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Reading this topic from "The Art of Electronics" : Integrators

How does RC keep V << Vin?

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    \$\begingroup\$ It is not a matter of RC keeping ("forcing") V << Vin. You should interpret this as: RC should be large, then when V << Vin we can use the formula. If RC is small and/or V isn't much smaller than Vin, we cannot use the formula. \$\endgroup\$ – Bimpelrekkie Mar 18 '19 at 12:34
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    \$\begingroup\$ It doesn't say RC keeps V << Vin. It says "we" keep it this way. \$\endgroup\$ – Dmitry Grigoryev Mar 18 '19 at 12:41
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    \$\begingroup\$ For those without access to the text the OP refers to, the fuller context can be read here at the end of page 26 of the older 2nd edition. \$\endgroup\$ – jonk Mar 18 '19 at 19:33
  • \$\begingroup\$ To the OP: All that's being said here is that if you can arrange things so that the capacitor stays relatively uncharged, then the charging current will closely approximate a constant. If, on the other hand, you allow the capacitor to charge "for a long enough while" then the magnitude of the current will be seen to change over time and can't be considered a constant, anymore. In that case, you wind up with no longer a constant but instead a continually varying current and, ultimately, it involves exponentials/logarithms and stuff gets more complicated. \$\endgroup\$ – jonk Mar 18 '19 at 19:37
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If \$RC\$ is very large, current is very small, \$\frac{dV}{dt}\$ goes to zero, thus \$V\$ does not change.

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  • \$\begingroup\$ Thank you sir! @Scott Seidman \$\endgroup\$ – Exia Mar 19 '19 at 3:22
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If RC is large, then V will grow more slowly than it would if RC was small. So over the course of the integration, V will stay smaller, making it easier to keep V much less than Vin. It's a rather dense sentence, and it's easy to get confused in it.

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