DC amp. sensor - digital 24VDC output.

Mr. Jankly,

You can put a shunt resister in series to the load. It will offer a millivoltage drop, whenever current flow through it. If the bulb or any load gets open circuited, there will be no flow of current and thus no voltage drop across the shunt.

So the millivolt can be fed to a differential op-amp (may be LM-741) and connect one relay to the output of the opamp to operate alarm circuit.

Thank you for your comment. A shunt, op-amp and 12V relay, I believe it will work. I have not heard about this op-amp before, seem to be a nice thing. But as I have to monitor many circuits - I was hoping to find a cheaper system, I have to sell this system approximately 100-200 pieces./year for a comparable price and make a profit out of it. A shunt for every single bulb is a bit to expensive. I was thinking if there is somewhere some kind of an amp-sensor a clamp or a ring that will sense if there is any DC current running in the circuit at all. I don´t need analog monitoring, only something to tell the PLC with a +24VDC input if there is any DC current running or not - "No current=bulb fail".

As I mentioned before the price is an issue in this case, I am competing with another company with a solution for this problem.

Put a 24 volt bulb across the fuse and mount it on the front panel with a legend saying 'fuse blown' or something...

Works pretty well and is cheap!

John.

This is not about blown fuse, that is no problem, what I need to monitor is the bulb-failure, or any other load-failure i.e. open circuit.

Okay, if you need a simple and cheap load failure.

How about using a reed relay with a low resistance wound coil, to only drop ½ a volt or whatever, in series with the load, so any load holds the relay closed, when the load fails the relay opens?

What you do with the relay contacts is up to you?!

Cheap, simple, reliable, what more could you want?

John.

Would a current mirror such as http://www.zetex.com/3.0/3-3-2b.asp?rid=120 work?

See this app. note: http://www.maxim-ic.com/appnotes.cfm/appnote_number/746/

Or High-side current sensing in Google.

You beat me to it again!!!! I wanted to write that!! All the other answers are simply too complicated for what is required.....

I gave you a GA for speed, not for the answer!!!

What will the voltage drop be?

What is the Z of the coil?

What is the wattage rating of the coil? Can it handle a 50W load?

If you add a low ohm load in series, arent you adding load? and using more power?

Greetings Elecroman,

Please forgive me as I am unfamiliar with the reed relay you are referring to. The factory made, store bought relays I am failure with will not handle a 50W load in the coil circuit.

The input module jangkly is using is a Sinking DC input, so a relay or switch is a better solution than a direct input from a shunt. Contact closure is all that is required to satisfy the input as the module supplies the 24V used for current sink input.

BMXDDI3202K Modicon M340 Discrete DC Input, (32) 24 VDC, Sink (positive logic), IEC Type 3, 1-FCN type connector (40 points)

Ahh no problem Crossfire... the reed relay I mentioned is just a glass envelope with two contacts inside in a vacuum (very reliable), the contacts make a connection when in a magnetic field of a certain number of ampere-turns and so the coil can be wound to suit the application.

I've used them in all sorts of automotive sensing applications, for instance to sense when the cigar lighter is actuated - as the cigar lighter can easily take 15 amps, the coil around the reed relay can just be one turn of the supply wire - so very easy to fit, without breaking the supply wire etc...

John

Good answer, well put.

You can actually switch a reed relay with a few turns of the wire going to say a lamp. Just the current of the lamp will activate the relay, almost unnoticeable in reality.......

Naturally, the thickness of the wire needs to be correct for the current as do the number of turns for activation....but they do not need much.

Thank you Electroman, this seem to be a good solution. I was really looking for some unit that is ready to go -plug and play, but they all seem to break the 100$ barrier ae. Using reed relay´s I would need to make a circuit board and solder 6 relay´s on it and wire-connectors for the wires from the light switch and to the lights and for the wires for the switch´s on the relay. then put this in a box or some other suitable protection. This will take me a few hours and my time is money. Maybe it is possible to have this circuit made in china, Japan or somewhere over there for less. Do you have any recommendations for a company that will make a circuit like this? about 20 units pr. year.

If you only need 20 a year I doubt if going to China would be economically feasible!

However, my business is used to making short production runs, whether its out-sourced or not, I'm sure I could quote a reasonable price...

It will depend so much on the environment they will be used in as well as safety aspects etc...

Contact me if you want me to look at a quote for you.

John www.marchdesigns.com

Try wiring a 8 Ohm resistor in series with the lamp. Now measure the voltage across the resistor with the light in the socket and energized, it should be around 5Volts with a 15 watt bulb that has 38.4 ohms and a current draw of .625A.

The addition of the 8 Ohm resistor will bring the loop total up to 46.4 so the lamp may not be as bright as before, but it should still light up.

A higher watt bulb will yield a higher voltage across the shunt, so you may need to add a small solid state voltage regulator to the output of the shunt to protect the PLC input channel.

The resistor is a shunt and will satisfy a 5Volt discrete input to a PLC. Write the logic so that it compares the input to the output, if both are high than all is good. If output is high and input is low than you alarm that a possible power supply or lamp faults.

If you must have the input high when the shunt is in a low state, use an interposing relay and a separate power supply to the N.C. contact and the input.

It may be a cheap fix.

Yes this will work, but as you mentioned it will affect the voltage to the bulb´s and the bulb´s in this case are navigation lights on a boat so I don´t like to dim them down. Another thing in your solution is that my PLC is asking for 24VDC discreet input, this is Telemeanique Modicon M340 - BMX DDI 3202, the min voltage is 11VDC to ignite the input on this computer.

I thank you all for your comments but I think I found the best solution on www.a-a-c.com they produce DC-current detector with a internal relay contact.

Then you either did not read post #4, or you did not understand it!!!

That is the cheapest fix, the simplest fix and the one that uses less power than the others, basically no power at all!!!

It reminds me of the old proverb:-

You can lead a Horse to water, but he must drink on his own!!!

Ah well Andy,.... "some words fell on stoney ground..." huh?

John

Dead right.

The reason being I feel may be that the younger generation cannot imagine that such a simple fix is possible, without electronics, without extra power supplies etc etc....

I have used such relays in the 60s and 70s together with a small beeper to allow me to know if:-

1) Car Brake lights are both going on, or should I say, if one is on and the other is blown then beep.

2) Power is being applied to the brake lights when the pedal is depressed (to catch a blown fuse or similar). Beep if no power to brake lights.

It could not detect both lamps blowing at the same time, but as that is highly unlikely, I accepted that!! Nowadays cars have detection systems already built in!!!

These look large and rather expensive.

True, the price for those bastards is 120 - 150$ that is far to expensive for my application. They are also rather clumsy but I have enough space for 6 peaces.

What about of a Photo-resistor/diode or similar v. cheap light sensor which will watch Real Function of a Bulb, i.e. light production? Be sure to make logic with situation where Bulb is OK but is not energized (by PLC Output).