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# Low voltage cut-off circuit

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#### MisterBeppe

##### Junior Member level 2
Voltage reduction of a battery pack

Hi, here I have a battery pack composed by 16 batteries (Sanyo Eneloop, Ni-Mh, 1.2 volt - 2000mAh). Since the nominal voltage of each battery is 1.2 volt, from the output of the pack (these batteries are connected in series) there are 19.2 volt.

Well, I need this battery pack to feed a pair of portable loudspeakers: the specs of these loudspeakers are: 12 volt - 200 mA. So, I can adopt two different ways: Convert the 19.2 volt to 12 volt using a step down buck converter like the following (which is switching, not linear) or modify my battery pack just to use 10 batteries in series, but I have some doubts about the last solution: when the batteries are fully charged they provide more than 1.2 volt for each battery, so I can't have exactly 12 volt from the output.

Pro and cons of each solution?
Can I go with the buck converter?

Many thanks.

Re: Voltage reduction of a battery pack

Yup! Use the buck converter. I've used that module before and it's a nifty (+easy) way to power all sorts of things from variable, high battery voltages.

Your current requirements (200mA) are well within its capabilities and operating at such a (comparatively) high input voltage means the batteries will be only lightly loaded (~130mA. Buck converters are [approximately, and efficiency notwithstanding] constant power devices).

Re: Voltage reduction of a battery pack

Thank you for the clear reply, thylacine1975!

However, considering that the nominal voltage of the battery pack will be 19.2 volt, which will be the minimum voltage in volt for considering the battery pack exhausted? Somewhere I also read that is recommended to not go under some value, to prevent some damage for the batteries.

Re: Voltage reduction of a battery pack

Your amplified speakers probably work from alkaline cells that are 1.6V each when new so 10 of them provide 16V.
The speakers also probably work from a "12V" wall wart that might produce 16V when the music plays softly.
Your Ni-MH rechargeable cells are up to 1.5V fresh from the charger so 10 of them will produce 15V.
Then you can use 10 battery cells and a voltage regulator is not needed.

Do you have a charger that will charge 10 Ni-MH cells in series properly?

Re: Voltage reduction of a battery pack

Looks like brilliant value for money, the only suggestion I can make is to include a bypass switch as if the module is set to 12V, it needs 13.5V input, so if your batteries are dieing, bypassing the module will get you some extra use from the batteries. If you were clever, you could do this automatically.
Frank

Re: Voltage reduction of a battery pack

Your amplified speakers probably work from alkaline cells that are 1.6V each when new so 10 of them provide 16V.
The speakers also probably work from a "12V" wall wart that might produce 16V when the music plays softly.
Your Ni-MH rechargeable cells are up to 1.5V fresh from the charger so 10 of them will produce 15V.
Then you can use 10 battery cells and a voltage regulator is not needed.

Do you have a charger that will charge 10 Ni-MH cells in series properly?

Hi Audioguru,

My amplified speakers are not provided to works with internal batteries; just the external wall wart. I used a multimeter to check the exact voltage from the wall wart: I I measured a value of 12.70 volt.

As battery charger I bought a Technoline BC 700, which claims to be a "smart charger":
https://www.technoline.eu/details.php?id=1400&kat=15

Kind regards.

- - - Updated - - -

Looks like brilliant value for money, the only suggestion I can make is to include a bypass switch as if the module is set to 12V, it needs 13.5V input, so if your batteries are dieing, bypassing the module will get you some extra use from the batteries. If you were clever, you could do this automatically.
Frank

Hi Frank, unfortunately I'm not very skilled: can you suggest me a practical ways to implement this solution?

Thank you.

Re: Voltage reduction of a battery pack

You have a very good charger.

My charger is from the Energizer battery manufacturer and it is a stupid simple power supply, a resistor and a timer.
It does not detect the charge of cells so if they already have a charge then its timer cooks them.
If the power fails when its is timing then the timer starts again when the power is restored which also cooks the cells.
But its timer shuts off the charging when it times out, unlike another stupid charger I have that is simply a power supply and a resistor that overcharges battery cells.

Re: Voltage reduction of a battery pack

You have a very good charger.

However, as chuckey said, I was thinking to implement a "battery saver" circuit, which would disconnects the battery pack when the voltage of the battery pack will be equal to 16 volt. I've read that is recommended to not discharge Ni-Mh batteries under 1 volt for each battery: so, considering that I have 16 Ni-Mh in series, the lowest voltage to considering them exhausted is 16 volt, am I right?

I've seen many "battery saver" schematics, but most of them are designed to work with 12 volt battery pack, like this, which is rated for 12 volt:

But my battery pack will operate at a nominal voltage of 19.2 volt.
Would be also useful replace the buzzer with a relay (so I should also implement a transistor to drive the relay) which can disconnect the battery pack when the "alert" voltage is reached.

Can someone suggest me how to adapt this schematic for my needs?

Many thanks.

Re: Voltage reduction of a battery pack

The lousy old (the 741 opamp was designed 47 years ago) and relay will drain your battery. The output of the 741 does not go high enough to turn off the buzzer.
Instead, you need a very low current comparator and Cmos flip-flop to turn off a Mosfet that disconnects the battery from its load. The LED and buzzer should operate for only a few seconds then also turn off.

The circuit you show will detect a low battery and disconnect its load which causes the battery voltage to rise high enough for the battery to be connected which drops the battery voltage again which ....., so it disconnects and connects and disconnects and connects over and over. The circuit needs some hysteresis to prevent it.

MisterBeppe

### MisterBeppe

Points: 2
Re: Voltage reduction of a battery pack

The lousy old (the 741 opamp was designed 47 years ago) and relay will drain your battery. The output of the 741 does not go high enough to turn off the buzzer.
Instead, you need a very low current comparator and Cmos flip-flop to turn off a Mosfet that disconnects the battery from its load. The LED and buzzer should operate for only a few seconds then also turn off.

Many thanks for the hints. So I will reject that circuit based on the 741.

However, on the web there are a lot of "battery saver" schematics (a lot of them rely on the 741 and on the LM358), but unfortunately I don't have the knowledge to understand which is better and really functional. Can you address me to some schematic which could be valid/functional? I've also seen that on Ebay there are a lot of "already assembled" battery saver circuits; maybe there is something that would cover my needs?

Kind regards.

Hi,

Since I need to protect from the under voltage a battery pack composed by 16 Ni-Mh batteries (Eneloop AA, 2000mAh), which provides a nominal voltage of 19.2 volt (1.2 volt x 16 batteries), I've looked on Google for some "battery saver" circuit; most of them rely on old components (like the Op-Amp 741) or will take an high consumption of batteries. An user told me that I have to rely on a very low current comparator and a Mosfet, to avoid the drain of the battery pack. Well, after another search on Google I've found this schematic:

Description here: **broken link removed**

This circuit claims a very low consumption, and is rated to check a Li-Ion battery (3.7volt - 4.1 volt). Well, considering that I have a 19.2 volt battery pack composed of 16 Ni-Mh batteries, and since I've heard that each cell shouldn't be discharged under 1 volt to avoid any damage, I have to set this circuit to act when the output voltage is equal to 16 volt, right?

Please note that the battery pack is connected to a switching buck converter (this one: ) because I have to feed a pair of portable loudspeakers (12 volt - 200 mA).

Unfortunately I'm not very skilled and I have no idea how to modify the battery protection circuit to achieve my needs.

Can I have an help from someone?

Many thanks.

Do you like carrying around 6 extra battery cells plus a buck converter? If you need 12V then why don't you use 12V instead of 19.2V? When the 12V battery runs down to about 10V its output current will be so low that you will hear the amplified speakers complaining and you will turn it off. Then you don't need a low battery voltage protection circuit.

Do you like carrying around 6 extra battery cells plus a buck converter? If you need 12V then why don't you use 12V instead of 19.2V?

Well, I was thinking to the first solution (19.2 volt from the battery pack and a buck converter set to 12 volt) since the battery pack will start to provide more than 19.2 volt (when the batteries are fully charged maybe will provide more than 19.2 volt, let's assume 1.4 volt x 16 batteries = 22.4 volt), then the output will decrease to the nominal voltage (19.2 volt), then, again, to the "discharge" value of 16 volt.
In the meantime, during this process from the high value to the lower value, the buck converter will always assure me 12 volt for a longer time (while the battery pack will slowly discharge).

Instead, if i use 10 batteries, initially I will have 14 volt (fully charged, as before: 1.4 volt x 10 batteries = 14 volt), until 10 volt. I'd like to be sure to keep the value always to 12 volt.

Please consider that I am not very expert, so tell me if my assumption is right or wrong

EDIT: Otherwise I could implement a simpler but functional solution: using a digital voltmeter like this one to check the residual voltage:

To avoid an higher current draw due this voltmeter (these kind of digital voltmeter, however, should draw about 20mA), I can use a simple switch to exclude it. And since with 2000mA from the battery pack, I am sure to enjoy at least 2 * (or less) hours of usage; well, after an hour I can turn on the voltmeter to try to estimate the time to charge again the batteries. What do you think about?

*= Maybe I was wrong: here I have seen that with 2000mAh from the batteries and a current draw of 200mA I could enjoy also 7 hours of music, right?

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I'd like to be sure to keep the value always to 12 volt.
Why? The amplifier was designed to be used in a car where the voltage is from 10V to 14.4V. The amplifier can produce more output power and more heat with the higher supply voltage. You never said which amplifier IC and did not say the speaker impedance. Mono or stereo?

I have seen that with 2000mAh from the batteries and a current draw of 200mA I could enjoy also 7 hours of music, right?
Nope. The battery life calculator assumes the battery is constantly driving LEDs or a motor. Do you listen to an amplified continuous tone at full power? Then the calculator will be close to being correct.
But music is never constantly full blast (maybe acid rock music).

Who says the current is only 200mA? Then the total power is 12V x 0.2A= 2.4W. If the amplifier is class-AB analog then it produces an output power of 1.5W and 0.9W of heat.
A class-AB amplifier with a 12V supply can have an output swing of about 10V peak-to-peak which is 3.54V RMS. Then the output power into an 8 ohm speaker will be (3.54V squared)/8 ohms= 1.6W so I was close.

Music has a average power that is about 1/5th its maximum power. The maximum power is 2.4W then the average power is 0.48W and the average current is 0.48W/12V= 40mA. The idle current of the amplifier will be about 20mA so the total average current will be 60mA. A charge on the 2000mAh battery will last for 2000mAh/60mA= 33.3 hours.

Audioguru,

Wow! Your reply was very interesting! You provide me a very comprehensive explanation! I guess that you are very expert and skilled about audio stuff

However, quoting myself, I never talked about a car amplifier, but about "portable loudspeakers", which looks very similar to these:

these loudspeakers only works with an external adapter. I've bought them in 1995; they are tiny but the sound is very crystalline and also have a good dynamic. Unfortunately the label on the back is lost, and I can't remember the exact model nor the other specs of these active loudspeaker; I just know the specs of the power adapter (on the label I can read: 12 volt - 200mA).

My previous messages:

Well, I need this battery pack to feed a pair of portable loudspeakers: the specs of these loudspeakers are: 12 volt - 200 mA.

My amplified speakers are not provided to works with internal batteries; just the external wall wart. I used a multimeter to check the exact voltage from the wall wart: I I measured a value of 12.70 volt.

From the wall wart I just measured a voltage of 12.70 (without the load), so this is the reason why I was thinking to get the same value using the buck converter.

Your little "computer speakers" will operate fine from ten AA Ni-MH cells in series for about 33 hours or more for each charge.

Also if, when fully charged, the battery pack will provide 14 volt? The power adapter, instead provide 12 volt.

Anyway: the solution of the buck converter, since you disadvised it, which disadvantages will bring?

Also if, when fully charged, the battery pack will provide 14 volt? The power adapter, instead provide 12 volt.
Ten AA Ni-MH cells will be 14V only for a few minutes when they are fully charged. This battery will be about 13V for most of a discharge then when its voltage has dropped to 12V it drops very quickly when discharged more.

Anyway: the solution of the buck converter, since you disadvised it, which disadvantages will bring?
It and the 6 extra battery cells are not needed.

Ok: tomorrow I will try following your suggestion.
I will let you know.

Thank you.

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