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I am looking for something even simpler than this, with low power consumption (so no relay):

https://youtu.be/8GQrTZb9T_c?t=158

I need a circuit which can be built from basic components which can cut the battery completely off in case of the voltage dropping below 3.5 V, let's say.

I got a solar panel plus a charging circuit to charge a small 2000 mAh battery. At night there is another circuit which turns on and blinks some LEDs. The problem is that this circuit has a too high a standby current even during daytime, so it never lets the battery charge.

Any suggestions are welcome. I'm sure I can order many off-the-shelf circuits from Ebay to do this, but I would like to do it on the weekend.

Screen capture schematic from linked video:

enter image description here

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    \$\begingroup\$ I need a circuit which can be built from basic components <-- what are basic components? \$\endgroup\$
    – Andy aka
    Aug 27, 2022 at 16:55
  • \$\begingroup\$ I assume you want a high impedance circuit that "measures" the source voltage (relaxation oscillator is perhaps one of the higher impedance forms of circuitry possible here) and a means of enabling a low-impedance switch (semiconductor -- likely MOSFET) when the source voltage goes above some point. Takes a few very simple parts -- resistors, very small-valued capacitors, some general BJTs and a MOSFET. That's the direction I might try, first. \$\endgroup\$
    – jonk
    Aug 27, 2022 at 19:02
  • \$\begingroup\$ You mention low power but your resistor divider alone is 30 mW in your schematic. Have you considered a dedicated IC for the job instead? There are many COTS solutions for it on the market. \$\endgroup\$
    – winny
    Aug 27, 2022 at 19:22
  • \$\begingroup\$ Buy a TP4056 module - choose the one with low voltage cutout.This meets your need for a few dollars. Note that 3.7V is mean cell voltage for an about 3 to 4.2 v range \$\endgroup\$
    – ATCSVOL
    Aug 28, 2022 at 3:24
  • \$\begingroup\$ A single LiIon cell has about about 3V to 4.2V range and 3.7V average. \$\endgroup\$
    – Russell McMahon ♦
    Aug 30, 2022 at 10:51

1 Answer 1

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The circuit you show consumes a lot of current since it uses a bipolar transistor and the LM431 needs 1mA to reliably regulate.

The easiest way is to use a purpose-built Li-ion battery protection chip such as the ubiquitous DW01. enter image description here

They're about 5 cents each in small quantity from suppliers such as LCSC, even cheaper on the domestic market in China. There are also very many manufacturers, probably more than for, say, the TIP42, so it should qualify for "jellybean" status.

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  • \$\begingroup\$ I already have this battery charger circuit, how else do you think I charge the battery. This don't do justice either because the load is constantly draining the battery. This dummy circuit is simply disconnects the battery under a critical voltage and reconnects it when it detects charging. Now the problem is that as soon as it reconnects it the load is draining it as well so it cannot charge the battery up. Obv solution would be to use a bigger solar panel but I try to keep this project low budget. \$\endgroup\$
    – user106458
    Sep 1, 2022 at 9:13
  • \$\begingroup\$ What I need here is a circuit which connects the load when the battery is fully charged and disconnects it when a critical low level what I define is reached eg 3.5V. Your circuit will connect it on 3.7V right away and keep it connected. \$\endgroup\$
    – user106458
    Sep 1, 2022 at 9:17

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