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I recently bought online three solar panels rated 6v x 500 mA for some circuits I had in mind. What I got instead were 12v cells, and I am certain they are rated 250mA, since the seller had those too.

They were all the same price, so, instead of complaining about it (and wait some extra time for the exchange shipping process), I decided to go along and rethink what to do with my circuits. After all, it is (sort of) more juice.

Most of my ideas are limited to 5v circuits, and a 7805 regulator would do the trick nice and smooth. I just wonder whether the 7805 can supply more current than it has available (if necessary), or it would take a DC/DC converter to achieve this.

Let's say, in a full on sun condition with the cell at its peak (12v x 250mA), would the 7805 be able to supply 5v x 400mA?

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    \$\begingroup\$ Linear regulator won't. Get a switching (buck) converter, it will get you efficiency of 80-90% \$\endgroup\$
    – Eugene Sh.
    Commented Feb 23, 2016 at 18:34
  • \$\begingroup\$ 6*0.5 = 12*0.25 so it's not more juice, it's the same amount at best. Less, if you have to convert the voltage down. \$\endgroup\$
    – pipe
    Commented Feb 23, 2016 at 18:41
  • \$\begingroup\$ I know, I kinda expressed myself poorly. \$\endgroup\$
    – Giordano Bruno
    Commented Feb 23, 2016 at 19:03

2 Answers 2

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In a linear regulator, Iin = Iout. Since Vin > Vout, the remaining power is dissipated in the pass device. If you use a buck switching regulator, your efficiency will improve dramatically because in a buck, Pin = Pout (excluding converter losses).

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The 7805 is a linear regulator and can only provide as much current out as goes into it. A switching regulator can do what you want though.

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