I have an Arduino board with a pump attached to it, and for powering both, I'm using a 18650 battery cell with a DC-DC Boost Converter. I plan to leave it working even when I'm not home, and I know that if the battery over discharges, there will be trouble, so I would like to design an over-discharge protection board.

I can see from this video that it is not a big deal to build one, however, I don't have a battery alarm buzzer. Is there any way to build such a protection board without that part?

Or is there any simpler way to build such a circuit with the components that I have: a relay (JQC-3F(T73)), some diodes (N4007 type), capacitors(22, 47 and 100uF) a potentiometer, a bunch of resistances (from 1mΩ to 51 kΩ) and transistors (PNP-8550, NPN-8050 and IRF520).

I'm attaching the scheme given by the person that made the video: enter image description here

  • \$\begingroup\$ I don't have a battery alarm buzzer Hmm, why not buy one? They can be found on Ebay (you should have figured that out yourself). The essential function of the buzzer is that it detect the over discharging, not sure why you'd think we can just leave it out. Making something from scarp parts you haev lying around will result in a much worse solution compared to using this module which is designed for that task. It is a cheap module, buy one. \$\endgroup\$ – Bimpelrekkie Sep 28 '17 at 8:59
  • \$\begingroup\$ I did that already, but the thing that I'm building is a watering system for my plants and since I'm leaving soon I need to build it now, and the shipping takes a lot of time. I agree about the already made ones. I have ordered that too. I will get it in two weeks. :-| \$\endgroup\$ – Physther Sep 28 '17 at 9:04
  • \$\begingroup\$ Anyway, since you suggest that you only have one 18650 cell to protect, why not use a module designed for protecting a single cell, like: ebay.com/p/… Buy it, connect it, done. \$\endgroup\$ – Bimpelrekkie Sep 28 '17 at 9:06
  • \$\begingroup\$ It's not a question about ordering components, it's a question about how to build a circuit without that component, which still stands. That's a judgmental "-1", not a professional one, but let it be. I would like an answer, not points. \$\endgroup\$ – Physther Sep 28 '17 at 9:13
  • \$\begingroup\$ Yes that is judgmental because you're not clear about what solution you are looking for. An answer would be to use a DW01 based protection circuit but you do not have the DW01, you need to order it. You are vague about the components which you do have, that does not even include any active components (like a transistor). \$\endgroup\$ – Bimpelrekkie Sep 28 '17 at 9:22

Since you have a 5 V relay (which appears to have a 2.2 V drop off voltage) you might be able to do this:


simulate this circuit – Schematic created using CircuitLab

This will only work if:

  • the relay will latch on with a little bit less than 5 V (most relays will)
  • the relay will switch off at around 3 V

You have to close sw1 to start the thing up. You could use a piece of wire instead of a switch.

The relay coil will keep itself powered as long as the battery has enough voltage. When the battery voltage drops too low, the relay will fall off and switch off everything.

Do note that this is still a "hack" solution, when powered the relay will draw current and thus slowly (in a couple of days perhaps) discharge the battery.

  • 1
    \$\begingroup\$ And if the relay holds in at 1.5V you have killed the battery. Test the relay dropout voltage before doing anything else. \$\endgroup\$ – user_1818839 Sep 28 '17 at 11:34
  • \$\begingroup\$ Okay. So I did some testing before implementing the above diagram by following the instructions on this website: na.industrial.panasonic.com/blog/… and apparently, at 2.2 V, I get continuity between COMM and NO, which I believe means that the Drop Out Voltage is lower than 2.7V, which is the lowest limit of Li-ion. Is this interpretation correct? \$\endgroup\$ – Physther Sep 28 '17 at 12:59
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    \$\begingroup\$ OK, so the relay switches off at 2.2 V which is less than 2.7 V which would be better. Yes your interpretation is correct. The lower voltage of 2.7 V is not a hard limit but more a safe choice. Some Li-Ion cells do not suffer damage when discharged below 2.7 V, others might suffer damage. But since you also have the 1N4001 we can use it to increase the voltage where the relay drops off. I'll update the schematic. The diode should add around 0.5 - 0.7 V to that 2.2V so you'd get 2.7 - 2.9 V drop off. For test: just add the diode in series with the relay coil. \$\endgroup\$ – Bimpelrekkie Sep 28 '17 at 13:05
  • \$\begingroup\$ One extra request, please! Can you write which port is the NO and which one is COMM? It's the first time I work with relays and it would be very helpful \$\endgroup\$ – Physther Sep 28 '17 at 13:11
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    \$\begingroup\$ The part that you cannot see in the picture (a micro controller with an ADC running some software) is the actual "clever"part. The buzzer does just that: buzz, so you hear a sound. The voltage reader is just a simple LED display. It is the combination of those with the micro controller and software that makes everything "clever". \$\endgroup\$ – Bimpelrekkie Sep 28 '17 at 15:53

You probably don't need this answer anymore but maybe for future reference.

Available on the Arduino, actually the atmega chip is a bandgap voltage. Basically the chip produces a fairly stable and linear(ish) voltage reference of about 1.1v (it will vary a bit due to temp etc)

This allows you to compare your supply voltage against this reference, without any external components at all just using your Arduino. So a software only solution can be implemented. I've used this on attiny85's.


Forget doing it with the handful of components you have. You need a fairly accurate voltage reference, a comparator, and a switch. Each of these you can test separately and know what you're getting.

OK you could use one of your transistors and the relay as the switch. And you could use the forward voltage of four diodes as a crude and inaccurate voltage reference, and roll your own comparator using a long-tailed pair (differential amplifier) made out of more of those transistors. I'll not go into the details here, I'd need to write a book chapter to cover it properly, including what you'd need to do to maintain accuracy across supply voltage and temperature. But that gives you some useful search terms if you insist on doing further research.

But it'll consume a lot more power, be less accurate, and a lot more work than the purpose-built board. Which probably provides overcurrent and overcharge protection too.

Alternatively, use an analog input channel on the Arduino to read the voltage, make the decision and hold in the relay. Getting it started when the Arduino is off is an interesting problem though : maybe a pushbutton across the relay switch as in Bimpelrekkie's answer. Hold it long enough for the Arduino to boot and drive the relay coil if the measurement voltage is good.

  • \$\begingroup\$ Thank you for the answer. Wow, that sounds like heavy work for me, as I'm just a beginner. I like the Arduino part though. It sounds easier to implement. The pushbutton thing has to be turned on manually, which is inconvenient for automation, but this would happen only when the battery is flat, correct? \$\endgroup\$ – Physther Sep 28 '17 at 13:06
  • \$\begingroup\$ Same as the other answer in that respect. But the Arduino could turn off that output for other reasons inc. crashes, or crash without turning the battery off. \$\endgroup\$ – user_1818839 Sep 28 '17 at 13:10
  • \$\begingroup\$ Thank you, @Brian. I guess I will have to wait and get the right board for this. I thought there would be an easier option. \$\endgroup\$ – Physther Sep 28 '17 at 13:29

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