Quote:
Originally Posted by ulysses  which means that making R or C higher will make f lower, and vice versa. Your shunt resistor will be in parallel with the actual load impedance, and serve to lower the effective load impedance seen by the capacitors. |
But if we didn't use a shunt in the first place, the input impedance or R in this case can't be raised. Is this right?
So, to force it lower, one would have to raise the cap value.
Using 8488Ω (not sure thats the correct Zeq) one would need approx .376uf on each leg