MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 1 / 23
Impedance matching and
Smith Chart : Basic
Considerations
Γi
0
0.5
1
Γr
1
By K-C Chan & A. Harter
1. Introduction
2
2. Power transfer with RF signals : reminders
2
3. Impedance Smith Chart : reminders
4
4. Admittance Smith Chart
13
5. Equivalent impedance resolution
16
6. Matching impedances step by step
18
7. Matching illustration on Maxim mixer MAX2680
20
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 2 / 23
1. Introduction
a. When dealing with practical implementation of radio applications, there are always
some tasks that appear nightmare for many of us : need to match the difference
impedances of the interconnected blocks : antenna to LNA, RFOUT to antenna, LNA
output to mixer input, etc… The matching task is required mainly for a proper
transfert of signal and energy from a "source" to a "load". The load can be then a
source for a next block, etc…
b. At high radio frequencies, the spurious (wires inductances, interlayers capacitances,
conductors resistances, etc… are important non-predictible contributors for the
matching network building. Above few tenth of MHz, theortical calculations and
simulations are often not enough anymore. And in-situ RF lab measurements along
with tuning work have to be considered for obtaining final correct values. The
computations values are required to set-up the type of the structure and target
components values.
c. There are many possible ways to "do impedance" matching :
i. Computer simulations : complex to use since such simulators are dedicated
for many stuffs and not only for impedance matching ! One has to be
familiarized with many input data to be entered on right formats and be expert
to find the useful data among the tons of results coming out ! In addition, one
has to recognize that circuit simulators are usually not pre-installed on the
notebook.
ii. Manual computations : tedious due to the length ("kilometric") of equations
and the complex nature of the numbers to be manipulated.
iii. Feelings : these can be acquired only one has already many years devoted
only for the RF domains. In short term, this is for super-specialist !
iv. Smith Chart. This present article concentrates on the Smith Chart method : the
main target here is to refresh its construction background and summarize
practical ways to use it by some practical illustration such as finding matching
network components values. Of course matching for maximum power transfer
is not the only stuff one can do with Smith charts : optimization for best noise
figures, quality factors impact, stability analysis, etc,… can be effectivelly
covered. These aspects will be handled in an other arcticle (ref : Alphonse
Harter).
2. Power transfer with RF signals : reminders
a. Before to efficiently introduce the Smith Chart utilities, it is recommended to have a
short refreshment on wave propagation phenomenons when Ics wires between ICs on
a design are in RF conditions (above 100 MHz). This can be true at many
cicumstancies such as in RS485 line, between a PA and an antenna, between a LNA
to a donwconverter mixer, etc…
b. Maximum power transfer
It is well known that for getting a maximum power transfer from a source to a load,
the following condition must happen :
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 3 / 23
Source impedance = complex conjugate of load impedance, or :
Rs + j.Xs = RL – j.XL
Zs
Rs
E
Xs
XL
ZL
RL
In this condition, the energy transferred from the source to the load is
maximalized. In addition to an efficient power tranfer, this condition is required if one
wants to avoid reflection of energy from the load back to the source; this is
particularly true for high frequency environments like video lines applications or RF
& microwave networks.
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 4 / 23
3. Impedance Smith Chart : reminders
a. It appears as a circular plot with a lot of interlaced circles on it. It has been "invinted"
by a certain Mr Phillip Smith at the Bell Lab in the 1930's. When correctly used,
matching impedances with apparent complicate structures can be made without any
computation. The only effort to do is to be able to read and follow values along
circles.
b. The Smith chart is a polar plot of the complex reflection coefficient (also called
gamma and symbolized by Γ), or more mathematically defined as : the 1- port
scattering parameter s or s11
c. How is the Smith chart constructed ? We start from the load where the impedance
must be matched. Instead of considering its impedance directly, one expresses its
reflection coefficient ΓL. which is an other manner to characterize a load (such as
admittances, gain, transconductances, etc… The ΓL is more useful when dealing with
RF frequencies.
So : we know the reflection coefficient is defined as the ratio between the reflected
voltage wave and the incident voltage wave :
ZO
ΓL =
Vrefl
Vinc
Vinc
Vrefl
ZL
The amount of reflected signal from the load is depending on the degree of mismatch
between the source impedance and the load impedance. Its expression has been
defined as follow :
ΓL =
Vrefl Z L − Z O
=
= Γr + j. ΓI
Vinc Z L + Z O
(equ 2.1)
Since the impedances are complex numbers, the reflection coefficient is a complex
number too.
In order to reduce the number of unknown parameters, it is useful to freeze the ones
that appear often common in the application. Here Zo (the characteristic impedance) is
often a constant and a real industry normalized value : 50 Ω, 75 Ω, 100 Ω, 600 Ω,
etc…
We can then define a normalized load impedance :
z = ZL/Zo = (R + jX) / Zo = r + jx
(equ 2.2)
With this simplification, we can rewrite the reflection coefficient formula :
ΓL = Γr + j.Γi =
Z L − ZO
( Z L − Z O ) / Z O z − 1 r + j.x − 1
=
=
=
Z L + ZO
( Z L + Z O ) / Z O z + 1 r + j. x + 1
(equ 2.3)
Here one can see the direct relationship between load impedance and its
reflection coefficient. Unfortunnally the complex nature of the relation is not
practically useful. The Smith chart is a kind of graphical representation of the
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 5 / 23
above equation. In order to build the chart, the equation must be re-written in
order to extract standard geometrical figures (likes circles or stray lines).
By reversing the (equ 2.3) :
z = r + jx =
1 + ΓL 1 + Γr + jΓi
=
1 − ΓL 1 − Γr − jΓi
(1 + Γr + jΓi ).(1 − Γr + jΓi ) 1 − Γr2 − Γi2 + j.2.Γi
r + jx =
=
(1 − Γr − jΓi ).(1 − Γr + jΓi ) 1 + Γr2 − 2.Γr + Γi2
(equ 2.4)
(equ 2.5)
By equalling the real parts and the imaginary parts of (equ 2.5) we obtain 2
independent new relations :
r=
1 − Γr2 − Γi2
1 + Γr2 − 2.Γr + Γi2
(equ 2.6)
x=
2.Γi
1 + Γ − 2.Γr + Γi2
(equ 2.7)
2
r
By developing (equ 2.6) :
r + r.Γr2 − 2.r.Γr + r.Γi2 = 1 − Γr2 − Γi2
(equ 2.8)
Γr2 + r.Γr2 − 2.r.Γr + r.Γi2 + Γi2 = 1 − r
(equ 2.9)
(1 + r ).Γ − 2.r.Γr + (r + 1).Γ = 1 − r
2.r
1− r
Γr2 −
.Γr + Γi2 =
r +1
1+ r
2
2.r
r
r2
1− r
2
Γr2 −
.Γr +
+
Γ
−
=
i
2
2
r +1
1+ r
(r + 1)
(r + 1)
(equ 2.10)
2
r
2
i
(equ 2.11)
(equ 2.12)
(Γr −
r 2
1− r
r2
1
) + Γi2 =
+
=
2
r +1
1 + r (1 + r )
(1 + r ) 2
(equ 2.13)
(Γr −
r 2
1 2
) + Γi2 = (
)
r +1
1+ r
(equ 2.14)
Parametric equation of the type (x-a)² + (y+b)²=R²
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 6 / 23
This above relation is a parametric equation, in the complex plan (Γr,Γi), of a circle centred at
coordinate (r/r+1 , 0) and having a radius of 1/(1+r).
The points situated on a circle are all the impedances
characterized by a same real impedance part value.
For example, the circle R=1 is centred at coordinate
(0.5;0) and having a radius of 0.5. It include the point
(0,0) which is the reflection zero point (the load is
matched with the characteristic impedance). A shortcircuit as load gives according to the formula a circle
centred at coordinate (0.0) and having a radius of 1.
Γi
r =0 (short)
r =1
0.5
0
Γr
1
r =∞
(open)
For a open-circuit load, the circle is degenerated as
a single point (circle centred at coordinate 1,0 and
having a radius of 0) : this correspond to a
maximum reflection coefficient of 1 : all the
incident wave is totally reflected)
Some particular cases to be noticed :
•
•
•
•
•
All the circles have one unique same intersecting point at the coordinate (1,0)
The 0 ohms circle (r=0; no resistive part) is the biggest one
The infinite resistor circle is reduced to one point at (1,0)
There is no negative resistors (if r <0, then we are in face of amplifier or oscillators)
Changing a resistor is just to select an other circle corresponding to the new r value
By developing (equ 2.7) :
x + x.Γr2 − 2.x.Γr + x.Γi2 = 2.Γi
(equ 2.15)
1 + Γ − 2.Γr + .Γ = 2.Γi / x
2
Γr2 − 2.Γr + 1 + Γi2 − Γi = 0
x
2
1
1
Γr2 − 2.Γr + 1 + Γi2 − Γi + 2 − 2 = 0
x
x
x
1
1
(Γr − 1) 2 + (Γi − ) 2 = 2
x
x
(equ 2.16)
2
r
2
i
(equ 2.17)
(equ 2.18)
(equ 2.19)
Again, this is also a parametric equation of the type (x-a)² + (y+b)²=R²
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 7 / 23
This above relation is a parametric equation, in the complex plan (Γr,Γi), of a circle centred at
coordinate (1 , 1/x) and having a radius of 1/x.
The points situated on a circle are
all the impedances characterized
by a same imaginary impedance
part value x. For example, the
circle x=1 is centred at coordinate
(1;1) and having a radius of 1. All
circles (constant x) include the
point (1,0)
Γi
0
0.5
1
Γr
In difference with the real part circles,
x can be positive or negative : this
explain the duplicate mirror circles at
the bottom side of the complex plan.
All the circles centres are placed on
the vertical axe intersecting the point
1.
1
•
•
•
•
•
•
Some particularities to be noticed :
The zero-reactance circle (thus a pure resistive load) is just the horizontal axis of the
complex plane.
The infinite reactance is a degenerated circle to one point situated at (1,0).
All constant reactance circles have the same unique intersecting point at (1,0)
Positive reactance’s (inductors) are on circles on the half upper part of the chart, while
negative reactance’s (capacitors) are on the half bottom part of the chart
Changing a reactance is just then to take an other circle of the family corresponding to the
new value x.
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 8 / 23
Now, to complete our Smith Chart, we superpose the 2 circles families and one can see that
every circle of a family does intersect ALL the circles of the other family. Thus knowing the
impedance in the for of r + jx, the plot gives immediately the corresponding reflection coefficient : one
has only to find the intersection point of 2 circles : one circle corresponding to the value r and one
circle corresponding to value x. The opposite operation is also trivial : knowing the reflection
coefficient, find the 2 circles passing both on that point and read the values r and x corresponding to
the found circles.
Γi
0
0.5
1
Γr
1
Now we can do following tasks (the list is not complete) :
•
•
•
•
•
•
Situate any impedance as a spot on the Smith Chart
Find the reflection coefficient Γ for a given impedance (and knowing the characteristic
impedance)
Find the impedance when knowing Γ
Convert impedance to admittance and vice-versa
Find equivalent impedance
Find components values for a wanted reflection coefficient (in particular find elements of a
matching network)
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 9 / 23
Since the Smith Chart resolution technique is basically a graphic method, the precision of the solutions
depends a lot on the graph definitions. Here below we joined a typical Smith Chart used by all the RF
engineers :
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 10 / 23
•
Example 1 : situate on the Smith chart and by considering a characteristic impedance of 50
ohms, the here below listed impedances.
Z3 = j200 ohms
Z4 = 150 ohms
Z1 = 100 + j50 ohms Z2 = 75 –j100 ohms
Z5 = open-circuit
Z6 = short-circuit
Z7 = 50 ohms
Z8 = 184 –j900 ohms
First thing to do : normalize first the impedance (division by the characterise impedance) :
z2 = 1.5 –j2
z3 = j4
z4 = 3
z1 = 2 + j
z6 = 0
z7 = 1
z8 = 3.68 –j18
z5 = ∞
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 11 / 23
•
Example 2 : Direct extraction of reflection coefficient Γ on the Smith Chart.
Solution : Once the impedance point well situated on the plot (intersection point of a constant
resistor circle and of a constant reactance circle), just read the rectangular coordinates :
projection on the horizontal axis will give Γr (real part of the reflection coefficient) and the
projection on the vertical axis will give Γi (imaginary part of the reflection coefficient).
Constant
resistor r
Γi
Constant
reactance x
Γr
•
0
1
Example 3 : Take the 8 cases handled in example 1 and extract their corresponding Γ directly
on the Smith Chart.
Γ1 = 0.4 + 0.2j
Γ2 = 0.51 - 0.4j
Γ3 = 0.875 + 0.48j
Γ4 = 0.5
Γ5 = 1
Γ6 = -1
Γ7 = 0
Γ8 = 0.96 - 0.1j
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 12 / 23
Working with admittance
The Smith Chart is built by considering impedance (resistor and reactance). Adding elements
in series are trivial : they correspond to moves along circles. Unfortunally, summing elements
in parallel needs some additional consideration in the Smith Chart.
We know that by definition, Y = 1/Z and Z = 1/Y. The admittance is expressed in Mho or Ω-1
(in the older time as Siemens or S). Of course, since Z is complex, Y is also complex.
Y = G + j.B where G is called “conductance” and B the “susceptance” of the element.
One has not to conclude that G=1/R and B=1/X !!! which are wrong !
By normalising by y = Y/Yo :
y = g + j.b
What happen with the reflection coefficient ?
Γ=
Z L − Z O 1 YL − 1 YO YO − YL 1 − y
=
=
=
Z L + Z O 1 YL + 1 YO YO + YL 1 + y
One can see the expression for Γ is opposite in sign compared to its expression with z :
Γ(y) = -Γ
Γ(z)
Thus, if we know z, by inverting the signs of Γ, we found a point situated at the same distance
to (0,0) but in the opposite direction. The same result can be obtained by making a rotation of
180° angle around the centre point.
Constant
conductance g
circle
z
Γi
180°
-Γr
Γr
-Γi
y
Constance
susceptance
b circle
Of course, while Z and 1/Z do represent exactly a same component, the new point appears as a
complete different impedance (different point in the Smith chart, different reflection value,
etc…). That is because the plot is an impedance plot and the new point is in fact an
admittance. The value read on the chart has then to be read as mho.
This method is OK for making conversion, but not enough to make circuit resolution when
dealing with elements in parallel.
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 13 / 23
4. Admittance Smith Chart
Above, we have seen that every point in the impedance Smith Chart can be converted into an
admittance counterpart by making a rotation of 180° around the origin of the Γ complex plane.
Thus, an Admittance Smith Chart can be obtained by rotating the whole Impedance Smith
Chart by 180°. This is great since one has not to build an other chart : the same can be used !
Automatically, the intersecting point of all the circles (constant conductance’s and constant
susceptances) is at the point (-1, 0). With that plot, adding element in parallel becomes trivial.
One can also mathematically demonstrate the Admittance Smith chart construction :
ΓL = Γr + j.Γi = =
1 − y 1 − g − jb
=
1 + y 1 + g + jb
By reversing the (equ 3.1) :
(equ 3.1)
y = g + jb =
1 − ΓL 1 − Γr − jΓi
=
1 + ΓL 1 + Γr + jΓi
(1 − Γr − jΓi ).(1 + Γr − jΓi ) 1 − Γr2 − Γi2 − j.2.Γi
g + jb =
=
(1 + Γr + jΓi ).(1 + Γr − jΓi ) 1 + Γr2 + 2.Γr + Γi2
(equ 3.2)
(equ 3.3)
By equalling the real parts and the imaginary parts of (equ 3.3) we obtain 2
independent new relations :
g=
1 − Γr2 − Γi2
1 + Γr2 + 2.Γr + Γi2
(equ 3.4)
b=
− 2.Γi
1 + Γ + 2.Γr + Γi2
(equ 3.5)
2
r
By developing (equ 3.4) :
g + g.Γr2 + 2.g.Γr + g .Γi2 = 1 − Γr2 − Γi2
(equ 3.6)
Γr2 + g.Γr2 + 2.g.Γr + g .Γi2 + Γi2 = 1 − g
(equ 3.7)
(1 + g ).Γ + 2.g .Γr + ( g + 1).Γ = 1 − g
2.g
1− g
Γr2 +
.Γr + Γi2 =
g +1
1+ g
2
2.g
g
g2
1− g
2
2
Γr +
.Γr +
+ Γi −
=
2
2
g +1
1+ g
( g + 1)
( g + 1)
(equ 3.8)
2
r
2
i
(equ 3.9)
(equ 3.10)
(Γr +
g 2
1− g
g2
1
) + Γi2 =
+
=
2
g +1
1 + g (1 + g )
(1 + g ) 2
(equ 3.11)
(Γr +
g 2
1 2
) + Γi2 = (
)
g +1
1+ g
(equ 3.12)
Parametric equation of the type (x-a)² + (y+b)²=R²
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 14 / 23
This above relation is a parametric equation, in the complex plan (Γr,Γi), of a circle centred at
coordinate (-g/g+1 , 0) and having a radius of 1/(1+g).
Γi
g =0 (open)
g =∞
(short)
-1
0
The points situated on a circle are all the admittances
characterized by a same real part (conductance)
value. For example, the circle g=1 is centered at
coordinate (-0.5;0) and having a radius of 0.5. It
include the point (0,0) which is the reflection zero
point (the load is matched with the characteristic
admittance). An open-circuit as load gives according
to the formula a circle centred at coordinate (0.0) and
having a radius of 1.
Γr
For a short-circuit load, the circle is degenerated as
a single point (circle centred at coordinate -1,0 and
having a radius of 0) : this corresponds to a
maximum reflection coefficient of -1 : all the
incident wave is totally reflected)
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 15 / 23
By developing (equ 3.5) :
b + b.Γr2 + 2.b.Γr + b.Γi2 = −2.Γi
(equ 3.13)
1 + Γ + 2.Γr + .Γ = −2.Γi / b
2
Γr2 + 2.Γr + 1 + Γi2 + Γi = 0
b
2
1
1
Γr2 + 2.Γr + 1 + Γi2 + Γi + 2 − 2 = 0
b
b
b
1
1
(Γr + 1) 2 + (Γi + ) 2 = 2
b
b
(equ 3.14)
2
r
2
i
(equ 3.15)
(equ 3.16)
(equ 3.17)
Again, this is also a parametric equation of the type (x-a)² + (y+b)²=R²
This above relation is a parametric equation, in the complex plan (Γr,Γi), of a circle centred at
coordinate (-1 , -1/b) and having a radius of 1/b.
Γi
-1
The points situated on a circle
are all the admittances
characterized by a same
imaginary part value b. For
example, the circle b=1 is
centred at coordinate (-1;-1)
and having a radius of 1. All
circles (constant b) include the
point (-1,0)
0
Impedance matching and Schmitt Chart : practical aspects consideration
Γr
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 16 / 23
5. Equivalent impedance resolution
When solving problem where elements in series and in parallel are mixed together, one can
use the same Smith Chart and rotate it at any instant when conversions from z to y or y to z are
previously performed before to rotate the chart.
The best way to expose the method is to do this through an example :
Let’s consider the network here below (the elements are normalized with Zo=50 ohm):
Serie reactance (x) is positive for inductance and negative for capacitor. While susceptance (b)
is positive for capacitance and negative for inductance.
x = 0.9
Z=?
x = -1.4
x=1
b = 1.1
r=1
b = -0.3
The circuit has to be cut in different portions such as :
x = 0.9
x=1
x = -1.4
b = 1.1
b = -0.3
r=1
Z=?
A
B
C
D
Z
•
•
•
•
•
Step 1 : We start from the end, which is here a resistor of 1. Since this is in series with an
inductor of 1 ohm, the spot on the Smith chart comes immediately : intersection circle r=1 and
circle x=1. We mark this point by A
Step 2 : The next element is an element coming in shunt (parallel) : we will have to switch into
the Admittance Smith Chart (rotation by 180° of the whole plane). Before to do this rotation,
first, convert the previous point into admittance : this gives the spot A’.
Step 3 : Rotate the plane by 180° : we are now in the admittance mode. The shunt element can
be added by going along the conductance circle by a distance corresponding to 0.3. This has to
be made in the anti-clock wise direction (negative value). After doing this, we obtain point B.
Step 4 : We have an element to be inserted in series. This means we need to switch to the
impedance Smith Chart. Before to do this, reconvert the previous point into impedance (it was
an admittance). This gives the spot B’.
Step 5 : Rotate the plane by 180° : we are now in the impedance mode. The series element is
added by circulating along the resistance circle by a distance corresponding to 1.4. This has to
be made in the anti-clock wise direction (negative value). After doing this, we obtain point C.
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 17 / 23
•
•
Step 6 : Again, there is a shunt element, same operations as already described above
(conversion into admittance followed by plane rotation of 180°. Then apply move by distance
1.1 in the clock-wise direction, since the value is positive, along the constant conductance
circle). We obtain point D.
Step 7 : Recon version into impedance prior to add the last element (series inductor). We
obtain the asked value : z situated in the intersection of circle resistor 0.2 and circle reactance
0.5 : thus z = 0.2 +j0.5. If the system characteristic impedance is 50 ohms, then Z = 10 + j25
ohms
Impedance matching and Schmitt Chart : practical aspects consideration
MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 18 / 23
6. Matching impedances step by step
This is the reverse operation of finding equivalent impedance of a given network. Here, the
impedances are fixed at the 2 access ends (often the source and the load). The task is to design
the network to be inserted between them in order to insure impedance matching. Apparently, it
looks like as easy as finding equivalent impedance, but the problem is that there may exists
infinite ways (i.e. matching networks) to realize a same matching. Other inputs can be
provided or fixed : filter type structure, quality factor, limited choice of element values, etc…
The approach is to add series and shunt elements on the chart until the desired impedance is
achieved. Graphically, it appears to find the way to link 2 spots on the Smith Chart.
The best method to explain the approach is to illustrate the tasks with the help of a practical
example :
Matching network
ZS = 25 - j15 Ω
Vs
ZL = 100 – j25 Ω
L=?
Z* =
25 + j15 Ω
C=?
Here the problem consists to match a source impedance Zs to a ZL load at the operation
frequency of 60 MHz. The network structure has been fixed as a low pass L type.
An other manner to see the problem is to say how to force the load to appear like an
impedance of value = Zs* (complex conjugate of Zs).
Step 1 : First normalize the different impedance values. If not given, choose the value in the
same range than load/source values (question of readability on the Smith chart). Here we take
Zo = 50 ohms
Thus
zs = 0.5 – j0.3 and z*s = 0.5 + j0.3
zL = 2 – j0.5
Step 2 : Situate the 2 points on the chart. We mark A for zL and D for Z*.
Step 3 : The first element connected to the load is a capacitor in shunt : conversion in
admittance (see previously described methodology). We obtain a point A’.
Step 4 : Detect the arc portion that the next point will appear after the connection of the
capacitor C. Since we don’t know the value of C, we don’t know where to stop. We know the
direction (C in shunt means in the clock-wise direction in the admittance Smith chart). The
point B will be an admittance.
Since the next element is a series element, the point B has to be back converted in impedance.
Thus a point B’ will be obtained. This point B’ has to be situated in the same resistor circle
than D. Graphically, there is only one solution from A’ to D but the intermediate point B (and
hence B’) needs to be localized by “test-and-try” until one falls on the point D.
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MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 19 / 23
Step 5 : after having found the points B, B’, we can then measure the lengths of arc A’-B and
arc B’-D. The first gives the normalised susceptance value of C and the second gives the
normalised reactance value of L.
After apprx 15 minutes , one found the points B and B’ :
The arc A’-B measures b = 0.78 and thus B = 0.78 * Yo = 0.0156 mho
Since ω.C = B, then C = B/ω = B/(2.π.f) = 0.0156/(2.π.60.106) = 41.4 pF
The arc B’-D measures x = 1.2 and thus X = 1.2 * Zo = 60 ohm
Since ω.L = X, then L = X/ω = X/(2.π.f) = 60/(2.π.60.106) = 159 nH
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MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 20 / 23
7. Impedance matching illustration on Maxim mixer MAX2680
Practical application on MAX2680-MAC2681-MAX2682 which are down converters mixers
able to handle input RF signal on the range 400MHz to 2.5GHz into output IF frequencies
from 10 MHz to 500 MHz. The external matching networks on the RF input and IF output
ports determine their operation modes. Their structures and elements are various and quasi
infinite. The applications nature (i.e. Q-factor, needs to pass or filter out DC, range limits of
components, etc…) will fix the best choices.
In their datasheet, a typical matching structure has been proposed with elements values for
some “standardized” frequencies :
• Input RF at 400 MHz, 900 MHz, 1950 MHz and 2450 MHz
• Output RF at 45 MHz, 70 MHz and 240 MHz
The question many of us are asking or will ask is : how to proceed if the application used other
frequencies ?
The above highlights on impedance matching will help us to handle the issue :
Let us first consider the proposed IF output matching network :
Zs
C2
Load
ZL
50 Ω
L1
MAX2680
MAX2681
MAX2682
R1
IF Matching
network
Zs is the IF output stage internal impedance (Thevenin equivalent) that needs to be matched
with the external line of 50 Ω. The Zs has been given for the 3 different devices and for the 3
standard IF frequencies :
Zs
MAX2680
MAX2681
MAX2682
IF = 45 MHz
960-j372
934-j373
670-j216
IF = 70 MHz
803-j785
746-j526
578-j299
IF = 240 MHz
186-j397
161-j375
175-j296
The datasheet contains a table with elements values for the same above defined 3 standard IF
frequencies :
IF Matching
L1
C2
R1
IF = 45 MHz
390 nH
39 pF
250 Ω
IF = 70 MHz
330 nH
15 pF
Open
IF = 240 MHz
82 nH
3 pF
open
Let us verify on the Smith chart the data for the MAX2680 at IF = 240 MHz.
• First step : Choose the characteristic impedance : here Zo = 50 Ω, compute the
normalized values zL and zs*. Here we have zL = 1 and zs* = 3.72 + j7.94
• Step 2 : Plot the zL and zS* on the chart
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MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 21 / 23
•
Step 3 : Start from zL, turn anti-clock wise (because capacitor in series thus negative
reactance) of a quantity still to be determined. The end point must be situated on the
constant r=1 circle in a such way that the L1 shunt (anti-clock wise rotation in the
admittance chart) will end to zs*. Optionally, shunt R1 must be added if the
intermediate point is to far for reaching zs*.
The operations are illustrated in the following Smith Chart.
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MAXIM France – Field Application Engineering – Technical article – Jan 2000 – KCC/AH – Page 22 / 23
The arc A-B gives the value of the capacitor C2
The arc C-D gives the value of the inductor L1
It is not necessary to add the shunt resistor R1 in this case.
Value of arc A-B measured on the plot : r = 4.5 units, thus Z = 50 * 4.5 = 225 ohms
Thus C2 = 1/(225*ω) = 1/(225*2*π*f) = 1 / (225*2*π*240 106) = 2.94 pF
Value of arc C-D measured on the plot : s = 0.3 units, thus Z = 50/0.3 = 167 ohms
Thus L1 = 167 / ω = 167 / (2*π*f) = 167 / (2*π*240 106) = 110 nH
The same manipulations for MAX2681 give same results since the impedance is quasi
identical.
For the MAX2682, an identical set of operations on the Smith Chart give 3.7 for the arc BC
and 0.37 for the arc CD. These results for C2 = 3.6 pF and L1=89 nH
Values are thus very close for a same IF frequency on the 3 products MAX2680, MAX2681
and MAX2682
At IF = 70 MHz, the arc lengths AB and CD are very close to the cases at IF = 240 MHz, these
end then with C2 = 12 pF and L1 value is 310 nH
At IF = 45 MHz, the shunt resistor is required if we want to avoid too big inductance. The
“price” to pay is a little bit a degradation of the Q factor. The shunt resistor allows a decrease
in constant g circle in the Admittance Smith chart. One can observe both the AB and the CD
arcs are shorter than the situation without the shunt. After the manipulation on the chart, one
can measure arc AB = 1.65 (resulting in a 42 pF value for C2) and arc CD = 0.5 (resulting on
an inductor value L1 of 351 nH). The case at 45 MHz is illustrated in the hereafter Smith
Chart.
After having placed the Zs and the ZL spots on the chart, start from ZL (point A : r=1, x=0) and
we know the first element is a series capacitor : thus next point is B situated somewhere on the
circle r=1. The point B transformed into admittance must give a point C situated 180° phase.
As explained above, it will be added a shunt resistor : converted into conductance, it
corresponds to subtract a value. We choose (apprx) 250 ohms (0.2 mho normalized
conductance). It is a shift of point C into D along a constant susceptance circle starting from C.
One D is obtained, its recon version into impedance must fall into Zs (point E).
We measure arc AB : 1.65 meaning 82.5 ohms at 45 MHz and giving C2 = 42 pF
We measure then arc CD : 0.45 meaning 111.1 ohms at 45 MHz and giving L1 = 390 nH
By considering also the need to use standardized components values, one find the data given
in the datasheet.
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Impedance matching and Schmitt Chart : practical aspects consideration