I'm designing a circuit to charge mobile battery. My source is giving a DC output of 15V, 100mA, whereas I need DC 6V, 450mA for charging. Is there any way to boost DC current?

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    \$\begingroup\$ No. The power you want out of this converter is more than you are putting in. As Scotty would say Ye no canna cheat da laws of physics. \$\endgroup\$ Feb 23, 2013 at 21:10
  • \$\begingroup\$ Can you describe the 15V/100mA source? It's impossible to say how to upgrade the source to boost the current without knowing what it is. Is it just a transformer with a rectifier? Or is it a regulated supply? Linear or switching? \$\endgroup\$
    – Kaz
    Feb 24, 2013 at 2:35
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    \$\begingroup\$ @OlinLathrop - [Nitpick] I'm pretty sure he said 'Ye canna change da laws of physics.' \$\endgroup\$ Feb 24, 2013 at 2:35
  • \$\begingroup\$ Hi, what are you trying to charge? \$\endgroup\$
    – hassan789
    Feb 24, 2013 at 5:12
  • \$\begingroup\$ What does "for charging" mean? If you can trade off less current, then you could still conceivably charge your mystery load, but it would take longer because you can't provide the full 450mA. \$\endgroup\$ Jun 14, 2016 at 21:20

2 Answers 2


Think of the problem in terms of power. Power = current * voltage. The maximum power input of your system is 15V * 0.1A = 1.5 Watts. Suppose you design a Buck regulator to step the voltage down to 6V. The regulator is not 100% efficient (80% is not an unreasonable number). 80% of 1.5W is 1.2W.

Your desired output is 6V * 0.45A = 2.7W, more than twice what your regulated source can provide. You can't get something from nothing. You're going to need a power source that outputs more current (and/or at a higher voltage) such that the regulator can output at least 2.7W after conversion losses.

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    \$\begingroup\$ According to most claims I've read, xfmrs are generally ~80% efficiency, but DC-DC converters are supposed to approach 95%. Of course, either way you go, the OP's desired 6V*0.45A output would require (an impossible) 180% efficiency. \$\endgroup\$ Jun 12, 2016 at 14:43
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    \$\begingroup\$ Yeah, I've seen anything from 80% to 95% for pretty typical DC-DC converters. The math in this case happens to work out to rounder numbers with 80% efficiency. \$\endgroup\$
    – Joe Baker
    Jun 14, 2016 at 20:08

Current can be boosted for short durations. Capacitors could be used to supply adequate power for less than the 100% duty cycle. The batteries would be charged in spurts rather than under continuous power.


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