LED lights when circuit broken

[Rod@Hillhead]

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

 

[Rod@Hillhead]

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

 

[George Herold]

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.

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.

 

[Phil Allison]

"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

 

[Rod@Hillhead]

"Rod@Hillhead"

Thanks for your reply, I should explain more. I want a simplecircuit
to detect when a switch is open. Normally it is closed. If I connect
anLEDin 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 theLEDin 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 theLEDwhen it is open.

** Errr  -   is a slowly blinkingLED OK ??

...   Phil

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

 

[amdx]

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

What else is the switch supplying and what is the voltage?
AC or DC?

 

[P E Schoen]

"Rod@Hillhead" wrote in message

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

Try:
http://www.jaycar.com.au/images_uploaded/LM3909.PDF
But they are hard to come by and expensive:
http://www.ebay.com/sch/i.html?_trksid=p3984.m570.l1313&_nkw=LM3909&_sacat=0&_from=R40

Or you could use a 50 cent PIC:
http://www.mouser.com/ProductDetail...P4KuaKjx0gRSxmwZDO/NjWmScBt/abu/1AOaulPRldA==

Paul
www.pstech-inc.com

 

[[email protected]]

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

Use an LM3909 cct across the switch.

 

[P E Schoen]

wrote in message

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:
http://touchstonesemi.com/wp-content/uploads/2012/07/ts1001ds r1p1 - dec 11.pdf
http://www.digikey.com/product-detail/en/TS1001IJ5T/TS1001IJ5TCT-ND/3622637

You can also use an LMC555 which works down to 1.5V.
http://www.digikey.com/product-detail/en/LMC555CMM/NOPB/LMC555CMM/NOPBCT-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

 

[Michael Black]

wrote in message


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:
http://touchstonesemi.com/wp-content/uploads/2012/07/ts1001ds r1p1 - dec 11.pdf
http://www.digikey.com/product-detail/en/TS1001IJ5T/TS1001IJ5TCT-ND/3622637

You can also use an LMC555 which works down to 1.5V.
http://www.digikey.com/product-detail/en/LMC555CMM/NOPB/LMC555CMM/NOPBCT-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.

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

 

[ehsjr]

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

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:
http://quantsuff.com/LED2.htm

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

Ed

 

[[email protected]]

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

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.

 

[Jasen Betts]

"Rod@Hillhead" wrote in message


Try:
http://www.jaycar.com.au/images_uploaded/LM3909.PDF

But they are hard to come by and expensive:

http://www.ebay.com/sch/i.html?_trksid=p3984.m570.l1313&_nkw=LM3909&_sacat=0&_from=R40

yeah, I heard LM3909 was End of life

Or you could use a 50 cent PIC:
http://www.mouser.com/ProductDetail...P4KuaKjx0gRSxmwZDO/NjWmScBt/abu/1AOaulPRldA==

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.

http://www.digikey.com/product-detail/en/TLC551CP/296-10333-5-ND/380508