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Low battery indicator

April 16, 2011 Electronics / DIY electronics 158 997 20

This indicator will inform you in time that your battery is low. The main advantage of the proposed scheme is the clarity of operation. In other words, the signal LED lights up immediately without a smooth dimming. Plus, the device can quite accurately track a given threshold.

The scheme is simple and, with proper assembly and working radio elements, it starts working immediately without configuration. You only need to adjust the threshold for the trimmer resistor R2.

This circuit is designed for a 12 volt battery.

Also, the circuit can be redone for batteries with a voltage of 6 - 4 volts. For this, it is necessary to take the resistors R1, R2 with a resistance of 10 kOhm, and the resistor R4 - 560-100 Ohm.

PS: You can build more than one circuit for the battery, but say three. That is, the first LED will work at a voltage of, say, 11. The second at 10.5 volts, and the third at 10 volts. Thus, you will more fully own information about the state of your battery.

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Comments (20)

#1 Veent Guests April 16, 2011 20:42
0

yes yes five points all done by yourself
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#2 Dan-star Guests April 20, 2011 17:13
0

Is there a column from a hand-held radio (such a small one with 2 buttons) can it be connected to the phone? Or will the mobile burn?

Or the mobile will burn. The column says: 0.4 volts, 8 (omega) 271-1 sx. Help
Pts like this site =)
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#3 feelloff Guests April 20, 2011 19:40
0

No, it will not burn.
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#4 Rybak27 Guests May 26, 2011 14:03
1

why do you write here? no need to litter!
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#5 BABA_ZINA Guests June 7, 2011 14:50
1

tell me what parts need to be taken to do something 4.8v
thanks in advance!
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#6 feelloff Guests June 7, 2011 17:35
3

Try to read the article.
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#7 Vvlad Guests August 28, 2011 14:46
3

Maybe someone will lay out a circuit for 3 volts, och necessary?
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#8 SEREGAKONAN Guests April 7, 2012 10:22
0

and kt315i or with another index can be used?
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#9 feelloff Guests April 7, 2012 15:28
0

[b] SEREGAKONAN [/ b], it is possible.
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#10 mageffect Guests September 2, 2012 08:32
0

Please tell me, can I connect a relay to the indicator to disconnect the load?
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#11 1goga1 Guests May 8, 2013 20:06
1

"batteries with a voltage of 6 to 4 volts. To do this, you need to take the resistors R1, R2 with a resistance of 10 kOhm, and the resistor R4 - 560-100 kOhm."
Why R1, R2 at 10 kOhm? Maybe R1, R3? And R2 is 560-100 kOhm. Enlighten the student I did not catch up. Thanks in advance
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#12 dreno Guests July 22, 2013 22:02
0

And the board is a piece from the board of some tape recorder or TV? And from below we cut the necessary tracks and soldered jumpers, right? Cool invented, no need to poison the new board :)
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#13 dreno Guests July 22, 2013 10:16 PM
0

For some reason, there are 220k resistors, 56k - 2pcs on the board, 390 - 2pcs. And the circuit is with completely different resistors: \
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#14 Denoff Guests September 1, 2013 22:22
0

I simplified the scheme a bit. Excluded R5 and C1, R4 replaced with 300 Ohms and instead of R2 set a constant resistance of 1.3 kOhm. The rest of the details left unchanged. It fulfills at 10.48v.
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#15 Sergey Guests September 12, 2013 22:49
0

Is it possible to redo the circuit so that when [u] reaches [/ u] let's say 12v the LED lights up?
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#16 eklips89 Guests September 27, 2013 19:04
0

C1 skoka farad?
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#17 dreno Guests January 13, 2014 9:03 p.m.
0

To comment under number 14. DenOFF, at R4 = 300 Ohm, the current through the standard LED will be offhand ~ I = 10/300 = 33mA, the LED may burn out.
By the way (a little off topic), an overestimated current is typical for all car LED lamps sold that are designed for exactly 12 volts, while the voltage on a running car is 14 volts, which leads to failure of LEDs over time: fellow:
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#18 serj Guests July 25, 2014 00:53
0

tell me. how to refine this circuit so that its operation logic is reversed - at normal voltage the LED is on, and when it is low it goes off, make a semblance of protection against excessive battery discharge (preferably with the nominal value of the parts)
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#19 Artyom Guests September 30, 2015 09:59
1

I made a schematic, set up one to light up at 10.8V and 11.48V goes off, the other lights up at 9.42V to 9.96V. Thus, when the voltage drops to the threshold of turning on the LED, it will not go out in the first case up to 11.47 V in the second 9.95 V. Is it possible to somehow reduce this range?
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#20 radar Guests February 15, 2019 09:39
0

Regarding such a scheme, I’ll say that it is quite suitable for devices that consume high currents, for example, for mobile speakers, screwdrivers, etc. applications, moreover, in the case of battery power, it cannot be connected directly to the battery (it will drop to zero). For low-power devices, 10-15 mA, which the LED consumes (and the circuit will consume a little more), can be critical in terms of the duration of the device, if they are comparable to it in terms of current consumption. Therefore, it is better to use at least an indicator from the old cassette. having a current of FULL arrow deviation of about 40-100 microamps, which is more than 100 times less than the current consumed by the LED.
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