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When looking into the datasheet of an voltage regulator, they often recommend adding a capacitor from the ADJ pin to ground, to improve ripple rejection.

enter image description here

It can also often lead to an significant increase in PSRR when doing so, according to their performance graphs:

enter image description here

But why? It seems kinda unintuitive to me. Adding a huge capacitor at the feedback pin makes me think that the changes in the output voltage will take longer time for the regulator to measure(i.e slower control loop), because the capacitor has to be charged/discharged first. But this doesn't seem to be the case.

What am i missing?

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  • \$\begingroup\$ Add the schematic to your question. The internal use of the transistors, resistors, etc plus the aluminum on-silicon routing (crosstalk) determines the PSRR. Some regulators operate on 1uA total current, and large external capacitors are the only (low current) way to hold crucial internal nodes "quiet". \$\endgroup\$ – analogsystemsrf Mar 19 at 14:37
  • \$\begingroup\$ @analogsystemsrf: I don't really have any schematic other than the reference design as shown on the first picture. This was more like a theoretical analysis on why it works like this. Sure that makes sense, but why does it not make the regulator slower? \$\endgroup\$ – Linkyyy Mar 19 at 14:43
  • \$\begingroup\$ What's stopping you from modelling with Functional Block Diagram in TI spec 7.2 to understand it? \$\endgroup\$ – Sunnyskyguy EE75 Mar 19 at 15:05
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That regulator senses voltages with respect to its output pin, it tries to keep its ADJ pin 1.2v below the output.

Consider what the output ripple means. It not only means the output ripples with respect to the ground. It also means the ground potential is rippling with respect to the output pin.

Without that capacitor, the ripple arrives at the regulator ADJ pin through the R2/R1 potential divider, so it only rejected with that low gain.

With Cadj, the ripple on the ground terminal arrives unattenuated at the ADJ pin, so there is much more loop gain available to reject it. Study the rejection graphs carefully. You'll find the difference with and without Cadj is equal to the R2/R1 attenuation ratio.

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  • \$\begingroup\$ Yes that does make sense. Thanks for explaining. \$\endgroup\$ – Linkyyy Mar 19 at 20:24

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