ECE 6130 LUMPED ELEMENT IMPEDANCE MATCHING
Reference: Bowick, RF Circuits, pp. 66-75 (Handout)
Also: Handout sources for SMT caps, inductors
The L-Network:
Four different forms of the L-Shaped Network. LowPass (A,B) and High Pass (C,D):
Purpose of the Elements:
SHUNT Element : Transform large ZL to smaller value with real part equal to the real part of Zs.
SERIES Element: Resonate with (cancel) the imaginary part of the impedance.
Design steps:
Qs = Q of series leg
Qp = Q of parallel leg
Rp = Real part of parallel leg
Rs = Real part of series leg
Xs = Imaginary part of series leg
Xp = Imaginary part of parallel leg
Xs and Xp can be either capacitive or inductive, but must be of opposite type.
Example:
Match a 100-ohm load (Rp) to a 50-ohm line (Rs) at 1 GHz. Do not allow DC power to reach the load.
Since we do not want DC power to reach the circuit, we need one of the high pass configurations such as ( C )above. High Pass configuration (D) would work equally well.
Calculate values of L and C:

PROBLEMS/LIMITATIONS with this method:
MATCHING COMPLEX IMPEDANCES
Two methods can be used to remove the load reactance:
Example:
Same example as above, but let the load have a reactive component:
ZL = 100 -j 25 ohms
Design the matching network ignoring the reactive component, EXACTLY like it was done above.
The XL = -25ohms is equivalent to a capacitor in series with the load resistance RL. The value of CL = 1/(w XL) = 6.4 pF
This gives either of the two circuits shown below:
ABSORPTION OF CL:
CL can now be "absorbed" in the HP-D configuration.
The combination of Cmatch and CL should equal C from the original matching network design.
That is :
C = Ctotal = 3.18pF = Cmatch in series with CL = Cmatch CL / (CL + Cmatch)
OR:
Cmatch = C CL / (CL - C) = (3.18pF)(6.4pF) / (6.4-3.18pF) = 6.32 pF
LIMITATION:
This only works if CL is greater than Ctotal.
Your book has a similar example where the load is modeled with a parallel C,R configuration.
TO CONVERT FROM SERIES TO PARALLEL LOAD ADMITTANCE:
Write the impedance of the above configurations:
Zp = (Rp) (1/jw Cp) / (Rp+1/jw Cp) = Rp / (1+ jw Cp Rp)
Zs = Rs + 1/ jw Cs
Equate: Zs = Zp
This gives two equations and two unknowns.
The unknowns are either Rs,Cs (if you have Rp,Cp) OR Rp,Cp (if you have Rs,Cs)
Solving gives
OR
For our case where Rs=100 and Xs = -25 ohms, we get Rp=106.25 ohms and Xp = -425 ohms.
This is a shunt capacitance Xp = - 1/(w C) where C = 0.374 pF
RESONANCE of CL:
If you cannot absorb CL, you may be able to resonate its effect away. This is done by canceling it out with an inductance. This is easier when the C and L are in parallel.
Then Lp = 1/(w 2Cp)=67.72H
The input impedance that needs to be matched is now ZL = Rp (which is different than the original RL).
Proceed with normal matching as in the first example. See also EXAMPLE 4.3 in the handout.
