IMPEDANCE

Matching
LOADED Q
 The Q of a resonant circuit was defined to be equal to the ratio of the c
enter frequency of the circuit to its 3-dB bandwidth

 The loaded Q of a resonant circuit is dependent upon three main factors

.
1. The source resistance (Rs).
2. The load resistance (RL).
3. The component Q as defined previous.
The effect of Rs and RL on loaded
Q
Rs = 50

Rs = 1000
Effect of Q vs. Xp.
Maximum Power Transfer
 In DC circuits, maximum power
will be transferred from a source
to its load if the load resistance
equals the source resistance
Maximum Power Transfer
 The source (Zs), with a series reactive component of +jX (an inductor), is
driving its complex conjugate load impedance consisting of a −jX
reactance (capacitor) in series with RL. The +jX component of the source
and the−jX component of the load are in series and, thus,cancel each
other, leaving only Rs and RL, which are equal by definition.
 Since Rs and RL are equal, maximum power transfer
The L Network
Simple Black Box Analysis

 Source 100-ohm
 Load 1000-ohm
 So, in this situation
 The available power from source would be
lost about 4.8 dB

To maximum power transfer

This is done by forcing the 100-ohm source to see 100 ohms when it lo
oks into the impedance-matching network. But how?
First Step.

 Simple place -j333-ohm capacitor is placed across the 1000-oh

m load resistor
 So we have
Impledance..

 Nowthat we have an apparent series 100−j300-ohm impedance

for a load,
To match the impledance

 All we must do to complete the impedance match to the 100-ohm sourc

e is to add an equal and opposite (+j300 ohm) reactance in series
Summary
 The function of the shunt component of the impedance-matchi
ng network is to transform a larger impedance down to a small
er value with a real part equal to the real part of the other termi
nating impedance (in our case, the 100-ohm source).
 The series impedance-matching element then resonates with or
cancels any reactive component present, thus leaving the sourc
e driving an apparently equal load for optimum power transfer .
For this condition:
XS R P
Qs  Q p  
R S XP
R P ( jX P )
R S  jXS 
R P  jX P

R P X2P R 2P X P
RS  2 XS  2
R P  X2P R P  X2P
RP
 1 QS  1 Q P
2 2

RS
Equation for design of the impedance-
matching

The quantities Xp and Xs may be either capacitive or inductive reactance but ea

ch must be of the opposite type. Once Xp is chosen as a capacitor, for example,
Xs must be an inductor, and vice versa.
Example 1
 Design a circuit to match a 100-ohm source to a 1000-ohm load at 100
MHz. Assume that a DC voltage must also be transferred from the sour
ce to the load.

The need for a DC path between the

source and load dictates the need for
an inductor in the series leg,
Solution
Solution (con’t)

Final circuit
DEALING WITH COMPLEX LOADS

 Real world input/output impledance

Transmission lines, mixers, antennas, transistor and most other sources
Two Basic Approaches in Handling Co
mplex Impedances
 Absorption
 To actually absorb any stray reactances into the impedance-matchin

g network itself.
 This can be done through prudent placement of each matching elem

ent such that element capacitors are placed in parallel with stray cap
acitances, and element inductors are placed in series with any stray i
nductances.
Two Basic Approaches in Handling Co
mplex Impedances
 Resonance
 To resonate any stray reactance with an equal and opposite reactanc

e at the frequency of interest.

 Once this is done the matching network design can proceed as show

n for two pure resistances in Example 1.

Example 2

 Use the absorption approach to match the source and load shown below
(at 100 MHz).
Solution

 The first step in the design process is to totally ignore the reactances an
d simply match the 100-ohm real part of the source to the 1000-ohm re
al part of the load (at 100 MHz)
 Goal
Example 3

 Design an impedance matching network that will block the flow of DC

from the source to the load in Fig. The frequency of operation is 75 M
Hz. Try the resonant approach.
Solution

 The need to block the flow of DC from the source to the load dictates t
he use of the matching network
 first, let’s get rid of the stray 40-pF capacitor by resonating it with a sh
unt inductor at 75 MHz.
 Now that we have eliminated the stray capacitance, we can proceed wit
h matching the network between the 50-ohm load and the apparent 600
-ohm load
THREE-ELEMENT MATCHING

The three-element Pi network. The three-element T network

The Pi Network

 The Pi network can best be described as two “back-to-back” L network

s that are both configured to match the load and the source to an invisib
le or “virtual” resistance located at the junction between the two networ
ks.
 The significance of the negative signs for −Xs1 and −Xs2 is symboli
c.
 They are used merely to indicate that the Xs values are the opposite typ
e of reactance from Xp1 and Xp2, respectively.
 Thus, if Xp1 is a capacitor, Xs1 must be an inductor, and vice versa.

Similarly, if Xp2 is an inductor, Xs2 must be a capacitor, and vice ve

rsa. They do not indicate negative reactances (capacitors).
 Now, we have
Example 4

 Using Figure below as a reference, design four different Pi networks to

match a 100-ohm source to a 1000-ohm load. Each network must have
a loaded Q of 15.
Solution

 From


 The Q for the other L network is now defined by the ratio of Rs to R

 Notice here that the source resistor is now considered to be in the shunt leg
of the L network. Therefore, Rs is defined as Rp, and
 Now the complete network design
 Remember that the virtual resistor (R) is not really in the circuit and, therefo
re, is not shown. Reactances −Xs1 and −Xs2 are now in series and can si
mply be added together to form a single component.
 The only constraint is that Xp1 and Xs1 are of opposite types, and Xp2 and
Xs2 are of opposite types.
 Therefore, to perform the transformation from the dual-L to the Pi network,
the two series components are merely added if they are alike, and subtracted
if the reactances are of opposite type.
 Which one to choose?
 Depend on any number of factors including:
1. The elimination of stray reactances.
2. The need for harmonic filtering.
3. The need to pass or block DC voltage.
The T network

 The design of the 3-element T network is exactly the same as for the Pi net
work except that with the T, you match the load and the source, through two
L-type networks, to a virtual resistance that is larger than either the load or
source resistance. This means that the two L-type networks will then have t
heir shunt legs connected together
Q value of T

 since we have reversed or “flip-flopped” the L sections to produce the T net

work, we must also make sure that we redefine the Q formula to account for
the new resistor placement, in relation to those L networks.
Example 5

 Using Figure below as a reference, design four different networks to m

atch a 10-ohm source to a 50-ohm load. Each network is to have a load
ed Q of 10.
Solution

 we can find the virtual resistance we need for the match

 From previously,

 Now, for the L network on the load end, the Q is defined by the virtual resis
tor and the load resistor. Thus,
Solution
THE SMITH CHART

 The chart was originally conceived

back in the 1930s by a Bell
Laboratories engineer named
Phillip Smith, who wanted an
easier method of solving the
tedious repetitive equations that
often appear in RF theory
Smith Chart Construction

 Step 1: The reflection coefficient of a load impedance when given a

source impedance can be found by the formula:

 In normalized form, this equation becomes:

 where Zo is a complex impedance of the form R+jX

 The polar form of the reflection coefficient can also be represented in
rectangular coordinates:

 So, we have

 If we draw the family of curves we have:

Two families of smith chart
Combined Together
Basic Smith Chart Tips: Important
 All the circles have one same, unique intersecting point at the
coordinate (1, 0).
 The zero circle where there is no resistance (R=0) is the largest one.
 The infinite resistor circle is reduced to one point at (1, 0).
 There should be no negative resistance. If one (or more) should occur,
you will be faced with the possibility of oscillatory conditions.
 Another resistance value can be chosen by simply selecting another
circle corresponding to the new value.
Plotting Impedance Values

1+j1

1-j1
Let try to read this
Example

 If Z =0.5+j0.7
ohm.
 Series capacitive
reactance of
–j0.7
Add Inductance
Remember

 Series capacitive reactance move downward

 Series inductor move upward
Conversion of Impedance to
Admittance
 Convert any impedance (Z) to an admittance (Y), and vice versa.
 This can be accomplished by simply flipping the Smith Chart over.
 it can be extremely useful in designing match networks with components like
series or shunt inductors and capacitors
 A shunt inductor causes rotation counter-clockwise along a circle of constant
admittance.
 So, a series capacitor, added to a load, causes rotation counter-clockwise alo
ng a circle of constant resistance, while a shunt capacitor causes rotation cloc
kwise along a circle of constant admittance.
Amittance VS Impledance

 an admittance is simply the inverse of an impedance

 where the admittance (Y) contains both a real and an imaginary part,
similar to the impedance (Z).
Circuit representation for admittanc
e.
Example

 Impedance Z =1+j1.
Notice that the two
points are located at ex
actly the same distance
(d) from the center of t
he chart but in opposite
directions (180◦) from
each other.
Impedance and Admittance coordinates,
Admittance Manipulation on the Chart
we begin with an admittance o
f Y =0.2−j0.5 mho and add a
shunt capacitor with a suscept
ance (reciprocal of reactance)
of +j0.8 mho.