LED lights when circuit broken

Hi, What is the simplest (cheapest, lowest energy) way to get an LED to light if the circuit from a 1.5v battery is broken? Thanks, Rod

Reply to
remcg
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Where is the LEDs power coming from?

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JF
Reply to
John Fields

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

Reply to
remcg

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.

Reply to
George Herold

"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.

** Errr - is a slowly blinking LED OK ??

... Phil

Reply to
Phil Allison

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

Reply to
remcg

Try:

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But they are hard to come by and expensive:
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sacat=0&_from=R40

Or you could use a 50 cent PIC:

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Paul

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Reply to
P E Schoen

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There are lots of ways to do it, but if you're using a SPST switch to 
turn a load on and off, then this might work for you: 
  
View using a fixed-pitch font. 

   +-------O--> |  
   |            | 
   |            O 
   |    Iled->  | 
   +-[R]-[LED>]-+-- V2  
   |+           |               
 [BAT]          |      
   |            |    
   +----[RL]----+ 

With the switch closed there'll be a miniscule voltage drop across it, 
so the LED won't light. 

With the switch open, current will flow through R, the LED, and the 
load, RL. 

You can limit the current through the LED by knowing the resistance of 
the load at the current you want to run through the LED, the voltage 
drop across the LED at that current, and then choosing R to allow that 
current in the circuit. 

Is that what you want to do?
Reply to
John Fields

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

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Reply to
Rich Webb

Use an LM3909 cct across the switch.

Reply to
robertgush

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:

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20dec%2011.pdf
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37

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

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BCT-ND/270765

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

Reply to
P E Schoen

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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

Reply to
Michael Black

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:

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

Interesting Joule Thief info here:

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

Ed

Reply to
ehsjr

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.

Reply to
robertgush

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yeah, I heard LM3909 was End of life

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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.

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Reply to
Jasen Betts

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