Fail Lamp Detection.

Seems horribly overcomplicated...

A simple non latching push button to connect the 24v DC to the lamp to test it would surely suffice?

If you add more 'stuff' you add to test circuits, then you end up needing test circuit test circuits... so on ad infinitum

Del

Del...with due regard to your concern, the query is to annunciate the failed lamp to a remote location.

A lamp test (button) is an option which you find in most panels..but the OP wants to know how to annunciate it to a remote control panel without pressing the lamp test button.

Cheers.

The 120Ω resistor is to hold a sable voltage drop across the coil of the relay. In series with the lamp the coil is being held energized while the lamp is on. Contacts open to the alarm. If the lamp blows creates an open circuit. Alarm relay closes alarm goes off.

Hi Ozzb and all..

Thanks for your explanation. In fact i am also wondering,why in two cases, the original circuit uses a 6V relay in one instant and a 12v in another instant to sense the lamp failure.?

How would one calculate that the value of the resistor should be 120 ohm? I know it cannot be very high because the lamp would lose its brightness intensity. The power of the lamp is 45W , 24V.

Joy to u and shalom

They probably used what was easilly available, or a 'standard' part from their stores...the resistor values were probably selected simply to 'make it work' rather than as a result of some feindishly clever calculation.

Del

OK. Now for some sums:

(24V)² / 45W = 12.8 Ω filament resistance at this voltage.

Dropping 6V across the resistor, and making unrealistically no allowance for the variation of lamp resistance with applied voltage, nor the resistance of the cable

18V / 12.8Ω = 1.41A drawn by the lamp with this relay in series with it.

Relay + resistor in parallel: 6V / 1.41A = 4.3Ω

Resistor draws 6V / 120Ω = 0.05A

Therefore relay is drawing 1.41A - 0.05A = 1.36A and has a resistance of 6V / 1.36A = 4.4Ω, more or less.

Therefore Del's suggestion that the installer just happened to use a resistor that was hanging around is probably correct, and it is more than likely now that it is there as a voltage spike eliminator than anything else. The selection has probably been made on a practicality and safety basis rather than anything more esoteric.

Note that a silicon rectifier, like the 1N4001 (cheap as chips), could have been used instead of the resistor provided the rectifier was arranged reverse-biased; this is a common way of preventing high voltage spikes appearing at relay terminals on DC circuits.

Ooooooh! More! More!

If the filament of a signalling lamp is detected in this way, the fracture of a filament, which causes the detector relay to change state, can also be used to change the aspect of a signal to one that is more restrictive (being an important consideration in railway safety). For a green/double-yellow/single-yellow/red 4-aspect colour light, for example:

• If the green filament breaks the signal in rear can be wired to show double-yellow at best, as well as the signal in question.
• If one of the yellow filaments breaks the signal itself will show single yellow, no problem, and the detector circuit can be made to override the signal in rear to show at best double-yellow.
• If both yellows fail or the red fails then the signal in rear can be made to show red and those trackside handsignalmen with their coloured flags and hand lamps then take over until the lamp changer arrives with more lamps for the fixed installation.

Hi PW,

While i appreciate your explanations, especially on the 120 ohm resistor used as a "sucker" to bypass the self generated emf by collapsing magnetic field, i am wondering how the user selected the 120 ohm resistor.. I know it will also work without the use of the bypass resistor. By how would u go around selecting the 120 ohm resistor?

Shalom..

1. What is the running current and rated voltage of the lamp?
2. What is the running current of the lamp at 18V, allowing for any series resistance presented by cables feeding it?
3. What is the operating current of the relay at 6V?
4. Apply Ohms Law and Kirchoff's Law.

Hi PW ,

Thanks for the detailed explanation, Makes sense to me now! And Del the Cat was prophetically correct all along huh! Once again thanks for taking your valuable time to answer. It was enlightening.

Joy for u and shalom...

Moo Moo.... 6/3

A possible solution:

Hi Qqberci..

Thats a neat circuit..

Thanks

You could use a small ferrite cored hand built CT whose primary (only a couple of turns) can be put in series with the 24 Vdc supply and the lamp. If the lamp blows there will be a spike at the secondary (a few tens of turns) which can then be processed digitally to provide the required signal at a remote place.

Hi all,

So there must be something wrong with the second part.. in which a 2.2 ohm resistor was connected across the coil of a 12v relay to power up the 24v 45W lamp!, because the calculations doenst make sense with 2.2 ohm resistor in place.

Thanks.

This is a simple circuit which I have used, sorry it is on the small side cannot seem to enlarge it. 12V or 24V take your pick. Also if you wire a solid state beeper via a diode you can fit any number of these indicators and use one beeper for them all.

This method does not interfere with the lamp circuit and supply to the indicator circuit can be from a totally separate supply if required.

The simplest way is to buy a normally closed reed relay.

Wrap say 20 turns of wire closely around it (some come with a plastic bobbin already in place!) and place it in series with one of the leads going to the lamp.

Check the relay to see if the points are now open when the lamp is on and closed when the lamp is off. If the relay does not close, add a further 20 turns till you get reliable operation. Make sure the wire is thick enough to handle the lamp current!!

Add alarm circuit to the relays contacts so that alarm goes off when relays contacts are closed!

Reverse everything if you can only get a normally open version!!

Hi, to all,

It does not seem quite right to me, the small six volt relay coil would have a resistance of around forty Ohms, drawing only 0.15 A. The resistor is 120 Ohms, drawing say 0.05 A total draw 0.2 Amps. Did I see that the lamp was 24 V 45W. ?

Even at 18V it would draw over 1Amp, this would be certain death for the relay.

To stay alive the lamp would need to have a resistance of no more than 90 Ohms,or so, making it around 3.6 W.

Dave.R.