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I am working on a digital voltmeter circuit using ICL7107 IC. Having trouble powering it up. There is a -5 V supply to pin 26. I used a 7805 regulator for +5V supply but wondering how to supply the -5V. a) Can I use the output of 7805 to somehow convert to negative? b) is there an IC for this?( my max current rating is 3.34Amps). Note: I tried using 7905 with 7805 but no use the 7805IC goes burning hot. As soon as I removed the connections between two it behaved properly(only7805). Please suggest some idea to achieve this.

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marked as duplicate by PeterJ, Daniel Grillo, Dave Tweed Apr 10 '16 at 17:31

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    \$\begingroup\$ What do you mean by: "my max current rating is 3.34Amps"? \$\endgroup\$ – EM Fields Apr 10 '16 at 10:16
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Aside from the obvious general answers of using a 7660 charge pump or making a dual supply with a transformer and 7805/7905, you can make a charge pump with a single gate buffer and drive a couple ceramic caps and a BAT54 dual Schottky. Very tiny and very cheap- it will give you -3V or better, which is typically good enough, depending on your range. The actual current drawn on the minus supply is about 1mA.

You can drive the buffer with the clock output of the chip, so this is a solution that is specific to the this particular situation.

If you didn't follow what I wrote, you can find a circuit using inverters in some versions the datasheet, just modernize it by using the single gate buffer and dual Schottky diodes.

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