I am trying to generate -8.5V from my 5V battery. I found a suitable IC for my purposes. The datasheet of the IC (circuit schematic is available below) says

use sealed line noise proof method

(for pin 10). This IC will be mounted on a PCB and I'm not sure how to do what they are asking for.

enter image description here

Can someone guide me through this? What exactly am I suppose to do?

  • \$\begingroup\$ This isn't even remotely the chip you need to generate -8.5V from a +5V input. \$\endgroup\$ – Marcus Müller Aug 6 '18 at 17:37

Looks like a poor translation to me.

In English, that should be a shielded wire between the wiper of the pot and the IC pin 10. The shield should only be connected to Vdd.


That's probably a translation error.

Without the figure, I would have assume they meant shielded cabling, with the figure, it looks like they mean a twisted pair.

But twisted pair makes less sense, since one of the conductors is just going to Vdd.

So, that's probably a "shielded line", with the shield connected to Vdd. Do NOT make the connection at the pin.

So, bad translation, and misleading schematics: do you really want to use this IC?

If you wanted a voltage tripler: If you don't want overly high accuracy or low ripple, a simple step-up converter with an Opamp comparing 1/3 of the output voltage to the input voltage would do the same.

If you really just wanted a stable -8.5V supply (not voltage) from your +5V: This isn't at all the chip you want. You want an inverting switch mode power supply IC. There's literally thousands of them. Ask for example TI.com's power supply designer wizard for what you need.


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