0

Hi I am a newbie and hopefully my question is simple to answer.

I have a single magnetic reed switch in a track of a model railway ahead of a point/turnout. The reed switch is connected with 16V AC and a diode. The model trains have a magnet that triggers the reed switch as it passes. The two inputs of the turnout motor connect to the reed switch and the transformer. That is fine for triggering a thrown switch to closed or visa versa by switching the polarity wiring of the diode. However, I want the turnout to alternate between "thrown" and "closed" so that each train that passes goes in alternate directions.

I know that this would be easy to achieve if I placed a reed switch on each track after the turnout (each with opposite polarities). But for various reasons (less wires) I need to use a single reed switch before the turnout.

I am assuming some form of flip flop relay (maybe an H bridge?) is what I need but I am getting confused about the best way to trigger or wire it (and how to avoid it bouncing back and forth).

Thanks

Rich Lam
  • 1
  • 1
  • Sounds like a bad idea to me. You're creating a situation in which the turnout can move while the train is still passing over it. – Dave Tweed Oct 21 '23 at 18:15
  • No the reed switch is before of the turnout so it switches before the train enters the turnout. – Rich Lam Oct 21 '23 at 18:18
  • Ah, OK -- so you're also moving the magnet to the other end of the train(s). So what you want is an edge-triggered flip-flop using relays -- see the second diagram in that answer. – Dave Tweed Oct 21 '23 at 18:28
  • Thanks. The magnet is in the engine of each train. I tried to see how the circuit in the diagram would help my situation but it is beyond me. I will do some more reading on edge triggered flip flops. – Rich Lam Oct 21 '23 at 18:43
  • You can get impulse latching relays, which change state each time you apply power to the coil. Or you can buy modules that do this by combining a flip flop with a regular relay. – Finbarr Oct 21 '23 at 21:13
  • Thanks! That sounds like precise what I need. I think this might be it "6-24V Flip-Flop Latch Relay Bistable Self-locking Low Pulse Trigger Module. Trigger mode: Low pulse trigger This is a bistable (self-locking) relay module, Trigger once, the relay pull (and holds);Trigger again, the relay releases (and holds)". – Rich Lam Oct 22 '23 at 02:05
  • Also it sounds to me like I would need two reeds ahead of the turnout. The first is the impulse relay to switch the diode polarity used for the second reed switch which would trigger the turnout motor. I am worried that if I just had one reed switch and the train magnet passed over for 1 second it would keep flip flopping. Or is there a way to stop any bouncing and still use one reed switch? Thanks for your answer that is just what I was looking to find out. – Rich Lam Oct 22 '23 at 02:13
  • I probably should clarify that the turnout motor just needs a pulse to move from 'closed' to 'thrown' (it may burn out if power was left on). So my real follow on query is whether I could wire a single reed (no diode) to the Trigger on the impulse relay and also wire the same reed to each load side of the relay (with alternate diode polarities). In that case, would the impulse relay trigger and flip in time for the pulse to run through the load side of the relay? – Rich Lam Oct 22 '23 at 02:29

1 Answers1

0

A single reed switch, an impulse / ratchet relay and an electrolytic capacitor are required.

enter image description here

The impulse relay, getting set / reset in alternate passes, will generate capacitor charge / discharge pulses to set / reset the points.

The only question is the availability of a 16 V ~ impulse relay.

vu2nan
  • 18,175
  • 1
  • 16
  • 47
  • Thanks. I'm sorry (being a newbie) but I don't quite follow how using only one diode would allow the point motor to flip flop - and how the capacitor could feed the alternate polarity to the point motor. The point motor has two connections - and two diodes are typically used to control the polarity of the AC fed into one connection (eg using a manual switch to flip flop if it were done manually). The other side of the AC is fed to the other connection. I have found DC impulse relays that handle 6-24V - maybe I could use a bridge rectifier to convert the trigger signal into DC? – Rich Lam Oct 22 '23 at 11:10
  • Just thinking about it, I should use a low voltage DC signal to run through the reed switch for the trigger side of the impulse relay and then separate 16V AC power supplied to the load side. So finding the relay should not be a problem. – Rich Lam Oct 22 '23 at 11:34
  • Hi Rich, The diode is for half-wave rectification of the 16 V ~ source. The capacitor charge and discharge pulses, of opposite polarity, are fed to the point motor by the impulse relay contacts. In this design two diodes are not required. – vu2nan Oct 22 '23 at 17:52
  • Yes, Rich, The two diodes have been used to switch the points with half-wave rectified voltages of opposite polarities. You may use a bridge rectifier.You may also try the '6-24V Flip-Flop Latch Relay Bistable Self-locking Low Pulse Trigger Module'. – vu2nan Oct 22 '23 at 18:06
  • Thanks very much I think I have almost understood how it works. You have shown the "discharge"/reset state in the diagram. When the impulse relay flips the capacitor is then connected to the half wave rectified AC to charge and (from what I see) that rectified AC also flips/sets the motor. However, in the charge/"set" position isn't the rectified AC flowing continuously through the motor (rather than being a pulse) until the next reset? That is the part confusing me now. It needs to be a pulse to avoid burning out the motor. – Rich Lam Oct 22 '23 at 19:57
  • Anytime Rich! The charging current is DC (half-wave rectified) not AC. Hence there is no question of continuous current. – vu2nan Oct 23 '23 at 04:10
  • Thanks again for your patient explanations. I understand that it is not strictly continuous as it is half wave rectified AC (so pulsing DC). However I am still concerned that the continuous pulsing DC would burnout the motor in the charge/set position. For manual control the switches used are momentary switches so the pulsed DC only flows for a second or two. I understand in the discharge position the size of the capacitor would determine the length of the pulse. But when in the set position I understand the rectified AC would flow until it resets. Am I missing something? – Rich Lam Oct 23 '23 at 05:13
  • Hi Rich, In the case of AC the polarity changes every half cycle. As the polarity alternates, capacitors are repeatedly charged and discharged allowing continuous alternating current flow. However, in the case of half-wave pulsating DC, the current flow ceases when the capacitor is fully charged. – vu2nan Oct 23 '23 at 05:38
  • OK. I had assumed that in the charge/set position the capacitor would be 'continuously' charged with the half wave AC and continuously drained by the motor - so that the half wave AC would effectively just flow through the capacitor and the motor while in the charge/set position. But I have asked enough questions, best way to understand now is to test it out. Thanks again for your help. Much appreciated. – Rich Lam Oct 23 '23 at 05:45
  • Hi Rich, Have fun! – vu2nan Oct 23 '23 at 05:56