Spehro Pefhany wrote:
> The output acts like a *current* source.

That's the same as saying it has inifinite impedence.  As I said, I agree
that the capacitors are in series for calculating the load capacitance in
this case.

However I don't agree that the driver has infinite impedence.  It has CMOS
high/low output drive transistors, and will look roughly resistive.  And, I
think there is a good chance that this resistance is about the same or lower
than the impedence of the capacitor on the pin at that frequency.

Do you at least agree that if the driver has finite impedence, then just
adding the two capacitors is not valid?


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