Given a max load of 1pF, V = 1.2v and a frequency of 400MHz, how to calculate the max Current?

Answer 1

The impedance of a pure capacitor is 1/j(2*pi*f*C)

j is the imaginary operator (means that the phase of the current will be 90 degrees to the voltage applied)
f is the frequency in Hz
C is the capacitance in Farads
pi is 3.141...

The magnitude of the impedance is simply 1/(2*pi*f*C)
332 Ohms

1.2 V / 332 Ohms = 3.6 mA /_ 90 degrees

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