Track circuits

[David TaylorParticipant @davidtaylor63402]

[David TaylorParticipant @davidtaylor63402]

Hi,

I'm having trouble understanding the track circuiting article from Beer Heights Light Railway.

I think relays work by having current flowing through a coil, moving the internal switch from one contact to another when it is energised.

I think the article saying that when no wheels are bridging the rails, current flows down one rail, through the relay – energising it – and back through the other rail. When a wheelset bridges the rails the current can flow more easily through that so doesn't reach the relay, which de-energises.

Now onto the current leakage through the sleepers and ballast. On a wet day more current can conduct through those, so less current reaches the relay – perhaps not enough to energise it, hence the variable resistor near the power source. More current must be allowed into the circuit in this case.

In the hysteresis paragraph it talks about the on/off voltages for the relay. I don't understand that bit. Is the relay energised based upon current flowing or voltage over it?

If I have the above wrong, can someone explain it in terms a 10 year old can understand?

[DiodeDickParticipant @diodedick]

Electro-mechanical relays (typically) comprise a coil wrapped round an armature. The frame supporting the armature has a movable bit which can change over the contacts, and sometimes move a "flag" which indicates whether the relay is on or off. Current passing through the coil creates a magnetic field which attracts the movable bit and changes over the contacts. The coil has a nominal voltage, 50volts DC, or whatever and the maker's data sheet will specify the tolerances on the nominal figure. During system commissioning "Pick up" and "Drop off" voltages will be recorded. Pick up is usually about 80-85% of nominal and drop off maybe 60%. The current passing through the coil is dependent on the voltage and is substantially less when the relay is "energised" because the moving bit is drawn onto the exposed end of the armature, closing the magnetic circuit. This characteristic was exploited by the Laycock overdrives fitted to 1960's cars. A fair bit of current was required to energise the relay, and end hence engage the overdrive, but comparatively little to maintain it. The other side of this trait is that if sufficient current is not available to properly pick up the relay, the magnetic circuit will not be closed, full current will be drawn and the coil will burn itself out. That is how washing machine door lock solenoids fail during supply "Brownouts"

Those with nothing better to do in the workshop will find lots more online about the Clifford and Snell D2600 relays which were the industry standard for the protection circuits of large power plant turbines of yesteryear.

Dick

[duncan webster 1Participant @duncanwebster1]

The current passing through a relay cool is dependant on the impedance of the coil and the applied voltage only. For DC setup the impedance is just the resistance, for AC then coil inductance comes into it.

[DiodeDickParticipant @diodedick]

Yes, I know the difference between internal resistance of a battery and source impedance of an ac circuit, but in line with the request in the original post, I was trying to keep the the explanation within terms that a 10 year old would understand.

Dick

[Robert Atkinson 2Participant @robertatkinson2]

Beer Heights LR uses DC track circuits so lets stick with that and not confuse the OP with AC.

A relay is operated by current. They are an electromagnet operating a switch. The force from an electromagnet is proportional to the number of turns of wire in the coil and the current flowing through it. So lots of turns need less current. Most applications actually operate on voltage not current. For a given voltage the current depends on the resistance in the circuit (Ohms law I =V/R). The wire in the relay coil has resistance and the designers of the relay choose a wire size so the required number of turns to operate the switch (at current x) have the correct resistance to cause the correct (x) current to flow at the required voltage.
With the relay de-energised the moving part that is attracted to the electromagnet (armature) to close the switch is at it furthest away from the electormagnet so the attractive force is weakest .Typically switch is held OFF by a spring. This will be at it's weakest point when off. This helps balance the weak force from the electormagnet in this position. The spring force increases in a linear matter as the armature moves to the ON positin (closer to electromagnet) but hte magnetic force increases with the inverse square of the distance. Thus in the ON position the magnetic force is much higher than required to over come the spring force. So onec the relay is ON it takes a lot less current (or voltage) to keep it on. This is hysteresis. For a simple relay once it is on it will stay on even if the current (voltage) drops to half the value required to trun it on in the first place.

Robert.

[Ian BurksParticipant @ianburks19652]

I haven't see the article you refer to however, you are correct that the principle of operation is that a proportion of the track circuit current is diverted via the locomotive & carriage wheelsets. If correctly setup, the track relay should drop indicating that the track section is no longer clear.

It is also true that part of the current will leak via the ballast & sleepers. As you say,this depends on the weather, the amount of oil & cinder in the ballast, the amount of water dripping from the roof of tunnels, etc. Remember also that Concrete sleepers are full of rebar. This all needs to be taken into account when setting up the track circuit. Typically the ballast resistance (on a standard guage track) could be between 2 Ω when wet to 100Ω when frozen for a track section up to 1000ft long.

However a track relay is a sensitive device. Its DC resistance might be only of the order of 5-10 Ω. Typically this type of relay may require 100-140mA to energise. It is the current flow that creates the magnetic field but, from Ohms law, you can see that the relay will energise or "pick" with only around 0.5V across its coil.

Once the armature has pulled in, the air gap in the magnetic path is smaller & therefore it will remain energised even if the current reduces slightly. This is the Hysteresis. Typically the pickup voltage might be 0.5V but the drop-away voltage might be 0.35V.

When testing a track circuit, a variable resistor is connected across the rails & reduced until the track relay drops. This is known as the "drop shunt" value. Then the test resistor is set to zero & increased until the relay just picks. This is known as the "pickup" or "prevent shunt". As a rule of thumb, the higher should be at least 133% of the lower. In addition, a shunt of 0.5Ω at any point along the track should drop the track relay. This will generally give a good compromise between reliably detecting a clear track while coping with varying weather & track conditions.

[David TaylorParticipant @davidtaylor63402]

Thanks all!

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