I am driving an audio amplifier with a 3.7V lipo.

With a latching circuit in front.

[![enter image description here][1]][1]

The problem is, that when I turn up the volume, the voltage drops under 3.7 volts (especially on bass) and the amplifier turns off for a couple of milliseconds.

The amplifier accepts 3.7 - 5 volt.

What is the simplest way to keep it above 3.7 volt? (I think that is the problem.)

I think it needs some sort of voltage regulator.

When I plug in the lipo direct into the amplifier(without the latching circuit) I can turn up the volume to max. But only as long the lipo is nearly fully charged.

Isn´t it possible to somehow put a large capacitor in, to buffer the short voltage drops a bit ?

  • \$\begingroup\$ The voltage drops below 3.7V because you're increasing the load when you increase the volume. The battery has an internal resistance, if the current increases so does the dropped voltage across the batteries internal resistance (Think Ohms law where R is ~80mOhm) \$\endgroup\$
    – user103993
    Commented Jan 11, 2019 at 10:38
  • \$\begingroup\$ The reason it works when the LiPo is nearly fully charged is because at full charge at LiPo is ~4.2V so it has some room to play with in terms of voltage drop \$\endgroup\$
    – user103993
    Commented Jan 11, 2019 at 10:40
  • 1
    \$\begingroup\$ Thanks, I got that, but i wonder why it working without the latching circuit , plugged directly. And so baddly with the circuit. Maybe because its on the breadboard(not solderd). But i need some way to keep the voltage above that 3.7. Until the battery protection circuit kicks in(and says iam empty). Even with usb 5v i can´t max the volume, with the latching circuit in front. \$\endgroup\$ Commented Jan 11, 2019 at 10:45
  • \$\begingroup\$ Because with your latch circuit you have a MOSFET to take into account aswell. Resistance of an IRF7319 is another ~60mOhm on top of your ~80mOhm internal battery resistance. Also in terms of max volume, do you have a figure for the voltage required to achieve that? A USB will only provide 500mA so you need to figure out what voltage = max volume then use Ohms law (where R is your speaker resistance) to find the current you need \$\endgroup\$
    – user103993
    Commented Jan 11, 2019 at 10:50
  • \$\begingroup\$ Ok, i thought the MOSFET doesn´t eat that much. Thanks , so what can we do against that voltage drop. what kind of part we need here? Would be nice to keep the voltage above 3.7! \$\endgroup\$ Commented Jan 11, 2019 at 10:55

2 Answers 2


As has been mentioned in the comments, your issue is that the voltage is dropping due to load and the battery voltage dropping.

As your amplifier can take 3.7-5V, and you want it to operate off a battery, you're options are either to change the battery, or change the load.

You could do a new battery set up, using two cells in series to have a voltage of 5-8V ish (depending on cells chosen). then use a buck converter (LDO if you want to keep it as simple as possible) to keep voltage around 4V. This would be the easiest, and probably best method.

Or you could use current battery with a boost converter, to boost the voltage to a constant 4V or so. Assuming the battery can provide the power, and your boost converter is good enough, this will do the job cheaper than the first option. But this is more complicated and more to go wrong. Designing a good boost converter for a sensitive circuit (such as audio) is a bit tricky.

Or you could limit the load by reducing your volume limit, using lower power amplifier and speaker set up. Which you clearly don't want to do as you're asking this question.

  • \$\begingroup\$ Ok, 2 cells are no option because of the space.( and the charging get more complicated) The cell iam using now is an 650 mAh rated at 20c. Reducing volume isn´t an option. \$\endgroup\$ Commented Jan 11, 2019 at 11:03
  • \$\begingroup\$ You can buy a boost converter board, places like sparkfun sell them premade. However with a 650mAh battery how long do you realistically expect the speakers to play music for? An hour? \$\endgroup\$
    – user103993
    Commented Jan 11, 2019 at 11:10
  • \$\begingroup\$ Yes an hour . Maybe i update to 2000mAh. \$\endgroup\$ Commented Jan 11, 2019 at 11:19
  • \$\begingroup\$ If you increase the mAh of the cell the size will definitely increase \$\endgroup\$
    – user103993
    Commented Jan 11, 2019 at 11:22
  • \$\begingroup\$ A bit but thats ok, i don´t like 2 cellc becaus the charging get tricky than. So the option is a boost converter can you direct me to something? \$\endgroup\$ Commented Jan 11, 2019 at 11:35

As it seems that going from a 650 mAh ('16340' type cell?) => 2000 mAh ('18650'?) is not a (space) problem, may-be an attractive approach can be to use 4xAAA NiMH in stead. You gain voltage (from 3.7 V => 4.8V (max. 6V after fully charging)) and AAA's are available up to 1000 mAh. So: more voltage pushing you away from the sub-3.7 voltage range, as well as longer "sound time". If 6V would be too much you may add a Schottky diode in series. Can it be simpler?
B.t.w: why do you mention 3.7-5V acceptable for your amplifier, and 5-18 V range for the auto On/Off switch?

  • \$\begingroup\$ "If 6V would be too much you may add a Schottky diode in series" How did you calculate that a Schottky diode would drop 1V (i.e. 6V down to the 5V maximum quoted for the amplifier)? \$\endgroup\$
    – SamGibson
    Commented Jan 12, 2019 at 18:42
  • \$\begingroup\$ @SamGibson Sorry, I was a bit too fast. I more or less pre(!)sumed that 'USB'-type voltage limits would hold, i.e. 5.25V at least and 5.5V virtually always in practice. So, if one wants to be fully safe put in a Schottky+conventional diode in series => typ. 1.0-1.1 V at 1A. \$\endgroup\$
    – Kees_DCRF
    Commented Jan 12, 2019 at 19:34
  • \$\begingroup\$ Thanks for the update. Yes, we don't know what tolerance there is on the maximum 5V stated by the OP - let's assume the limits are as given, for now. The problem is that, if we use your updated 2 x diodes, we also have that 1.1V voltage drop at the lower end of the battery voltage range as well. So when total battery voltage drops below 4.8V (i.e. 1.2V per battery) for the 4 AAAs in series that you mentioned, it's already reached the minimum 3.7V for the device. Since approx 1.2V is a typical median for NiMh batteries during discharge, that's lots of battery capacity unable to be used :-( \$\endgroup\$
    – SamGibson
    Commented Jan 13, 2019 at 2:29
  • \$\begingroup\$ @SamGibson I tend to agree: that likely is a problem. \$\endgroup\$
    – Kees_DCRF
    Commented Jan 13, 2019 at 8:19
  • \$\begingroup\$ Thanks for the input and ideas, but 4xAAA NiMH are quit expensive,also the charging times are quit high. And u need some sort of holder for them what also drives the cost and space up. I ended up with using a 3.7 volt lipo 2000mah for now. I will using something like that. ablic.com/en/semicon/datasheets/power-management-ic/… to close the mosfets, when voltage drops to the critical level. \$\endgroup\$ Commented Jan 14, 2019 at 23:55

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