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Schematic for turning on LED when Bulb Blows

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RichD

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Hi Guys,

I have very limited electronics experience (Did GCSE Electronics 15 years ago) and I am looking to build a circuit that will light an LED when a bulb blows. I think they are 120v mains powered. Can anyone offer any advice or assistance on how this can be achieved please?

Thanks

Rich
 

off the top of my head, I'd suggest a neon bulb [and resistor] in parallel if the bulb is of medium or higher wattage as the simplest
 

Thanks for the reply. i should have put the background in my original post. basically i volunteer at a local hospital radioand our phones are muted so that they dont have an audible ring. instead there is a light that flashes. the problem is that when the bulb blows there is no way to tell without ringing the phone thus we could have patients calling and no one answers coz the bulb has blown. Each unit has two bulbs so what i am looking to do is light an led when a bulb blows, which would be signaled by an open circuit. the dificult part is the there will be 120v going through the bulb when the phone rings which will blow most components.
 

a neon bulb with a series resistor [across the bulb] is the most simple....................
 

the neon bulb and resistor have a much higher resistance, so they will tend not to light
 

the neon bulb and resistor have a much higher resistance, so they will tend not to light
I believe you are not understanding the operation of parallel circuits. The voltage across the neon bulb and resistor is the same whether the bulb is conducting or burned out, thus the neon bulb will always light.
 

I think I would look at an LED replacement for the original bulb. Not to check if the original has blown but to replace it with one that is far more reliable.

As Crutschow points out, assuming there isn't something dropping the voltage in series with the existing bulb, which is unliklely, the voltage will be the same regardless of the bulb working or not. It is probably more trouble than it's worth to monitor it's current or light output and illuminate an alternative instead.

Brian.
 

A neon + resistor across the switch would illuminate when the switch is off if the bulb is working and non illuminate when the switch is off if the bulb isn't working (and be off when the switch is on whether the bulb is working or not). However, I am not sure what would be the point of it!

Keith
 

Who said anything about wiring it across the switch?

I just did! The original question was
I am looking to build a circuit that will light an LED when a bulb blows
Wiring across the bulb certainly won't give the desired effect. Wiring across the switch can be used to indicate when a bulb blows in limited circumstances. It can also be used by wiring across fuses to show when a fuse blows.

Keith
 

Good day all
i think Keith meant a circuit like this
lamp blown.jpg
in a similar way , a LED circuit can be made as well
lamp blown led.jpg
hope this helps
Regards
 

Thanks for all the replies. Some great discussion here. In the examples above would the switch need to be closed for the LED to light? The switch is effectively part of a phone so the light flashes as the phone rings. I need the LED to be permanently lit when the bulb blows. Both the bulbs are behind a plastic cover so you cant see the bulbs until they have both blown the idea is that the LED can sit outside the casing and just glow when one of the bulbs blows. That way I can see at a glance when one of the bulbs has blown. and change the one blown bulb before they have both blown and we find we are missing calls.

- - - Updated - - -

Thanks for all the replies. Some great discussion here. In the examples above would the switch need to be closed for the LED to light? The switch is effectively part of a phone so the light flashes as the phone rings. I need the LED to be permanently lit when the bulb blows. Both the bulbs are behind a plastic cover so you cant see the bulbs until they have both blown the idea is that the LED can sit outside the casing and just glow when one of the bulbs blows. That way I can see at a glance when one of the bulbs has blown. and change the one blown bulb before they have both blown and we find we are missing calls.

Wont the second digram above light the led when the switch is open and deluminate when the switch is closed?
 

Good day Rich
yes it does , that was meant to be bulb OK LED
the LED circuit above can be used so LED lit when you have blown bulb , by connecting the LED and the resistor R1 to the collector of the optocoupler instead of the Emitter but however the LED then will lit when you get a call as well as the switch will close
 

betwix mentioned simplest solution to problem LED replacement for bulb.

deepmark circuits are wrong. LED will not lit when bulb is blown or broken. It will lit only when bulb is OK and OFF.

Without sensing the bulb current when it should be ON circuit according to RichD request can not be made.
 

Good day Rich
yes it does , that was meant to be bulb OK LED
the LED circuit above can be used so LED lit when you have blown bulb , by connecting the LED and the resistor R1 to the collector of the optocoupler instead of the Emitter but however the LED then will lit when you get a call as well as the switch will close
Whether the LED is in series with the collector or emitter of the optocoupler, it lights either way when the bulb is good.

To reverse it so the LED is OFF when the bulb is good and ON when bad, you could connect R1 in series with the optocoupler collector and connect the LED from collector to emitter of the optocoupler (optocoupler emitter to ground). To avoid flickering you likely will also need to add a capacitor from the cathode of the 4007 diode to pin 2 of the 33k resistor (try 1µF).

But note that this only works when both bulbs go bad. If you wanted to check each bulb individually, then you will need to isolate the two bulbs somehow. One way would be to connect each bulb to it's own diode bridge and use one opto isolator for each bulb. Connect the two optocoupler transistors in series with R1 and with the LED across both transistors so that either optocoupler turning off will light the LED.
 
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