Understanding the AVCC pin wiring on ArduinoLeonardo (low-pass filter?)

Question

The Arduino Leonardo has an interesting difference with previous boards: the AVCC pin of the ATMEGA32U4 is connected to +5V through a MH2029-300Y ferrite beads and to GND through a 1uF capacitor.

In ArduinoUno and ArduinoMega2560, this pin was simply connected to VCC. According to the atmel datasheet, this pin should be connected to a Low-Pass filter when the ADC converted is used.

Arduino Leonardo schema

Is this really a low pass filter? If yes, how does it work? How would you call that type of circuit? How do you model it?

Answer 1

The inductor plus capacitor form a frequency-dependent voltage divider.

\$ \dfrac{V_{OUT}}{V_{IN}} = \dfrac{Z_C}{Z_C + Z_L} \$

For DC and low frequencies the impedance of L1 (\$Z_L\$) is low, and that of C1 (\$Z_C\$) high, so the input voltage won't be attenuated much. At high frequencies it's the other way around: \$Z_L\$ is high, and \$Z_L\$ is low. The attenuation is high, and the higher the frequency the higher the attenuation. So this is indeed a low-pass filter.

The inductor they used is not a good one, however. It's a high frequency EMI suppressor, targeted at frequencies of tens of MHz. (The used type has an impedance of 30 Ω at 100 MHz.)

The impedance curve shows a 0.5 Ω/MHz slope, so at 100 Hz the reactive part of the inductance is negligible.

What is actually needed is suppression of low frequency noise, like 100 Hz ripple from the power supply. Then this inductor is pretty useless, and it's like just having the capacitor.

For low frequencies inductors can be impractically large, then a resistor instead of the inductor would have been a better choice. The datasheet says AVCC shouldn't be lower than VCC - 0.3 V, but I couldn't find how much current AVCC uses. That won't be much, say 10 µA maximum. The cutoff frequency of an RC filter is

\$ f_C = \dfrac{1}{2 \pi RC} \$

So if we use a 15.9 kΩ resistor with the 1 µF capacitor, we have a 10 Hz cutoff frequency, and the frequency response will look like this:

The 10µA through 15.9 kΩ is a 159 mV drop, so that's within spec. A 100 Hz ripple will be attenuated by 20 dB, that's 1:10, which isn't much, but VCC should have been decoupled properly already, so the 20 dB is just extra. Above 1 kHz noise will be reduced by at least 40 dB, that's a factor 1:100.

Answer 2

Yes, it's a low pass filter. The ferrite bead is a lossy inductive component that acts somewhat like a frequency dependent resistor, with L dominating at low frequencies and R dominating at high frequencies. The capacitor shunts higher frequencies to ground. Combined they form an LC filter with low Q, which does not experience the sharp resonant peak (as long as the cutoff is in the resistive region of the bead) that can cause problems in a "standard" LC filter.
However as Steven notes this part is not particularly suitable for this application, since it is a high frequency part and the ADC is relatively low frequency. It would be of use in something that requires filtering at much higher frequencies, like an FPGA design, high frequency ADC, etc.