Question

Match a load impedance of 100 ‐ j100 Ω to a 50 Ω transmission line using a parallel inductor (next to
the  load)  and  a  series  capacitor.  Calculate  component  values  at  1  GHz  using  Smith  Chart.  Show  all
calculations on Smith Chart.

Answer 1

mpedance with 11.65nH parallel inductor 1.8pE series capacitor

Glven 2 = 10092 - 1-1002 Load impedance z = 5092 Characterstic impedance Let inductor which needs to be added in parallel has impedance of (*x) ohm. This inductor is added such that, resultant of load impedance and parallel inductor has 50 ohm as real part. There will be imaginary part present but it will be cancelled by series capacitor. ZL ind = _ind ix-2 2 + 1-2 i-x-(100-1-100) indi-x + 100-1-100 100-x-i + 100-X 2L_ind 100+ i-(x - 100) ZL ind (100-x-i + 100-x)-[100 - i-(x - 100)] [100 + i-(x - 100)]-[100 - i-(x - 100)] 21 ind =- 100-x? - 100xi + 20000-x-i x2 - 200-x+ 20000 We need 2 ind real part to be 50 ohm. hence 100x2 = 50 #2 - 200-x+ 20000 solving for x gives

x= 73.222 now x is inductor impedance at 1GHz 2-1-F-L = x solving for L L=245 = 27.7042 = 11.65mH Adding this inductor ZL_new impedance is now (100 - 1-100) -73.2-i L_new 100 - 1-100 + 73.2-i 2L new= new 7320 + 7320-1 100 - 1-(100 - 73.2) Solving it gives, 7320 + 7320-1 - simplify 100 - 1-(100 - 73.2) + 49.991789696815895147 + 86.5977996387466598991 ZL new = (50 + 86.6-1) 22 Now imaginary part of above calculated impedance is canceled by capacitive reactance. so it shall be same in magnitude but with opposite sign i-86.6= 2-A-1GHz-Cmatch i Solving for Cmatch

1 match -= 1.84pF 2-7-1GHz-86.6