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I'm thinking of designing a PCB based on this low battery indicator circuit:

It hooks up to your battery of choice and has a green and red LED, and uses a trimmer pot to adjust at what voltage want the red LED to turn on and the green off. I want to add a function where the red LED will start blinking when the battery life is extra low, or right about to die. If I can't make the existing red LED blink, i'm open to adding a third LED (a blinking one), and possibly having to add another pot to specify its voltage. If I were using a 7.4v battery pack, this would make the new setup: Green - 6.5v-8.4v; Red - 6.3v-6.4v; Blinking Red - 6.2v. The first two settings can be set with the existing potentiometer, I just need to add the third setting. Thanks!

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    \$\begingroup\$ you want to drain the battery faster once it starts to run low? \$\endgroup\$ – jsotola Dec 30 '17 at 19:47
  • \$\begingroup\$ Who is this blinking Red for? and when is it not useful? Perhaps a better solution is needed such as battery cutoff. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 30 '17 at 20:08
  • \$\begingroup\$ The battery will cut off at 6.1-6.2v by use of the protection circuitry. If someone is using a device running on batteries, then the blinking red will let them know they have that much time to finish what they were doing on the device. So green means they are fine, red means get to a stopping point, and blinking red means the device is about to die. Perhaps I don't understand the question? \$\endgroup\$ – GC7260 Dec 30 '17 at 20:22
  • \$\begingroup\$ The whole thing can be done for about \$10\:\mu\textrm{W}\$ overhead. Of course, the LED power would swamp this. Even with \$2\:\textrm{mA}\$ high efficiency LEDs, that is \$4\:\textrm{mW}\$, or 400 times as much. So I guess there's no real point in looking for a low power circuit. Since you want blinking, I'd probably start with a voltage to frequency converter concept and enable the blinking directly when the frequency is slower than some limit. If faster, use an added circuit that overrides the blinking. If still faster, override that (shutting it off) while now driving the green. \$\endgroup\$ – jonk Dec 30 '17 at 21:28
  • \$\begingroup\$ Of course, a single, very cheap MCU could do all of this with an ADC and some I/O. On the order of \$\mu\textrm{W}\$ of overhead power (added to the LED power), using sleep modes which still keep the I/O active but only periodically wake up to do an ADC measurement and adjust things. But that entrains a toolset, programming skills, etc. \$\endgroup\$ – jonk Dec 30 '17 at 21:33
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You can use a schmitt-trigger nand gate (in my experience most nand gates have schmitt-trigger input) and couple the output through an RC circuit to one of the inputs and use the other input to activate/enable it. the oscilation frequency will be 1/RC because (most) nand gates are set to have thresholds so that the charge/discarge time between thesshold voltages will be tau. there are many other ways to make an oscilating circuit but this is one of the simpler.

to detect the voltage level you could simply use an op-amp.

If you want to make a low battery indicator that behaves this way what I would do is I would use a small 8-bit mcu like an atmega328 for instance.

The atmega328p if I remember correctly can go down to almost 2v of supply voltage if you give it a low enough clock frequency (see the datasheet) and it would probably have a smaller footprint than your analog circuit.

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  • \$\begingroup\$ Not a great solution in terms of power drain unless you are going to implement a deep sleep mode. The '328p and similar ATMegas will draw about 0.2 mA running code with the A/D @ 1 MHz, and < 1 uA asleep. The biggest difficulty is that the supply voltage is over 6 V, so you end up needing a regulator or series diodes to set the MCU supply. \$\endgroup\$ – Jack Creasey Dec 30 '17 at 22:05
  • \$\begingroup\$ True, However looking at his circuit it looks like he is going to be drawing a lot more than 0.2 mA. besides I believe the atmega has a low-power mode, you can then let it wake up once in a while and measure the voltage. \$\endgroup\$ – Vinzent Dec 30 '17 at 22:14

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