IMPEDANCE Matching. LOADED Q The Q of a resonant circuit was defined to be equal to the ratio of the...

IMPEDANCEMatching

LOADED Q

The Q of a resonant circuit was defined t o be equal to the ratio of the center frequ

- ency of the circuit to its 3 dB bandwidth

The loaded Q of a resonant circuit is dep endent 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 load ed 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

- 100This is done by forcing the ohm 1 1 1 111 11 0 0 1 111 11 11

111 1111 111 11 1111111-1 1111111 1 etwork. But how?

To maximum power transfer

First Step.

Simple place -j - 333 ohm capa aaaaaa aaa a a a1 0 0

- 0 ohm load resistor So we have

Impledance..

aaaaaaa aa aaaa aa apparent 1series00−j - 300 ohm impedance aaa a load,

To match the impledance

A ll we must do to complete the imped ance match to - the 100 ohm source is

to add an equal and opposite (+j 300 ohm) reactance in series

Summary T he function of the shunt component of

the - impedance matching network is to tr ansform a larger impedance down to a s

maller value with a real part equal to the real part of the other terminating impeda - nce (in our case, the 100 ohm source).

The series - impedance matching elemen t then resonates with or cancels any reac

tive component present, thus leaving thesource driving an apparently equal load f

or optimum power transfer.

For this condition:S P

s pS P

X RQ Q

R X

P PS S

P P

R ( jX )R jX

R jX

P PS

P P

R XR

R X

2

2 2P P

SP P

R XX

R X

2

2 2

PS P

S

RQ Q

R 2 21 1

Equation for design of the impedance-matching

The quantities Xp and Xs may be either ca pacitive or inductive reactance but each

must be of the opposite type. Once Xp is c hosen as a capacitor, for example, Xs mus

t be an inductor, and vice versa.

Example 1

- Design a circuit to match a 100 ohm s ource to a - 1000 ohm load at 100 MHz.

Assume that a DC voltage must also b e transferred from the source to the lo

ad.aaa aaaa aaa a a a path between the s

ource and load dict ates the need for a

n inductor in the se ries leg,

Solution

Solution (con’t)

Final circuit

DEALINGWITH COMPLEX LOADS Real world input/output

impledance Transmission lines, mixers, antennas, transistor and most other sources

T wo B asic A pproaches in Handli ng Complex Impedances

Absorption To actually absorb any stray reactances i

nto - the impedance matching network its elf.

This can be done through prudent place ment of each matching element such tha

t element capacitors are placed in parall el with stray capacitances, and element i

nductors are placed in series with any str ay inductances.

T wo B asic A pproaches in Handli ng Complex Impedances

Resonance To resonate any stray reactance wit

h an equal and opposite reactance a t the frequency of interest.

Once this is done the matching netw ork design can proceed as shown for two pure resistances in Example 1.

Example 2

Use t he absor pt i on appr oach t o mat ch t he sour ce and l oad shown below 100(at MHz).

Solution

The first step in the design process is to totally ignore the reactances and si

- mply match the 100 ohm real part of t - he source to the 1000 ohm real part o

f the load (at 100 MHz) Goal

Example 3

Design an impedance matching netw ork that will block the flow of DC from the source to the load in Fig. The freq

uency of operation is 7 5 MHz. Try t he r esonant appr oach.

Solution

The need to block the flow of DC from the source to the load dictates the us

e of the matching network

- first, let’s get rid of the stray 4 0 pF capacitor by resonating it with a shun

t inductor at 7 5 MHz.

Now that we have eliminated the stra y capacitance, we can proceed with m

-atching the network between the 50 o -hm load and the apparent 6 0 0 oh

ml oad

- THREE ELEMENTMATCHING

- 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 networks that are both configured to match the load and the source to an invisible or “virtu

al” resistance located at the junction between the two networks.

The significance of the negative signs for −Xs 1 and −Xs 2 is symbolic.

They are used merely to indicate that the Xs values are the opposite type of reactance fro

m Xp aaa 1 Xp 2, respectively. Thus, if Xp aa a aaaaaaaaaa 1 Xs 1must b

e an inductor, and vice versa. Similarly, if Xp 2 is an inductor, Xs 2must be a capacito

r, and vice versa. They do not indicate nega tive reactances (capacitors).

Now, we have

Example 4

Using Figure below as a reference, de sign four different Pi networks to matc

- - h a 100 ohm source to a 1000 ohm lo ad. Each network must have a loaded

Q of 1 5 .

Solution

From

The Q for the other L network is now defi ned by the ratio of Rs to R

Notice here that the source resistor is no w considered to be in the shunt leg of th e 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, therefore, is not shown. Reactances −Xs 1 and −Xs 2 ar e nowi n ser i es and can si mpl y be ad ded together to form a single componen

t.

The only constraint is that Xp 1 and Xs 1 are of opposite types, and Xp 2 and

Xs 2 are of opposite types. Therefore, to perform the transformatio

- n from the dual L to the Pi network, the t wo series components are merely added

if they are alike, and subtracted if the re actances are of opposite type.

Which one to choose? D epend 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 exactlythesameas f or t he Pi net wor k except t hat wi

th the T, you match the load and the sou - rce, through two L type networks, to a vi rtual resistance that is larger than either

the load or source resistance. This mean - s that the two L type networks will then

have their shunt legs connected together

Q value of T

- since we have reversed or “flip flopped” the L sections to produce the T network, we must also make sure that we redefin

e t he Q formula to account for the new r esistor placement, in relation to those L

networks.

Example 5

Using Figure below as a reference, de sign four different networks to match

- - a 10 ohm source to a 50 ohm load. Ea ch network is to have a loaded Q of 10

Solution

we can find the virtual resistance we nee d for the match

From previously,

Now, for the L network on the load end, t he Q is defined by the virtual resistor an

d 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 C onvert any impedance (Z ) to an admittance (Y ), and vic

e versa. This can be accomplished by simply flipping the Smith C

hart 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 along a circle of constant resistance,

while a shunt capacitor causes rotation clockwise alon g a circle of constant admittance.

Amittance VS Impledance

an admittance is simply the inverse ofan i mpedance

where the admittance (Y) contains both a real and an imaginary part, similar to the impedance (Z).

Circuit representation for admittance.

Example

Impedance Z =1+j1.

Notice that t he two

points are lo cated at exa

ctly the sam e distance ( d) from the

center of th e chart but i n opposite d

irections (1 80◦) from e

ach other.

I mpedance and A dmittance coordinates,

Admittance Manipulation on theChart we begin with a

n admittance of Y 0= .2−j0

5 mho and add a shunt capaci

t or wi t h a suscept ance (recipr

ocal of reactanc e) of +j a08

ho.