how to calculate inductance and capacitance values

calculate capacitor series resonance

Hi Guys,

How to calculate the values of Inductors, capacitor and resistance in a series RLC circuit where the frequancy range is given from 450 MHz to 3500 MHz with 50 Ohm terminations on input and output. Please help me.

Thanks in advance

inductance s11 and r

The real part is R,it will not change with f.
From the resonate f,you can get the value of L*C.
At very high freq,C does not affect too much,you can get L,then C is out.
Or at very low freq, L does not affect too much,you can get C,then L is out.

calculate inductance with capacitance

thank you very much, it helped me.

calculate imaginary capacitance

if you have network analyzer , you can use S parameters to figure out the L,C,R

you can use S11 at very low frequency, in ehich the parasitic L is small and reactance is capacitive , then the imaginary part of it is C.

at very high freq teh C is out and L is dominant in imaginary part of S11.

by S21 you can measure the series R,

another complemantary methose is by looking at resonant freq.

by sweeping from 0.1 Ghz to 4GHz the first resonant freq is series resonant freq which represent a series R and C and L, second resonant freq which is normally higher , is parallel resonant freq which caused by parasitic L and C

Mehdi

imaginary part of inductance

In general this is a tough problem. However, I think Haitaoz gave the best advice; his method will work assuming that |1/wC| << |wL| at high frequencies and |1/wC| >> |wL| at low frequencies. If this is not true, you can try this method. You want to approximate Z12 at

Z12 = R + j(w*L - 1/w*(1/C))

As Haitaoz points out, R will be the real part of Z12. Note that there is a linear relationshipt between Imag(Z12) and (L, 1/C). So to obtain the L and C solve the least squares problem b = A*x where b is the imaginary part of Z12, A is nx2 matrix whose first column is "w" and second column is "1/w", and the solution vector x is x(1) = L and x(2) = 1/C.

P.S. solving for L in terms of nH and C in pF, keeps A much better conditioned than solving for L in H and C in F.