Smith Chart Part 2

Sheet 1 of 4

Smith Chart Tutorial Part 2 - Transmission Line Matching Solution of matching problems on a Smith Chart

Γ

Zo

Γ =

ZL

Z L − Zo Z L + Zo

If ZL = Zo then all the power is transmitted to the load. If ZL ≠ Zo then some power will be reflected ie mismatched. Matching add element to T.L to make Γ → 0.

jXm

Matching reactance

Matching susceptance

a

ZL

1 or YL ZL

b jBm

L1 Series Matching

L1 Shunt,Stub,parallel matching

Sheet 2 of 4

(a) Series Matching (1) Plot ZL = ZL/Zo on Smith Chart. (2) Draw VSWR circle through ZL. (3) Transform ZL to ZL’ by moving BACKWARDS from ZL to the r=1 circle. (4) Here ZL’ = 1 + jXL’ ∴read XL’ from Smith Chart. (5) Add matching element Xm = -XL’ at this point. The total impedance = ZL = 1 + jXL’ + jXm . = 1 + jXL’- jXL’ = 1 + j0 (centre of Smith Chart - matched).

Sheet 3 of 4

-jXL’

ZL

Z(in) = 1+j0

BACKWARDS by L/λg takes us to Zin.

L L/λg

XL’

ZL’ = 1+jXL’ ZL/Zo

Vmin

VSWR r=1

Series Stubs

XL = Inductive ZL/Zo

ZL’ = 1+jXL’

Short circuit

Open circuit

r=1

XL = Capacitive

Sheet 4 of 4

(b) Shunt Matching (1) Plot ZL. (2) Transform to YL (diametrically opposite point). (3) Proceed as for the series matching case, but using the Smith Chart as an admittance diagram.ie ⇒ g = 1 circle, read off BL’ , add Bm = -BL’ at distance L. NOTE +ve Susceptance is a shunt capacitance and -ve is a shunt inductance.

L/λg

ZL’ = 1+jXL’

YL/Yo

Vmin

B L’

VSWR r=1 ZL/Zo

Shunt Stubs

BL = capacitive YL/Yo

YL’ = 1+jBL’

Open circuit

Short circuit ZL/Zo

r=1

BL = Inductive