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I'm thinking of using a MAX680 charge pump IC to generate 10V from a 5V source, but I don't need the -10V output. The example here...

example circuit

...shows four capacitors, presumably two for the +10V and two for the -10V. As I only require the +10V output, can I simply omit the other two capacitors?

(NOTE: I believe there is a slight error in the schematic. The lower-left capacitor should presumably connect to C2+ and C2-)

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  • \$\begingroup\$ 1, why not just try it? 2, why not ask maxim? They would be happy to answer your question or point you to an alternative. \$\endgroup\$
    – Passerby
    Dec 25, 2017 at 6:05
  • \$\begingroup\$ also why not go with the 681 so you don't need external caps or use the recommended replacements as the 680 is obsolete \$\endgroup\$
    – Passerby
    Dec 25, 2017 at 6:11
  • \$\begingroup\$ [1] Because I haven't bought it yet. And [3] Because the 681 is twice the size. \$\endgroup\$ Jan 6, 2018 at 10:42

2 Answers 2

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This diagram from the data sheet MAX680 Datasheet shows clearly that the negative output is derived from the positive output.

MAX680 Datasheet

Therefore you can eliminate the capacitors for the negative output (C2, C4) if it is not required. Conversely, if only the negative output is required, you still need the capacitors for the positive output.

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Based on the idealized schematic on the datasheet, the two charge pumps are completely independent. Omitting the c2 and c4 caps will make the negative charge pump do nothing, as there is no path to switch.

But using a dual rail charge pump like this is... odd. unless you have some stock you need to use, you are better off using any number of single rail positive charge pump.

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  • \$\begingroup\$ Well, I've been trying to find a cheap, small (low pin count) through-hole charge pump IC, and I'm not having much luck. \$\endgroup\$ Jan 6, 2018 at 10:46

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