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This is a display IC for an iPhone 6. There was no display and all voltage outputs from the IC were 1.8 V (which is wrong). I finally traced it to shorted capacitor C1554.

What is its role? Is it a feedback? I am just trying to understand what happened with it being shorted.

enter image description here

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    \$\begingroup\$ Have you read the data sheet for the chip on the left? \$\endgroup\$
    – Andy aka
    Jan 25 at 18:31
  • \$\begingroup\$ I couldn’t find a data sheet for it \$\endgroup\$
    – ohmmy
    Jan 25 at 18:55
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    \$\begingroup\$ A poorly justified guess, at best - power management IC, mentions a switching node, so it could be a flying capacitor for a charge pump/bootstrapping. If you have any other evidence from the board or elsewhere on the schematic that might be able to prove or disprove this guess, please edit your post with it. \$\endgroup\$
    – nanofarad
    Jan 25 at 19:12
  • \$\begingroup\$ Looks like it's probably the charge pump capacitor to generate the -5V. \$\endgroup\$
    – GodJihyo
    Jan 25 at 19:12
  • \$\begingroup\$ Aye, most likely the flying capacitor for a charge pump to generate the negative supply rail, particularly as there's a pin named CPUMP too. \$\endgroup\$
    – Tom Carpenter
    Jan 25 at 19:20

1 Answer 1

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This is the Apple-proprietary "Chestnut" IC (and there are no datasheets to be found). From the part number, it's probably made by Texas Instruments for Apple.

I think the other commenters are correct and it's a flying capacitor for a charge pump to generate the negative rail. The positive boost rail is generated by a boost inductor (the 1.5uH inductor cropped off the left edge and connected to the SW pin). Since it's connected directly from the main Vcc to the chip, it can only be a boost converter.

Since there's only one inductor connected to the chip, that leaves a capacitor charge pump as pretty much the only possibility to generate a negative rail from the main Vcc (in this case, I think it's generated indirectly by an inverting charge pump fed back into the chip from the positive boost rail).

Then there are a few positive output LDOs, I2C communication and some other control and monitoring lines.

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  • \$\begingroup\$ Thanks. Are the ldo voltages generated from the main 6v boost voltage? \$\endgroup\$
    – ohmmy
    Jan 25 at 20:32
  • \$\begingroup\$ @ohmmy Yes, I believe so. \$\endgroup\$
    – Spehro Pefhany
    Jan 25 at 20:35

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