Hi folks

I Have been hunting around all day for an answer to my circuit problem. I am glad to have found this site as it seems an educational gold mine in this area.

This is probably a stupid question that I should already know the answer to but in my amateur circuit work i have never got round to learning this.

Basically for a number of years I have used this useful relay circuit to switch on a 9v door spyhole camera. I have the sensor leads going to one button which actuates the relay while the reset switches it off. I also have the 9vDC power adapter which powers the camera, powering the relay circuit. This has worked good for years however I have persistent problems forgetting this camera is on as there is no indicator to say so.

My question is HOW do I insert a small LED in this circuit to light when the relay is on and turn off when the reset is pushed. ( a power indicator light )

Any help on this would me very much appreciated.

 

Just connect the LED with a series resistor across the relay coil.
SG

 

Welcome to AAC!

Just put it on the timer output with a series current limiting resistor.

 

Or like this...

 

Thanks a lot for the replies guys, i will try this tomorrow. I noticed dl324 and dendad that both the descriptions use different Resistors, can I ask does this matter much, I am aware there is this ohms law stuff I am not to educated on to work all this out but does any this matter in this case?

 

Normally don't exceed 20 ma. You can use less current if the LED is bright enough for your application.
SG

 

both the descriptions use different Resistors, can I ask does this matter much, I am aware there is this ohms law stuff I am not to educated on to work all this out but does any this matter in this case?

The resistor value depends on the LED, desired LED current, and supply voltage. As mentioned, many LEDs specify a maximum continuous current of about 20mA, but you don't need to operate at that level for an indicator. You just want it to be bright enough. That depends on the lighting conditions where the LED will be.

For 20mA:
\( \small R = \frac{V_{OUT}-V_{LED}}{20mA}\)

 

Thanks ever so much for the replies guys I got it working perfect with a yellow LED and 390R, all fitted and working fine. I now understand a bit more about adding power indicator LEDs to my circuits

Much appreciated