Hi all. *newbie* (sorry)

I've had an idea in my head for a little while now, and figure it's about time I did something about it.

Basically, I want to create a circuit that will light up a secondary led/bulb when the primary bulb fails - (as an alert to the fact the main one has gone)

when I was younger, I remember having a project kit, and one circuit in the kit did a similar job to this - it sounded a buzzer when the two bits of tine foil closing the circuit are seperated (very basic alarm - used it to stop my brother going in my room!) any way, I think it was called a collapsing circuit, and I reckon this is what I need.

Is there anywhere you could point me to to get info on doing this?
many thanks,
David

 

Seems like you have to sense either the current or the light output from the primary bulb. Is it AC or DC? What is the voltage and the wattage?

 

It would be dc, but as for wattage/voltage this will vary, depending on the situation the curcuit is put in.
As far as a can remember, it was a really simple circuit using a couple of resistors of something.

 

This is very different from a light bulb. The two bits of foil constitute a crude switch, and for it to work there must be some sort of current-limiting resistance in the circuit. It is a piece of cake to detect the change in voltage across the switch or across the resistor.
An incandescent lamp, on the other hand, will generally have no additional current limiting resistance. The lamp is the only resistance in the circuit, so the voltage across it will not change appreciably when it burns out. The change in current must be detected, and this requires adding a series resistor and some other parts, or possibly using a Hall Effect switch, which can detect the magnetic field caused by the current.

 

If you have a look at the img I created (if it doesn't show up, it's at http://extrobe.4t.com/diag.GIF). bulb 'A' is the
main bulb, say the car bulb on a car. 'B' is an led (or the likes) that lights up when bulb 'A' has blown. 'C' I envisage being
something that detects the break in the circuit, and bypasses the power through to the LED.

Dont know if that makes it any clearer what I'm after, but it wont hurt )

 

The concept is simple. The devil is in the details. Making something simple that will work with any bulb and any voltage is probably impossible. What is your application?

 

ur idea same like mine .. but failed to work

http://www.electro-tech-online.com/v...?p=59895#59895

 

You might try a current transformer to detect current through the lightbulb. Current transformers have the advantage of not needing to connect to any of the wires in the circuit - you just wrap one of the wires around the transformer.

 

Except transformers require AC, and don't work with DC.

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Here's about the simplest circuit I can come up with that works over a wide range of supply voltages (3V to 36V) and lamp sizes. I don't actually have any lamp specs, so you might have to change some resistor values on the input side. The output drives about 18ma into the LED, relatively independent of the supply voltage. The LED forward voltage needs to be less than the minimum supply voltage by a few hundred millivolts. If one end of your lamp is grounded, this would have to be re-thought.

Attached Images

lamp_failure_indicator1.gif (6.3 KB, 443 views)

 

Sorry - got it into my head that this was for a standard light switch. Recently I've been doing a lot of AC lamp suff - I guess I have it stuck in my brain now.

 

thnx succesfully builded one ... Vcc = 24 Vdc and the 1ohm resistor is replaced with 0.5ohm 5watt resistor

 

Thanks for the feedback. What is the rating of the lamp you used?

 

LAMP DC24V 60 Watt ...

 

Hi,

This is the simplest one ive seen,
its was for vehicle lights,
intended for pairs of lights.

The one i saw was home made and only had six or seven turns
for each of its lamps.

I would guess a single might be made using a piece from a
broken magnet, but thats just a guess.

Regards, John

Attached Images

car_dead_bulb_indicator.jpg (12.0 KB, 343 views)