Are bypass caps reason why uC don't have flyback diode across ferrite bead in the sample schematics?

Question

Often the IC datasheet recommends putting a ferrite bead before AVCC pin. First I started wondering if I don't need a flyback diode but than I realized that capacitor is probably able to absorb the excess current.

For back of the napkin calculations I assumed that ferrite bead is an inductor so total energy stored before switching off is \$ CV_0^2 + LI^2 \$.

When the power is cut off all energy needs to be stored in capacitor so energy now is \$ CV_1^2 \$.

So if I want the voltage not to be above some \$V_t\$ I need to have capacitance of:

$$ C_{min} = \frac{LI^2}{V_t^2 - V_0^2} $$

So assuming \$L = \frac{X}{2 \pi f}\$, \$300\,\Omega\ @\ 100\,\mathrm{MHz}\$ ferrite bead, \$I = 20\,\mathrm{mA}\$, \$ V_t = 5\,\mathrm{V}\$, \$ V_0 = 3.3\, \mathrm{V}\$ it gives \$C_{min} \approx 14\,pF\$. Other way round standard \$ C = 0.1\ \mu\mathrm{F}\$ capacitor gives swing of \$0.3\,\mathrm{mV}\$.

Is this reasoning correct? I've made a few spherical cow assumptions and I'm just a beginner.

Answer 1

A typical ferrite bead impedance graph: -

This one peaks around 400 ohms and as you can see the impedance is virtually purely resistive. These are the eddy current losses in the magnetic material. Inductance has dropped off to virtually negligible amounts.

But it does have inductance at low frequencies (and very few losses) so it's worth calculating. At 1 MHz the losses can be assumed zero but the reactance is about 15 ohms (an estimate). So this provides us with the inductance i.e. L = \$X_L/2\pi F\$ = 2.4 uH.

At 20 mA this is inductance holds about 0.48 nJ of energy.

But, given that when you switch your supply off the 20 mA current flowing through the inductor gets benignly absorbed by power supply capacitors on the main power rail I don't think you need to worry.