impedance matching

Transcription

impedance matching

IMPEDANCE MATCHING
for
High-Frequency Circuit Design Elective
by
Michael Tse
September 2003
Contents
•
•
•
•
•
The Problem
Q-factor matching approach
Simple matching circuits
L matching circuits
π matching circuits
T matching circuits
Tapped capacitor matching circuits
Double-tuned circuits
General impedance matching based on two-port circuits
Immittance matrices and hybrid matrices
ABCD matrix and matching
Propagation equations from ABCD matrix
Michael Tse: Impedance Matching
2
Impedance Matching
• Impedance matching is a major problem in highfrequency circuit design.
• It is concerned with matching one part of a circuit to
another in order to achieve maximum power transfer
between the two parts.
max power transfer
Circuit 1
Circuit 2
space
max power transfer
3
The problem
Given a load R, find a circuit that can match the driving
resistance R¢ at frequency w0.
R¢
R¢
?
R
Obviously, the matching circuit must contain L and C in
order to specify the matching frequency.
4
The Q factor approach to matching
The Q factor is defined as the ratio of stored to dissipated
power
2p ⋅ (max instantaneous energy stored )
Q=
energy dissipated per cycle
In general, a circuit’s reactance is a function of frequency
and the Q factor is defined at the resonance frequency w0 .
X
w0
w
As we will see later, the Q
factor can be used to modify the
overall resistance of a circuit at
some selected frequency, thus
achieving a matching condition.
5
Low Q circuit
High Q circuit
X
X
w
w0
Definition:
Q=
w
w0
w 0 dB
w dX
= 0
2G dw w =w 0 2R dw w =w 0
B = susceptance
X = reactance
R = resistance
G = conductance
It is easily shown that
for linear parallel RLC
circuits:
Q = w0CR = R/(w0L)
6
Essential revision (basic circuit theory)
R
Resistance (Ω)
Z
IMPEDANCE
(Ω)
R
Resistance (Ω)
Y
ADMITTANCE
(S)
G
Conductance (S)
L
inductance (H)
jwL = +jX
reactance (Ω)
1 = –jB
jw L
susceptance (S)
C
capacitance (F)
1 = –jX
jw C
reactance (Ω)
jwC = +jB
susceptance (S)
7
Quality factor (Q factor)
Series:
X
1
G
Q= =
=
R RB B
L
wL
Q=
R
R
†
Parallel:
R
B
Q = = RB =
X
G
†
C
R
†
R
1
Q=
wCR
†
L
R
Q=
wL
R
C
Q = wCR
† L or C.
Higher Q means†that it is closer to the ideal
8
Series to parallel conversion
jX
Z = R + jX
Y=
1
1
R - jX
R
X
=
= 2
=
j
Z R + jX R + X 2 R 2 + X 2
R2 + X 2
R
1
1
R
X
=
2 +
ÊÊ R ˆ 2 ˆ
ÊXˆ
1+ Á ˜
jÁÁÁ ˜ + 1˜˜
Ë R¯
ËË X ¯
¯
1
1
=
+
Ê 1
ˆ
R(1+ Q2 )
jX Á 2 + 1˜
ËQ
¯
†
†
Ê
1ˆ
jX Á1+ 2 ˜
Ë Q ¯
R(1+Q2)
or j R'
Q
†
R
j (1+ Q2 )
Q †
†
†
=
Ê 1
ˆ
jRQÁ 2 + 1˜
ËQ
¯
9
Parallel to series conversion
jB
G
Z=
1
1
G - jB
G
B
=
= 2
=
j
Y G + jB G + B 2 G 2 + B 2
G2 + B2
Y = G + jB
†
†
G(1+ Q2 )
†
conductance (S)
Ê
1 ˆ susceptance (S)
jBÁ1+ 2 ˜
Ë Q ¯
Ê
ˆ
or j G' = j G (1+ Q2 ) = jGQÁ 12 + 1˜
Q
Q
Ë Q †¯
†
1
1
G
B
=
2 +
ÊÊ G ˆ 2 ˆ
Ê Bˆ
1+ Á ˜
jÁÁÁ ˜ + 1˜˜
ËG¯
ËË B ¯
¯
1
1
=
+
Ê 1
ˆ
G(1+ Q2 )
jBÁ 2 + 1˜
ËQ
¯
10
Example: RLC circuit (Recall Year 1 material)
Resonant frequency is w 0 =
R
L
C
Q factor is Q = R
1
LC
C
L
Z drops by 2 (3 dB) at w1 and w2.
1
Z=
(1/R) + jwC - ( j /w L)
Ê
ˆ
1
1
˜
w1,2 = w 0 ÁÁ 1+
±
2
2Q ˜¯
4Q
Ë
Bandwidth is Dw = w 2 - w1 =
w1 w0 w2
1
RC
Note: w1 and w2 are called 3dB corner
frequencies. Their geometric mean is w0. For
narrowband cases, their arithmetic mean is
close to w0.
11
Practical components are lossy!
=
C
RC
Q factor = QC = w0CRC
(unloaded Q factor)
=
RL
L
Q factor = QL = RL/w0L
(unloaded Q factor)
QLC = unloaded Q factor for the paralleled LC components
1
1
1
=
+
QLC QC QL
(easily shown)
12
Simple matching circuits
R¢
R¢
?
R
L matching circuit (single LC section)
p matching circuit
T matching circuit
13
Design of L matching circuits
Series L circuit:
Objective: match Yin to R’ at w0
L
Begin with
Yin
C
R
Yi n = jwC +
1
R + jwL
È
˘
R
w
L
˙
= 2
+ jÍw C - 2
2
2
ÍÎ
R + (w L)
R + (wL) ˙˚
Obviously, the reactive part is cancelled if we have
L
C= 2
R + w 20 L2
1
R2
- 2
where w 0 =
LC L
(#)
14
Thus, at w = w0, we have a resistance for Yin, which should be set to R’.
R 2 + w 20 L2
R¢ =
= R 1+ Q2
R
(
)
(*)
Here, Q is the Q-factor, which is equal to w0L/R (for series L and R).
So, we can see clearly that Q is modifying R to achieve the matching
condition.
Design procedure:
-Given R and R’, find the required Q from (*).
-Given w0, find the required L from Q = w0L/R .
-From (#), find the required C to give the selected resonant frequency w0.
15
Begin with
Shunt L circuit:
Zi n = jwL +
L
C
Zin
R
1
G + jw C
È
˘
G
wC
= 2
+ jÍwL - 2
˙
2 2
2 2˚
Î
G +w C
G +w C
Reactive part is cancelled when
C
L= 2
G + w 20 C 2
1 G2
where w 0 =
LC C 2
Finally, the matching condition requires that
R¢ =
1/G
R
=
1+ (w 0 C /G)2 1+ Q2
(*)
Design procedure is similar to the series case.
16
(#)
Other L circuit variations
C
L
Series:
C
R
C
L
Shunt:
R
L
C
R
L
R
Exercise: derive design procedure for all other L circuits.
17
General procedure for designing L circuits
Series L circuit (suitable for R’>R) :
jX1
R¢
R
jX2
Shunt L circuit (suitable for R’<R) :
jX2
R¢
jX1
R
R¢ = R(1+ Q2 )
Ê
1 ˆ
jR ¢
jX 2 = - jX 1ÁÁ1+ 2 ˜˜ = Q
Ë Q ¯
X
Q= 1
R
R
1+ Q2
jX1
jX 2 = = - j R¢Q
1
1+ 2
Q
B
R
Q= 1 =
G X1
R¢ =
18
Advantages of L circuits:
• Simple
• Low cost
• Easy to design
Disadvantages of L circuits:
• The value of Q is determined by the ratio of R/R’. Hence,
• there is no control over the value of Q.
• the bandwidth is also not controllable.
Solution: Add an element to provide added flexibility.
fi p circuits and T circuits
19
p matching circuits
Analysis by decomposing into two L
circuit sections:
jX2
R¢¢
jB3
R¢ + jX ¢
First section (from right):
jB1
R
jX’
Second section:
j(X2–R’Q1)
R¢¢
jB3
R¢ + jX ¢
R
X ¢ = X 2 - R ¢Q1
2
1+ Q1
B
Q1 = 1 = B1R
G
R¢ =
Q2 =
R¢
X ¢ X 2 - R ¢Q1
X
=
fi 2 = Q1 + Q2
R¢
R¢
R¢
R¢¢ = R ¢(1+ Q22 )
Q
Q
B¢¢ = B3 - 2 fi B3 = 2
R¢
R¢
20
Impedance transformation in p matching circuits
jX2
R¢¢
jB3
R¢ + jX ¢
jB1
R
R¢¢
R
1
1+ Q12
R¢
1+ Q22
Obviously, we have to set Q1 > Q2
if we want to have R”<R.
Likewise, we need Q1 < Q2 if we
want to have R”>R.
21
General procedure for designing p matching circuits
For R¢¢ < R
For R¢¢ > R
1.
Select Q1 according to the max Q.
1.
2.
Find R’ using R¢ = R /(1+ Q12 )
2.
3.
Get Q2 using Q2 =
Find R’ using R¢ = R¢¢ /(1+ Q22 )
R
2
-1
Get Q2 using Q1 =
¢
R
4.
Obtain X2 using X2 = R’(Q1 + Q2).
4.
Obtain X2 using X2 = R’(Q1 + Q2).
5.
B1 = Q1/R
5.
B1 = Q1/R
6.
B3 = Q2/R”
6.
B3 = Q2/R”
2
R¢¢
-1
¢
R
3.
22
T matching circuits
The analysis is similar to the p case.
jX3
jX1
jB2
R¢¢
R
R¢ + jX ¢
R¢
R
1+ Q12
The difference is that R is first raised
to R’ by the series reactance, and
then lowered to R” by the shunt
reactance.
The design procedure can be
similarly derived. (Exercise)
1
1+ Q22
R¢¢
23
General procedure for designing T matching circuits
For R¢¢ > R
For R¢¢ < R
1.
1.
2.
Find R’ using R¢ = R(1+ Q12 )
R¢
2
-1
Get Q2 using Q2 =
¢
¢
R
2.
Find R’ using R¢ = R¢¢(1+ Q22 )
R¢
2
-1
Get Q1 using Q1 =
R
4.
Obtain X1 using X1 = Q1R.
4.
Obtain X1 using X1 = Q1R.
5.
B2 = (Q1+Q2)/R’
5.
B2 = (Q1+Q2)/R’
6.
X3 = Q2R”
6.
X3 = Q2R”
3.
3.
24
Tapped capacitor matching circuit
Ê 1+ Q2 ˆ
p
C2 ÁÁ 2 ˜˜
Ë Qp ¯
C1
L
C2
R
R
1+ Q2p
Q factor
Qp = w 0 C2 R
25
C1
R’
Ê 1+ Q2 ˆ
p
C2 ÁÁ 2 ˜˜
Ë Qp ¯
L
Q1 = R’/w0L
R
R
R¢
1+ Q12 1+ Q2p
1+ Q2p
required
R¢
R
=
fi Qp =
2
2
1+ Q1 1+ Qp
R
1+ Q12 - 1
R¢
(
)
26
Ê 2 ˆ
1 1
1 Á Qp ˜
=
+
C C1 C2 ÁË1+ Q2p ˜¯
C
R’
R
1+ Q2p
L
For a high Q circuit, w 0 ª
1
LC
Also, we have the alternative approximation for Q1: Q1 ≈ w0R’C,
which is set to w0 / Dw .
Thus, we can go backward to find all the circuit parameters.
27
General procedure for designing tapped C circuits
1.
2.
3.
4.
5.
6.
Find Q1 from Q1 = w0 / Dw
Given R’, find C using C = Q1/ w0R’ = 1 / 2π DwR’
Find L using L = 1 / w02C
Find Qp using Qp = [ (R/R’)(1+Q12)–1 ]1/2
Find C2 from C2 = Qp / w0R
Find C1 from C1 = Ceq C2 / (Ceq – C2) where Ceq =
C2(1+ Qp2)/ Qp2
28
Advantages of π, T and tapped C circuits:
• specify Q factor (sharpness of cutoff)
• provide some control of the bandwidth
Disadvantage:
• no precise control of the bandwidth
For precise specification of bandwidth, use
double-tuned matching circuits.
29
Double-tuned matching circuits
Specify the bandwidth by two frequencies wm1 and wm2 .
transmission gain GT
wm1
wm2
w
There is a mid-band dip, which can be made small if the pass band is
narrow. Also, large difference in the impedances to be matched can be
achieved by means of galvanic transformer.
30
The construction of a double-tuned circuit typically includes a real
transformer and two resonating capacitors.
•
RG
C1
L11
M
•
L22
C2
RL
Transformer turn ratio n and coupling coefficient k are related by
n=
L1 1
k 2 L2 2
31
Equivalent models:
L22(1–k2)
n:1
•
RG
C1
•
L11
C2
ideal transformer
Ê1
ˆ
L11Á 2 -1˜
Ëk
¯
L2’
RG
C1
L11
C2 ’
RL ’
†
†
RL
Ê L11 ˆ
Á 2 ˜R2
Ë k L22 ¯
Ê L11 ˆ
Á 2 ˜C2
Ë k L22 ¯
32
Exact match is to be achieved at two given frequencies: fm1 and fm2.
L2’
RG
C1
L11
R1 R2
C2 ’
RL ’
Observe that:
• R1 resonates at certain frequency, but is always less than RG
• R2 decreases monotonically with frequency
So, if RL is sufficiently small, there will be two frequency
values where R1 = R2.
33
resistance
R2
R1
f
fm1
fm2
Our objective here is to match RG and RL over a
bandwidth Df centered at fo, usually with an allowable
ripple in the pass band.
34
General Impedance Matching Based on Two-Port
Parameters
Two-port models
i1
i2
+
+
v1
v2
–
–
Idea: we don’t care what is inside, as long as it can be modelled in
terms of four parameters.
35
Two-port models
i1
+
port 1
i2
+
v1
v2
–
–
z-parameters
(impedance matrix):
Èv1 ˘ Èz11
Í ˙=Í
Îv 2 ˚ Îz21
z12 ˘Èi1 ˘
˙Í ˙
z22 ˚Îi2 ˚
y-parameters
(admittance matrix):
Èi1 ˘ È y11
Í ˙=Í
Îi2 ˚ Î y 21
Èv1˘ Èh11
Í ˙=Í
Îi2 ˚ Îh21
y12 ˘Èv1 ˘
˙Í ˙
y 22 ˚Îv 2 ˚
h12 ˘È†i1 ˘
˙Í ˙
h22 ˚Îv 2 ˚
È i1 ˘ Èg11
Í ˙=Í
Îv 2 ˚ Îg21
g12 ˘Èv1˘
˙Í ˙
g22 ˚Îi2 ˚
†
h-parameters
(hybrid matrix):
†
g-parameters
(hybrid matrix):
†
†
port 2
v1 = z11i1 + z12i2
v 2 = z21i1 + z22i2
:
:
36
Finding the parameters
e.g., z-parameters
v1 = z11i1 + z12i2
v 2 = z21i1 + z22i2
z11 =
v1
v
= 1
i1 i = 0 i1 port 2 open -circuited
2
z12 =
v1
v
= 1
i2 i = 0 i2 port 1 open -circuited
1
†
z21 =
z22 =
v2
i1
v2
i2
=
i2 = 0
=
i1 = 0
†
v2
i1
port 2 open -circuited
v2
i2
37
Finding the parameters
e.g., g-parameters
i1 = g11v1 + g12i2
v 2 = g21v1 + g22i2
g11 =
i1
i
= 1
v1 i = 0 v1 port 2 open -circuited
2
g12 =
†
g21 =
g22 =
i1
i2
v2
v1
v2
i2
=
v1 = 0
=
i2 = 0
=
v1 = 0
†
i1
i2
port 1 short -circuited
v2
v1
v2
i2
38
Input impedance:
+
i1
i2
+
[Z]
v1
–
ZL
v2
–
Zin
v1 = z11i1 + z12i2
v 2 = z21i1 + z22i2
†
†
fi†
fi
v1
i2
= z11 + z12
i1
i1
v2
i1
= -z21 - z22
-i2
i2
fi
†
Ê z12 z21 ˆ
Z in = z11 - Á
† ˜
Ë Z L + z22 ¯
†
†
i2
Z in = z11 + z12
i1
i1
Z L = -z21 - z22
i2
39
Similarly, we can find the input impedance at any port in terms of any
of the two-port parameters, or even a combination of different twoport parameters.
We will see that the matching problem can be solved by making sure
that both input and output ports are matched.
ZG
i1
i2
[Z]
±
+
ZL
v2
–
ZIM1
ZIM2
matching: ZG = ZIM1
and ZIM2 = ZL
image impedances
40
The ABCD parameters (very useful form)
+
v1
i1
i2
[ABCD]
–
+
v2
–
Here, voltage and current of port 1 are expressed in terms of those of port 2. So,
this is neither an immittance matrix like Z and Y, nor a hybrid matrix like G and H.
Èv1˘ ÈA B˘È v 2 ˘
Í ˙=Í
˙Í ˙
i
C
D
Î 1˚ Î
˚Î-i2 ˚
Note: the sign of i2 in the above equation. This sign convention will
make the ABCD matrix very useful for describing cascade circuits.
†
41
+
i’
i1
[ABCD]1
v1
–
Èv1˘ ÈA1 B1 ˘È v' ˘
Í ˙=Í
˙Í ˙
i
C
D
Î 1˚ Î 1
1˚Î-i'˚
i”
+
v’
–
i2
[ABCD]2
Èv'˘ È A2
Í ˙=Í
Îi"˚ ÎC2
+
v2
–
B2 ˘È v 2 ˘
˙Í ˙
D2 ˚Î-i2 ˚
Since –i’ = i”, we have
Èv1†
˘ ÈA1
Í ˙=Í
Î i1 ˚ ÎC1
†
B1 ˘È A2
˙Í
D1˚ÎC2
B2 ˘È v 2 ˘
˙Í ˙
D2 ˚Î-i2 ˚
So, if more two-ports are cascaded, the overall ABCD matrix is just
the product of all the ABCD matrices.
†
42
To find the ABCD parameters, we may apply the same principle:
A=
B=
C=
v1
v2
=
i2 = 0
-v1
i2 v
i1
v2
=
2=0
=
i2 = 0
-i1
D=
i2 v
v1
v2
2
=
z11
z21
-v1
z z -z z
= 11 22 21 12
i2 port 2 short -circuited
z21
i1
v2
=
1
z21
-i1
z22
=
=
i2 port 2 short -circuited z21
=0
We can show easily that AD – BC = 1
†
if z12 = z21, i.e., reciprocal circuit.
43
Matching problem
ZG
i1
i2
+
v1
–
±
[ABCD]
+
ZL
v2
–
ZIM1
Input image impedance
v1 = Av 2 - Bi2
i1 = Cv 2 - Di2
†
fi
†
v1 Av 2 - Bi2
=
i1 Cv 2 - Di2
v2
A
+B
-i2
=
v
C 2 +D
-i2
AZ L + B
=
CZ L + D
Z in =
†
44
ZG
i1
i2
+
v1
–
+
[ABCD]
v2
–
ZIM2
Output image impedance
v1 = Av 2 - Bi2
i1 = Cv 2 - Di2
fi
v 2 = Dv1 - Bi1
i2 = Cv1 - Ai1
fi
because AD – BC = 1
†
†
†
†
†
v 2 Dv1 - Bi1
=
i2 Cv1 - Ai1
v
D 1 +B
-i1
=
v
C 1 +A
-i1
DZG + B
=
CZG + A
Z IM2 =
45
Under matched conditions,
ZG = ZIM1
fi
†
†
AZ L + B
Z IM1 = ZG =
CZ L + D
fi
Z IM1 =
and ZL = ZIM2
and
DZG + B
Z IM2 = Z L =
CZG + A
AB
†
CD
and
Z IM2 =
DB
AC
z11
y11
and
Z IM2 =
z22
y 22
Alternatively, we have
†
†
Z IM1 =
†
46
Note: image impedances are different from input and output impedances.
1.
Image impedances do not depend on the load impedance or the source
impedance. They are purely dependent upon the circuit.
Z IM1 =
2.
z11
y11
and
Z IM2 =
z22
y 22
Input impedance (Zin) depend on the load impedance. Output impedance
(Zout)†depends on the source impedance. For example,
Ê z12 z21 ˆ
Z in = z11 - Á
˜
Ë Z L + z22 ¯
Matching conditions:
• Source impedance equals input image impedance
• Load †
impedance equals output image impedance
47
Example
i1
Za
Zc
+
v1
–
port 1
We can easily see that
i2
+
Zb
z11 =
v1
= Za + Zb
i1 port 2 open -circuited
y11 =
i1
1
=
v1 port 2 short -circuited Z a + Z b Z c
z22 =
v2
i2
v2
–
port 2
y 22 =
i2
v2
= Zb + Zc
=
1
Zc + Za Zb
Thus, the image impedances are
Z IM1 = (Z a + Z b )(Z a + Z b Z†c ) and Z IM2 = (Z c + Z b )(Z c + Z a Z b )
†
48
Matching a cascade of circuits
1
ZIM1
2
Z’IM1 = ZIM2
i2
+
+
v1
v2
–
†
†
–
†
ZIM1
Z’IM2 = ZIM3
4
Z’IM3 = ZIM4
ZL
Z’IM4 = ZL
A wave or signal entering into circuit 1
from left side will travel without
reflection through the circuits if all ports
are matched.
Convention
i1
3
Propagation constant g
ZIM2
input power
v1i1
v1
e =
=
=
output power
v 2 (-i2 ) v 2
g
†
†
Z IM2
Z IM1
49
Propagation equations
v1i1
v
= 1
v 2 (-i2 ) v 2
eg =
In general,
†
Z IM2
Z IM1
†
= A+ B
Thus,
†
fi
if the 2-port circuit is symmetrical
v1 Av 2 - Bi2
B
=
= A+
v2
v2
Z IM2
†
†
v1
e =
v2
g
AC
=
BD
i1
= CZ IM2 + D =
-i2
eg =
D
A
A
D
(
(
AD + BC
AD + BC
)
)
v1i1
= AD + BC
-v 2i2
e-g = AD - BC
†
50
Combining eg and e–g, we have
eg + e-g
cosh g =
= AD
2
eg - e-g
sinh g =
= BC
2
Define
n=
†
We have
†
Z IM1
=
Z IM2
A
D
A = n coshg
B = nZ IM2 sinh g
sinh g
nZ IM2
cosh g
D=
n
C=
†
51
From the ABCD equation, we have
v1 = nv 2 cosh g - ni2 Z IM2 sinh g
v
i
i1 = 2 sinh g - 2 cosh g
nZ IM2
n
Dividing gives
†
v1
Z L + Z IM2 tanh g
2
Z in = = n Z IM2
i1
Z L tanh g + Z IM2
For a transmission line, ZIM1 = ZIM2 = Zo, where Zo is usually called
the characteristic impedance of the transmission line. Also, for a
lossless†transmission line, g = jL is pure imaginary, and thus tanh
becomes tan, sinh becomes sin, cosh becomes cosh.
v1
Z L + jZ o tan L
Z in = = Z o
i1
Z o + jZ L tan L
†
52
This is just the same transmission line equation. In communication,
we usually express L as electrical length, and is equal to
L = w l / v = 2p l / l
wavelength
frequency in rad/s
length of transmission line
velocity of propagation
So, we can easily verify the following standard results:
1. If the transmission line length is l/2 or l, then the input impedance
is just equal to the load impedance.
2. If the transmission line length is l/4, then the input impedance is
Zo2/ZL.
Impedance value for other lengths can be found from the equation or
conveniently by using a Smith chart.
53

impedance matching

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