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I've changed an 820 μF capacitor on the output of a 1.05 V linear regulator which feeds the PCH (Intel's Platform Controller Hub), or "southbridge", on a PC motherboard.    It's working fine but I want to confirm that I'm not stressing the new capacitor due to ripple current.

The replacement takes the rated impedance down from 36 mΩ to 28 mΩ.    (I determined gain/phase margins are fine.)

I'm thinking that in principle, the lower impedance could lead to higher ripple current (Ohm's Law).    The replacement does have a higher ripple current rating, 1.25 A instead of 1.14 A, but this isn't much higher.    Intuitively I would guess it's still fine—and it's not noticeably warm to the touch—but that's not an engineering answer.

I've got the (934 page) datasheet for the "Cougar Point" PCH.    From this and board schematics I've worked out a maximum load current of 8.3 A, but I don't know how much of this is in ripple, and at frequencies an e‑cap would "see."    The datasheet seems to be silent on both AC loading and bulk capacitor selection.    The 1.05 V is within spec from 1.00 V to 1.10 V, which doesn't preclude a way‑too‑high ripple current.

I've thought of using an oscilloscope to look at the ripple voltage across the cap, and use Ohm's Law to determine the current, given its impedance.    I only know the cap's impedance at 100 kHz, but that's another question.    Ideally I'd like the scope to plot \$\frac{dV}{dt}\$, but I don't think it will, so if the wave is jaggy, I'll have a headache.

Getting the chip to show me its worst-case ripple would not be easy, since it does many things, including integrated video, PCI‑E to three X1 slots and to onboard LAN and USB, 6 SATA ports, a bunch of USB ports, etc.    I could connect a few drives and run large file copies across them while also running some kind of GPU benchmark.    Maybe try a few different combinations of workloads.    I don't think I can reach the worst‑case load, but can at least get a "normal usage" maximum.

This is a lot of work, though, so I have to first ask:

  1. Is my concern valid, that lowering the ESR would lead to higher ripple current?

  2. If so, is taking measurements like this the best & easiest way to assure that the replacement is ok, or is there a simpler heuristic I could use?

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Is my concern valid, that lowering the ESR would lead to higher ripple current?

Yes.

If so, is taking measurements like this the best & easiest way to assure that the replacement is ok, or is there a simpler heuristic I could use?

You already mentioned that you touched the capacitor and it was not noticeably warm. I wouldn't dismiss this as "bad engineering". In fact I would take it as more or less definitive evidence that the ripple current is not excessive. Excessive ripple current will heat up a capacitor, and it is primarily this heat that causes the capacitor to fail prematurely.

That, and the fact that the ESR did not change that much, lead me to believe, with high confidence, that your capacitor is fine.

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You said linear regulator. In that case the capacitor will have almost zero current. A switching regulator uses the capacitor to store energy between cycles. A linear regulator should carry most of the current leaving the cap with little to do.

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  • \$\begingroup\$ Are you sure this applies even if the load makes large, rapid transients? The loop gain at 10 MHz is -70 dB; I assume that means transients in the 25 ns range would initially cause almost no change in pass transistor conductance, which in turn would mean an at least non-negligible discharge current if the jump is large. \$\endgroup\$ Apr 25, 2023 at 19:55
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Even in a switching regulator, the output capacitor ripple current is not given by the capacitor, but by the inductor and duty cycles.

In that case, a capacitor with lower ESR will reduce the power dissipation in the capacitor.

The only drawback can be reduced stability..If you have checked that this is fine, then there is zero reason for concern.

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