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I have got two 12 V 7.2 Ah lead-Acid batteries, which are connected in Series (26 V after charging). The batteries are connected to a step-up / boost module which have to be 2 V higher than the input. So the min. output is about 28 V

I want to use this to power an "mobile Amplifier", which is based on the compounds of an Logitech Z323 Speaker System. Now, my problem is, that the amplifiers input have to be between 14-18 V, but I can't get a lower output voltage from the boost module than the input voltage. In my opinion I could use a resistor combined with ohm's law. But the problem is, that the current-consumption of the amp (max 3 A) depends on the volume which is adjustable by a potentiometer.

Is there a way, to lower the output voltage to 14 V ignoring the actual current flow?

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2 Answers 2

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With the componenets on hand, it seems like it would be simplest to use only one battery at a time, perhaps with a switch to choose between them. You could configure the converter for around 15-16 volts out for something in the neighborhood of 12v in, which would seem to satisfy the rule (at least the one you mentioned) about its output vs input.

A resistor seems like a quite poor choice, as in addition to the variable load concern you mention, it would simply be turning your limited battery power into heat.

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  • \$\begingroup\$ It is one of the simplest solutions, but i didnt realized it... I think this should work fine for me. But this will decrease the devices runtime, so a switch would be a good thing. Otherwise, as Blup1980 mentioned, a buck-boost converter is a good choice, but I didn't found a cheap, powerful device which is suitable for my purposes available for shipping to germany. \$\endgroup\$
    – user20032
    Commented Mar 11, 2013 at 19:34
  • \$\begingroup\$ @J.E. If you wire the two batteries in parallel you get the full run-time, at the 12/14 Volt nominal voltage desired. \$\endgroup\$
    – Anindo Ghosh
    Commented Mar 12, 2013 at 4:45
  • \$\begingroup\$ Wiring them in parallel may not, however, be a good idea, as if one is stronger it can end up charging the other at a poorly controlled rate. \$\endgroup\$
    – Chris Stratton
    Commented Mar 12, 2013 at 5:43
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Remove the step-up and replace it by another DC/DC topology that accepts output voltages that "cross" the input voltage such as SEPIC, flyback, etc.

Then sets the output voltage of your DC/DC to 15V and that's it.

The output will be at 15V whatever the charge level of your batteries.

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