LED lights when circuit broken

Where is the LEDs power coming from?

Thanks for your reply, I should explain more. I want a simple circuit to detect when a switch is open. Normally it is closed. If I connect an LED in series with the switch then it will go out when the switch is open. However I want it to come on when the switch is open. If I connect the LED in parallel to the switch it will come on when the switch is open but my battery will run flat quickly. There must be a very simple solution to this, which will preserve the battery life when the switch is closed but light the LED when it is open. Rod

Well an LED + resistor in parallel with the switch might work.

You are basically stuck with the current drain of the LED, you can use a big R to limit current...(and light) or some more complicated pulsed gizmo.

George H.

"Rod@Hillhead"

Thanks for your reply, I should explain more. I want a simple circuit to detect when a switch is open. Normally it is closed. If I connect an LED in series with the switch then it will go out when the switch is open. However I want it to come on when the switch is open. If I connect the LED in parallel to the switch it will come on when the switch is open but my battery will run flat quickly. There must be a very simple solution to this, which will preserve the battery life when the switch is closed but light the LED when it is open.

... Phil

Hi Phil, Yes, a blinking LED would be better! Rod

Try:

Or you could use a 50 cent PIC:

Paul

You'll want an LED that doesn't have a lot of forward voltage drop (given that you're starting with just 1.5 V nominal) and these are swagged component values but something like this may be a place to start from:

_/ .-----------------o----o/ o-----o------------. | | | | | | .-. | | | | | | | | 1000 | | | | | '-' | | | | | | 1.5V >| | .-. --- 2n3906 |------------o | | - /| | Load | | | | | '-' | | | | | | | | | LED V -> .-. | | - | | | | | 10K | | | | | '-' | | | | | | .-. | | | | | | | | 10 | | | | | '-' | | | | | | | | | | '-----------------o--------------o------------' (created by AACircuit v1.28.6 beta 04/19/05

Use an LM3909 cct across the switch.

I already suggested that, but they are rare and quite expensive. They also draw about 550-750 uA. A PIC10LF320 draws only 25 uA. However, it is only rated to 1.8V minimum. You could build a relaxation oscillator that slowly charges a capacitor and then discharges it quickly through an LED. You can use a low voltage OpAmp such as the TS1001 which works from 0.65 to 2.5 V supply:

You can also use an LMC555 which works down to 1.5V.

Actually, it will be hard enough to light an LED on 1.5V unless you make a boost circuit. Perhaps something like a Joule Thief.

Paul

The National LM3909 was an LED flasher, ran off 1.5v. I don't recall an inductor, it did put a capacitor in series with the LED, so I guess it charged up and then discharged into the LED. Lots of secondary uses were found for the device, but it was intended as an LED flasher, and I think disappeared long ago from the catalog, not enough use for it.

Michael

Exactly.

He might be able to use John Fields' idea by putting the Joule Thief across the switch, like this:

/ +-----o o----------+---[Load]---+ | | | +---[Joule Thief]---+ | | | [Bat] | | | +--------------------------------+

If the load is low enough impedance the thief will run fine. He indicated that the battery ran down quickly when he put an LED across the switch and opened it, so presumably the impedance is low enough to allow the thief to run. If it isn't, he could go to a SPDT switch and run the thief this way:

Interesting Joule Thief info here:

Quantstuff did a lot of experimenting with the things, and his site shows a blinking thief as well as info on circuit efficiency.

Ed

Puzzling. Supply == 1v5, LED VF >= 1v8. How did the battery go flat? Unless it was assumed to be flat because the led didn't light? Maybe it did light initially as some primary cells initial voltage is above 1v5.

yeah, I heard LM3909 was End of life

that doesn't seem to be a 1.5V pic, and in any case for 1xAA operation you want chip that'll run on a 0.9V supply.