Low battery indicator

Can someone please provide me with a schematic to turn on an LED when a 9v PP3 gets low (~<20%), no microcontrollers, the simpler the better. I am hoping you can do this with a few diodes, transistors, resistors, and maybe some capacitors, but I can't think how this would work.

sort by: active | newest | oldest
1-10 of 18Next »
Goodhart6 years ago
Have you tried using an IC like the following for this? MC34064P 4.5v undervoltage Sensor
andy70707 (author)  Goodhart6 years ago
A 3-legged IC like that would be good, as I previously said, an 8-pin DIP would be too large for my application and would probably have to use flying wires as I am using stripboard, although it only does 4.6v. By then, the voltage regulator would have stopped drawing any power and the LED probably wouldn't even be on. The voltage regulator stops working at about 7v, so I want the low battery indicator to trigger around 7.8v. A 9v battery is considered 'dead' at around 7.2v anyway. As I said, I have already solved the problem using a transistor, 3 resistors and a potentiometer.
Oops. sorry for the late response then.
kelseymh6 years ago
Conceptually, I'd start with a joule thief circuit, so that the LED will work when the battery is low. Then, you add in a small relay in the normally closed (NC) configuration. You want a relay where the "minimum trigger voltage" is either adjustable, or is fixed at the value you want as a cutoff (e.g., 7.2V).

The idea is that as long as the battery is strong enough, it holds the relay open with some small power drain. Once the battery drops too low, the relay closes, and the joule thief kicks in to illuminate the LED.
andy70707 (author)  acmefixer6 years ago
I just breadboarded the circuit, and although it works in my emulation software, it doesn't work in real life. My transistors and resistors are all correct, but the LED is permanently on for both a 10v battery (full) and 7.2v (empty) 9v PP3. It does seem to dim slightly for the empty battery, but It should be on if it is low and off if it is anything about about 7.8v. Adjusting the 100k pot also has absolutely no effect.
andy70707 (author)  acmefixer6 years ago
I used a jumper for R3, and measuring Q1, I got 280mv-680mv when adjusting the pot, although the LED remains unaffected. I am using standard 1/4 watt, 5% tolerance resistors, all the exact value from the diagram except R5 I changed to 1.5k instead of 10 as when emulating it, 10k was too strong for it to light (even with 1.5k, it is still fairly dim). Shorting Q1 does make the LED go out, and they are 2N3904 diodes, all brand new.
andy70707 (author)  acmefixer6 years ago
Thanks for your help, I still couldn't figure out the problem with that circuit, but I built a similar one using only 1 transistor and it works, although I had a few problems, but changing the transistor seemed to work. I tested both transistors with a multimeter and they work fine, although only 1 would work in the circuit. They are both 2N3904 transistors, but made by different companies, so maybe the first two are just bad (they were free anyway).
andy70707 (author)  acmefixer6 years ago
I will try adjusting R3, and as for R5, I am not too bothered about battery drain as LEDs are only 25ma, and it will basically indicate that the user needs to replace the battery as the voltage is too low to charge, so it can use the remainder of the battery without any problems. I measured Q1 with an oscilloscope and is appears to be free from any kind of interference. As for Goodhart, I meant to say transistor, not diode. Sorry for the confusion.
A 2N3904 is s small signal NPN transister.
andy70707 (author)  acmefixer6 years ago
Thanks, that looks pretty good. I had a play around with it in some circuit emulation software, and at the moment it seems to do the opposite (dim the LED when battery is low), but hopefully I will be able to get it to work in reverse. Its a good start, so thanks!
1-10 of 18Next »