Question on power consumption on S32K1xx

if nothing else, the PI_matching capacitors on either end of the resonator will demand power.

Power (capacitor) === F * C * V^2

Notice the power scales linearly with the frequency.

Example: Power = 10MHz * (20 + 10pF) * 1volt_rms^2

Power = 1e+7 * 30e-12 * 1 = 30e-5 = 0.3 milliWatts, just to charge/discharge the capacitance.

Then the amplifier has quiescent current. And that circuit has junction capacitance which needs charge/discharge current.

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Often 10MHz oscillators use 22pF PI_network capacitors on each end of the crystal. And within the quartz is capacitance between the silver_plated contacts. I assumed 10pF. Hence the (20 + 10PF). I ignored th e22pF, so as to not pretend we have extreme accuracy here.

The 1 volt RMS is 2.828 volts (2 * sqrt(2) ), which is about the maximum voltage you'll see in a 3.3 volt VDD oscillator.

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To achieve that extra bit of phaseshift, sometimes needed for stable oscillation, the amplifier may be followed by a discrete resistor, so the PI network capacitor and the resistor provide crucial phaseshift.

IN that case, the resistor is always dissipating power.

Regarding losses of capacitors ---- there will be ESD structuree with unavoidable junction capacitances, and LOSSES within the substrate underlying those junctions.

Source: https://www.nxp.com/docs/en/data-sheet/S32K-DS.pdf

In short the chip will consume power depending on:

• What peripherals are enabled
• Current consumption of peripherals (use high impedance if possible)
• What clock speed chip is running at

The clock speed pf the S32L142 is determined by the PLL (and if you select an internal or external oscillator, it has an 8Mhz low power oscillator built in, or you can use an external oscillator from a range of 8-40Mhz). The S32L142 will then consume power at roughly 400uA/MHz, so the lower the clock speed the better power consumption (less switching of transistors is usually lower power in most digital devices). So set the PLL lower (which is configurable by software).

The S32L142 also has a low power mode, which uses a 128kHz internal oscillator. so if you can put the processor to sleep, it will use very little power.

Suppose, I use this 8MHz resonator between the pins 11 and 12, how to determine the current consumption due to this resonator?

The power consumption will be determined by the clock frequency of the SPLL (configurable by software) and then the s32 will use roughly 400uA/Mhz

EDIT: This is a much better diagram for your questions and the clocks in use, the document below also shows code on how to switch power modes on the S32k1xx
Source: https://www.nxp.com/docs/en/application-note/AN5425.pdf