Re: decoupling capacitor design calculation
We start with a simple formula about energy stored in a capacitor

but a capacitor has a series resistance (ESR),
that means we can not get back all the Energy we want back out of the capacitor and we will suffer a additional voltage drop.
So lets allow 0,05V drop at the ESR. This means our capacitor only is allowed to drop by 0,05V.
we recalculate and now we need 800uF not 200uF !
Also we know our µC will draw 1W of power with a 0,1V drop... this is - assume a 3,3VµC 1W/(3,3V-0,1V)=0,3125A of current. with an allowed drop of 0,05V at the ESR and 0,3125A current the ESR must be equal or lower than 0,05V/0,3125A = 500mOhm
So for our scenario you need a 800uF capacitor with a ESR &lt; 500mOhm to supply 1W for a µC running at 10MHz for 10clocks with a voltage drop of max. 0,1V
OK, lets simulate it
first a weak source which will drop at our load condition without the capacitor:


We see a drop by ~0,2V. Thats to much. so we connect our capacitor


and now we see what we expect. the voltage drop is reduced, current flows out of our capacitor (neg. direction)