Piezos in a high power system - How?

The 15 ohm shunt makes the piezo look like a current driven tweeter to the 2.2uF crossover capacitor. The small series capacitor attenuates the piezo tweeter. You could run an identical piezo in parallel.

The frequency of crossover is given by this equation:

f = 1/(2πRC)

So, for R =15 ohm and C = 0.0000022F we get f = 4823Hz
Thanks I get it now, the capacitive values of the piezos and the attenuation are so small it have no impact to the crossover. Then I learned something this week as well. Nice. I’ll save the Peavey filter for a rainy day.