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

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.

• Why are you expecting to see a flyback diode? Most ferrite beads do not behave much like an inductor. They are lossy at high sfrequencies. If your bead supplier provides a plot of impedance against frequency this should be obvious. – Warren Hill Jul 27 '18 at 8:01
• You've worked out the inductance from the impedance at 100MHz, but switching off the power done at DC. At DC the impedance of a ferrite bead is just a few Ohms, most of which is resistance anyway. Check the impedance v frequency charts. – Steve G Jul 27 '18 at 8:02
• @WarrenHill Because I expected ferrite bead to be kind of like inductor. – Maciej Piechotka Jul 27 '18 at 16:32
• Your assumptions about the shape of the cow do show excellent understanding of typical farm-property constraints in cow-gates. – analogsystemsrf Jul 29 '18 at 5:39

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.