motor until it completes movement to the requested position. The actual position is then checked against the requested position before a switch position indication is provided to the operator. This type of logic check is called a "switch correspondence" check.

For what its worth...I used to design TCS/CTC systems back in the day...hope that helps.

eT

Such contacts as you describe are not standard on a model railroad turnout. Some makes of switch machines do have extra contacts (Tortoise) while others do not (Atlas did not in the past). Various methods of controlling the switch machines would include circuitry to indicate the turnout position to make up for the lack of contacts.

Also, model RR turnout switch machines or control might have to include contacts to power the frog of the switch.

 

Please help...

I've now built the circuit and tried it out on the model railway and it works... almost!

I used a 2200uf capacitor and 1N4002 diodes, my power supply is 20v dc, please see the attached diagram.

When you flick the switch to the +ve the point motor switches and the capacitor charges, however the point motor is sluggish and only pulls the point about 3/4 across.

When you flick the switch the opposite way the capacitor discharges and the point switches back perfectly with a lot more strength.

If you try switching the point motor direct from the 20v supply it works perfectly in both directions, however this is not a solution since the motor only needs a short burst of power.

So it appears that whilst the capacitor is charging not enough current is getting through to switch the motor effectively.

Do you think the solution might be to increase the voltage or reduce the value of the capacitor?

Thanks

 

Further to my last post, since the point motor switches fine when the capacitor is discharging, could we design a circuit that uses the discharge of the capacitor to switch the point in both directions, not just one?

Please see my attached file (Point Switch2.jpg), I'm sorry this is rather simplistic, I'm sure it will need some modifications to work properly?

 

I dont think that circuit would work very well, I think you need 2 capacitors in place of the diodes. And also a 1 Meg resistor accross each cap to discharge the caps between switchings. Sorry, i dont have a quick schematic editor to make a schematic for you.

 

Do you think the solution might be to increase the voltage or reduce the value of the capacitor?

You could increase the voltage or increase the value of the cap (e.g. double it).

 

Other than your post, I have never seen a capacitive circuit like yours used for turnout switch machines. So, I cannot comment on how it works or how to improve it.

[Removed long description of inappropriate circuit.]
Edit: I forgot the original objective of this post.

Here is a page that describes several circuits for control of twin coil switch machines with a toggle switch. Look here for several circuits It does require a DPDT switch (not an SPDT), but the switch position still shows the position of the turnout

 

The energy stored in a cap is (1/2)*C*V^2, or half the capacitance times the square of the voltage.

So doubling the cap should give you twice the energy to move the coil, while doubling the voltage gives you 4 times the energy.

My only nagging question is if you swap the coils does it give the same performance? I mean if you wire so the discharge happens in the position it does not complete the change does it now work (and the problem go to the other side)? I'm curious the switch maching may need more oomph (a very techinal term you may not be familiar with, measures in units of elbow sweat) to go one way over the other.

If I was doing this I'd try to double the cap first. Doubling the voltage may help, although just increasing it to around 28V also doubles the energy.

Another thing: if your switches start to burn out you may need to add a small resistor in series to limit the very high peak current.

Come to think of it another thing that may be happening is the 20V power can't supply a large enough burst of current which is why the circuit works better in the discharge side. To get a bigger burst of current just add large cap(s) across the power supply itself. That may be better (cheaper) than increacing each and every cap on every switch.

 

Here is a schematic. Note there is a cap for each coil. That way, you're using stored energetic to move to both positions of the points.

 

Thank you for all your feedback

I am planning to build "Circuit 2 - with LEDs" as shown above, but there is one thing I don't understand.

The point motor coil must only receive a short burst of power, not constant power or it will burn out. When switch S1A is closed the coil will receive a burst of power from C1, but afterwards whilst the switch is still closed why is it not still connected to the incoming 20VDC? Do the diodes prevent this from happening?

 

The resistors, in addition to setting the recharge rate for the capacitors, also limit the current to the coils.

 

Thank you for all your feedback

I am planning to build "Circuit 2 - with LEDs" as shown above, but there is one thing I don't understand.

The point motor coil must only receive a short burst of power, not constant power or it will burn out. When switch S1A is closed the coil will receive a burst of power from C1, but afterwards whilst the switch is still closed why is it not still connected to the incoming 20VDC? Do the diodes prevent this from happening?

There are 4 diodes, two across the coil and two between the resistor and the coil.

The diodes across the coils are there to "catch" a huge spike the coil will generate when it is switched off fast. There is a chance this burst may pop the LEDs. It's cheap protection so I would keep it in. It works best when located directly across the coil itself.

The diodes between the resistor and the coil are there to keep the LED off when that sides coil is switched off. Otherwise it would still glow and become dim as the cap on that side recharges just after the other side it thrown. For a 1K resistor and a 2,000 uF cap that glow could take several seconds to extinghish. The diode makes the LED turn off instantaneously.

However... without the diode it will indicate that side is still charging and not ready to be used.

The 1,000 ohm resistor will limit the current once the cap is discharged. A maximum of 20V/1000 ohms = .02 amps will flow even after the cap is discharged. The actual is less and depends on the resistance of the coil itself; measure that with any ohmmeter. Once the switch goes the other side that resistor will recharge the cap.

If I guess (and it is a complete guess) the coil has 300 ohms of resistance the actual current will be 20V/(1000 + 300) = 0.015 amps, and the power in the coil is just 0.015^2 * 300 = 0.000225 * 300 = 0.0675 watts. That is a very safe amount of power to leave flowing continuously in the coil.

 

I have now built the circuit and it works perfectly, so problem solved!

Thanks again to everyone who replied and helped me out

 

The 1,000 ohm resistor will limit the current once the cap is discharged. A maximum of 20V/1000 ohms = .02 amps will flow even after the cap is discharged. The actual is less and depends on the resistance of the coil itself; measure that with any ohmmeter. Once the switch goes the other side that resistor will recharge the cap.

If I guess (and it is a complete guess) the coil has 300 ohms of resistance the actual current will be 20V/(1000 + 300) = 0.015 amps, and the power in the coil is just 0.015^2 * 300 = 0.000225 * 300 = 0.0675 watts. That is a very safe amount of power to leave flowing continuously in the coil.

I've been running some test on the circuit and it takes about 15 seconds for the capacitors to recharge, I'd like to reduce this time say to 5 seconds, but don't want to damage the coils by leaving too much power running through them when they are idle.

The resistance of the coils is only 4 ohms and the supply voltage is 19.4v. Therefore from ErnieM's calculation the current will be

Resistor 1K ohms
19.4/(1000+4) = 0.019 amps and the power in the coil 0.019^2 *4 = 0.0014 watts and 0.077 volts

If I reduce the value of the resistor, to also reduce the capacitor charge time, I presume we should get the following:

Resistor 470 ohms
Coil power 0.04 amps, 0.0067 watts, 0.16 volts.

Resistor 300 ohms
Coil power 0.06 amps, 0.016 watts, 0.25 volts.

Resistor 100 ohms
Coil power 0.18 amps, 0.139 watts, 0.74 volts.

Do you think therefore it would be ok to use a 300 ohm resistor without damaging the coils?