Question

Design an "L" type network using discrete components (inductors and / or capacitors) to couple a load of ZL = 150-j200 Ohms to a system with Zo = 100 Ohms. Choose 1 GHz as the center frequency.

a) Show the entire procedure step by step using the Smith Chart and clearly marking where you got the results from. Calculate the values ​​of the inductance and / or corresponding capacitances.

b) Simulate with ADS your circuit and clearly show the frequency for which you designed the coupling network. It must be clearly marked in the ADS plots.

c) Now use commercial components (standard values) of capacitors and inductors closer to your results of part b, and simulate again. Indicate if there is any difference between the two simulations. Justify your answer referring to the results obtained.

Answer 1

To Design an "L" type network using discrete components to couple a load of Z_L and a system with Z₀ which can be shown below as :

Where, Z_L = R_L + j X_L                                                                  { eq.1 }

Comparing with Z_L = (150 ) + j (200 )

Then, we get

R_L = 150    &    X_L = 200

(a) In this case, we have

R_L > Z₀

From an above figure, we get

Z_in = j X + [(j B) + 1 / (R_L + j X_L)]^-1                                                            { eq.2 }

Where, B = X_L(R_L / Z₀) R_L (R_L - Z₀) + X_L² / (R_L² + X_L²)                                      { eq.3 }

X = (1/B) + (X_L Z₀ / R_L)-(Z₀ / B R_L)                                                                                   { eq.4 }

Since R_L > Z₀, the argument is positive in the second square root of equation (3) and the value of 'B' must be real.

There are two possible solutions for 'B' in equation (3).

The value of X in equation (4) also has two possible solutions which depending on which 'B' from equation (3) is used.

Therefore, the value of inductance will be given by -

using a formula, we have

X_L = L = (2f) L

(200 ) = [2 (3.14) (1 x 10⁹ Hz)] L

L = [(200 ) / (6.28 x 10⁹ rad/s)]

L = 3.18 x 10^-8 H

The value of capacitance will be given by -

using a formula, we have

= 1 / L C

squaring on both sides & we get

4² f² = 1 / L C

C = 1 / [4 (3.14)² (1 x 10⁹ Hz)² (3.18 x 10^-8 H)]

C = 7.97 x 10^-13 F

C 0.8 pF