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Background info:

I have a small battery powered 25W amplifier/speaker combo. KustomSound CB25X, with two 6V 4.5Ah SLA batteries wired in series for 12V. Fully charged, the voltage seems to be close to 13V.

I also have a mixer - Yamaha MG06X, which runs off a 12V wall wart. The voltage off the wall wart is 12.3V.

The mixer is rated for 24W but it's unlikely I'll ever be using it at full power. Same goes for the speaker.

I also got myself a little preassembled LM2577 DC/DC switching voltage booster from online.

My question: Can I run the mixer off of the amp battery with the DC/DC booster? Can it tolerate a volt or so above its rated 12V e.g. when the battery is fully charged? (Because that DC/DC chip can't lower the voltage, unfortunately.)

Alternatively: can I wire the DC/DC booster across just one battery? Will that be bad for the amp or the batteries or the mixer? EDIT: I've been informed that's not a good idea, so ignore this paragraph.

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  • \$\begingroup\$ Drawing off only part of a battery (one or more cells) will imbalance the battery, potentially reducing its service life. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Jul 18, 2015 at 18:50
  • \$\begingroup\$ Okay, so that's a bad idea. Cool. Primary question: can an audio mixer tolerate a volt or two (maximum) over its input rating? \$\endgroup\$
    – Boloar
    Commented Jul 18, 2015 at 18:53

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Likely yes. 16V is the next size up capacitor rating, and most component have a tolerance for a single volt. Likely there is a regulator that drops the voltage inside anyway, and an inverter to provide a negative rail too. You could Crack it open to verify, but just to be safe put a standard 1n4001 diode in series with the input voltage, to bring it down about 0.7V. The same applies to undervolting it, slightly less than 12v is fine too.

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  • \$\begingroup\$ I went ahead, battery -> DC boost regulator -> mixer. Seems to be fine. The battery was 12.9V, voltage out of the regulator was 12.5V, though I'd set it to 12.0V. Only time will tell, I guess. XD \$\endgroup\$
    – Boloar
    Commented Jul 19, 2015 at 3:04
  • \$\begingroup\$ @Bolar a boost raises or steps up the voltage. Giving it a higher voltage than what it is set for will not work properly. \$\endgroup\$
    – Passerby
    Commented Jul 19, 2015 at 4:48
  • \$\begingroup\$ I know, I said that in the question ("the chip can't drop the voltage") I used 6V from one of the cells to set the output to 12V. So it should keep it steady as the battery discharges. ^_^ \$\endgroup\$
    – Boloar
    Commented Jul 19, 2015 at 7:53

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