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Oscillator Resonator Design Tutorial J P Silver
Email: [email protected]
1 ABSTRACT This paper discusses the design of various types of resonator that form the heart of any oscillator design.
The first section describes the different resonator types including lumped, coaxial, microstrip and dielectric. The following section deals with varactor diodes, including design equations, temperature & loaded Q performance. In the final section the definitions of loaded and unloaded Q are described with a worked example and design techniques on Q transformations.
2 INTRODUCTION The resonator is key to the design of an oscillator. The loaded Q determines the phase noise performance of the oscillator. The oscillator frequency will determine to some degree the type of resonator eg At microwave frequencies resonators can be coaxial or microstrip and at low frequencies the resonators are almost always made up of lumped components. This tutorial gives design data for various types of resonator.
3 RESONATORS [1] The resonator is the core component of the oscillator, in that it is the frequency selective component and its Q is the dominating factor for the phase noise performance of the oscillator. This section discusses the range of resonators, that can be used for an oscillator covering, dielectric, cavity, transmission line, lumped element and coaxial resonators.
3.1 LUMPED ELEMENT Lumped element resonators can be configured to form either a low, high or band pass filter, and the given number of elements is directly related to the Q and loss of the resonator. The simplest resonators can consist of just two elements an inductor and a capacitor ie:
3.2 TWO ELEMENT RESONATOR CIRCUITS Figure 1 shows a schematic diagram of a twoelement resonator. This circuit is seldom used in oscillators as the loaded Q will be very low as the source and load impedances will directly load the tuned circuit.
Q = LR
ω..2
Q = 2.R.Lω
Figure 1 Schematic of a two element, lumped resonator, together with loaded Q equations.
At resonance the transmission phase is zero and the network is loss less (except for the resistance of the inductor). The series resonator impedes signal transmission while the parallel network allows signal transmission. The main problem with such a simple resonator is achieving a required Q, for example if we want a Q of 30 we would need the following series inductor & capacitor at 1GHz:
0.05pF = 9477
91*21
= f21
= C
477nH = 1E9*2
30*50*2 = 2.R.Q = L
22
−
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
EE
Lππ
πω
Although the inductor is a realised value the capacitor could not be realised except in perhaps interdigital form. This could be used if the oscillator is designed for fixed frequency but the value is impracticable as a varactor in a voltage controlled oscillator. The situation can be improved by using more than two elements eg 3 or 4 as described in the next section.
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2 of 20
3.3 THREE ELEMENT RESONATOR CIRCUITS The diagram below shows a range of three element lumped resonators  Figure 2.
L
2L
2
C
L
X2XRX
RX
= Q
+=
Q RX
X X
LL
C L
=
= 2.
C
2C
2
L
C
X2XR
X
RX
= Q
+=
Q RX
X X
CC
L C
=
= 2.
X L & XCL C= =2 1
2π
π. .
. .f
f
Figure 2 Schematic diagram of a range of three element resonators together with equations to calculate the reactive components and loaded Q.
3.4 FOUR ELEMENT RESONATOR CIRCUITS Four element resonators are used most commonly in oscillators as the loaded Q of the resonator can be set independently of the resonant circuit so that sensible
component values can be calculated. Figure 3 shows a four element lumped resonator and Figure 4 shows an alternative configuration.
C shunt
C series L
Figure 3 Schematic diagram of a four element lumped resonator
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( )( )
Q unloaded L the is Q where
Q1
Q1
1 = Q where
12
R = X
:elyapproximat is Q loaded given a for reactance The.C of function a is Q Loaded
1 = L
:by given is f at resonate to inductance Required
resistance load tinput/oupu = R
1RR21
1C
:is L inductor series the withresonates whichecapacitanc Effective
u
uL
e
2/1
ocshunt
shunt
2series
o
o
2o
2o
e
series
−
⎟⎟⎠
⎞⎜⎜⎝
⎛−
++
=
−
L
eo
eo
shunto
oshunt
series
XQR
C
CC
C
ω
ωω
C shunt
C series
L
Figure 4 Schematic diagram of the alternative four element lumped resonator
( )
resistance load tinput/oupu = R
1R
2C
:is L inductor shunt the withresonates whichecapacitanc Effective
L.f2
1
= Ce
: inductor shunt resonate to eCapacitanc
admittance inductor shunt given a isB & Q unloaded L the is Q where
Q1
Q1
1 = Q where
X..21C 1
2R = X
o
2o
shunt
series
2
L
u
uL
e
cseriesseries
2/1
ocseries
+−=
⎟⎠⎞
⎜⎝⎛
−
=∴⎟⎟⎠
⎞⎜⎜⎝
⎛−
serieso
series
L
eo
CC
Ce
fBQR
ω
π
π
3.5 COAXIAL CABLE RESONATOR [2] A quarterwave coaxial resonator is formed by shorting the centre conductor of a coaxial line to its shield at one end, leaving the other end opencircuited. The physical length of the resonator is equal to one quarter the wavelength (90 degrees electrical length) in the medium filling the resonator. A diagram of a coaxial resonator is shown below in Figure 5.
λ /4
b
a
Figure 5 Schematic diagram of a coaxial cable resonator showing the critical dimensions.
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coaxcoax 41 length Resonator = ;
f
2.99E8 =
λελ
λ
λ
==r
air
air
The unloaded Q of the resonator is a function of the conductor losses, the dielectric losses and the physical dimensions of the coaxial cable ie:
112r
rD
C
DCU
Fm8.854x10=y;permitivit relative
;1 ie dielectric ofty conductivi =
..f.2 = factor) ssipationTangent/Di (Loss tan. = Q
by given is conductors the separates that dielectric the from oncontributi Q The
conductors the ofty conductivi = and ty permeabili = where
b1
a1
abLn....
2. = Q
by given is and conductors the inflow current to due lost energy to due is conductor from oncontributi Q The
Dielectric = D & Conductor = C e wherQ1
Q1
Q1
o
o
f
εε
ρσ
εεπσδ
σμ
σμπ
=
+
+=
3.6 DESIGN EXAMPLE OF A COAXIAL CABLE RESONATOR
The following example is for the design of a coaxial resonator to operate in an oscillator at 1GHz. The resonator is made from semirigid coaxial cable that contains a dielectric of PTFE, which has a relative permittivity of ~ 2.2 and a tanδ of 0.0004.
5.04cm = 36090.
2.21E9
2.99E8
= length Resonator
3.7 CALCULATION OF RESONATOR Q FACTOR
The Q factor of the resonator determines the phase noise performance of the oscillator. Loss in the coaxial cable from the conductivity of the sheath and the loss tangent of the dielectric will set the Q of the resonator. Most coaxial cables especially semirigid cables use copper as the conductor, therefore the equation for
the Q contribution for the conductor ie Qcc is given by: The dielectric of the cable also effects the Q of the resonator and is given by:
92.95 Q (0.000358) 3.58mm = b example above For3.58mm or 0.141" is cable rigidsemi
typical of diameter Overall
f8.398.b. =
Q unloaded to oncontributi Conductor = Q
cc
cc
=∴
The dielectric of the cable also effects the Q of the resonator and is given by:
6.98 2500
192.95
1 =
Q1+
Q1 =
Q1 unloaded Total
2500 0.0004
1 Q
10GHz @ 0.0004 ~ PTFE for tan
material dielectric of tangent loss tan.
1 = Q
unloaded to oncontributi loss Dielectric = Q
dcc
d
d
=+
==∴
δ
δ
Note the Qcc term dominates the overall Q factor of the resonator at this frequency.
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5 of 20
The table below shows (Table 1) design data for a range of common materials used in the construction of coaxial cables:
Material εr ρ tanδ Copper  1.56E8Ω.m  Gold  2.04E8Ω.m  Silver  1.63 E8Ω.m  Nylon 3.0 1091011Ω.m 0.012@3GHz PTFE 22.1 1E16 0.0004@10GHz
Polythene HD
2.25 >1014Ω.m 0.0004@10GHz
PVC flexi 4.5 1091012Ω.m
Table 1 Design data for a range of materials commonly used in the construction of coaxial cables. The parameters shown are relative permittivity (εr), resistivity ρ (1/ρ = conductivity) and tan delta (tanδ).
3.8 COAXIAL RESONATOR [3] A quarterwave coaxial resonator is formed, by plating a piece of dielectric material with a high relative permittivity using a highly conductive metal. A cylindrical hole is formed along the axis of a cylinder of high relative permittivity dielectric material. All surfaces, apart from the end surface, are coated with a good conductor to form the coaxial resonator. The physical length of the resonator is equal to one quarter the wavelength (90 degrees electrical length) in the medium filling the resonator. The diagram (Figure 6) below shows the key dimensions of a coaxial resonator.
λ/4
W d
End of resonator platedOuter surface plated
Inner surface plated
RC
L ≡
Figure 6 Schematic diagram of a coaxial resonator showing the key dimensions. Note the resonator is plated with silver except for one end to allow it to be grounded.
The expression for the unloaded Q of such a resonator is
( ) ⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛
dW.079.1.60 =Z Impedance Input
88.5 of withdielectric sivered a for 200 = 38.6 of withdielectric silvered a for 240 = k
mm in diameter inside = d mm, in diameter outside = where
d1
W14.25
dW.079.1Ln
.ok. =
rin
L
Ln
W
f
r
r
ε
εε
π
ε
πε
4.Zo.Q = Resistance
.103*2*4.25.
= eCapacitanc
mm in length Physical = 103.4.25
.8.Zo. = Inductance
8r
82r
Zox
x
l
ll
Below resonance, such shortcircuited coaxial line elements simulate highQ, temperature stable ‘ideal’
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6 of 20
inductors. They will only realise an ‘ideal’ inductor over a narrow range as shown in the diagram Figure 7.
X L
X C
S elf R eson an tF req u en cy
Freq u en cy →
‘Ideal’Inductance
R egion
Frequency
Figure 7 Frequency response of a coaxial resonator. The first region shows an area of inductance followed by a point of resonance followed by a region of capacitance. The resonator is usually used below the selfresonant frequency so that in a VCO the varactor can be used to resonate with the coaxial resonator.
In order to use the coaxial resonator as a ‘ideal’ inductor the resonator must be used below the selfresonant frequency.
3.9 DESIGN EXAMPLE OF A COAXIAL RESONATOR [4,5,6]
The following section describes the design of a coaxial resonator to be used in a varactor controlled oscillator at 900MHz. We need therefore to select a suitable resonator that is inductive at 900MHz. Assume an ‘ideal’ starting inductance of 4nH at 900MHz. The material chosen is a silverplated ceramic resonator with a relative permittivity of 38.6 from Transtech. It has a tab inductance of 1nH, a W/h ratio of 2.57, a width of 6mm and a characteristic impedance of 9.4Ω.
9.74mm = 9.415.1tan.
26036.0 =
Z
Ztan.
2 = resonator of Length
.900MHz at 15.1 is reactance whose3nH= 14 ie inductance
required the from inductance tab the subtract We
60.36mm = 6.38
00E8/3E = c/ = Wavelength
1
o
input1g
68
r
o
⎟⎠⎞
⎜⎝⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛
Ω
−
−
π
πλ
εf
long 0.161 = 0.60360.0973 is line coaxial the Therefore
1241MHz = 0973.01.
4800*6036.0 =
MHz 1.4.
=Frequency Resonant Self
415.7 =
0.002461
0.00614.25
0.002460.006.079.1Ln
.6E800240. = Q
=
d1
W14.25
dW.079.1Ln
.ok. = Q
g
g
λ
λl
of
f
⎟⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛
The part resonance could be tested to ensure that it occurs at the selfresonant frequency of 1.241GHz.
3.10 DIELECTRIC RESONATOR [7] At lower frequencies the length of W/d ratio of a coaxial resonator becomes too big to realise so a dielectric ‘puck’ is used instead. The dielectric resonator is often made from the same material as the coaxial resonators except that they are not plated with a lowloss metal. In addition they are mounted on planer circuits as shown below (figure 35) and are coupled to a transmission line without a direct connection. As with other resonators, standing TE waves will be set up within the resonator, which will be dependent on the physical dimensions of the cylinder. The diagram of a dielectric resonator is shown below in Figure 8
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7 of 20
a
b
Figure 8 Schematic diagram of a dielectric resonator showing the key dimensions.
The most common resonant mode in dielectric resonators is the TE01δ mode and when the relative dielectric constant is around 40, more than 95% of the stored energy are located within the resonator. For an approximate estimation of the resonant frequency in TE01δ mode of an isolated dielectric resonator, the following simple formula can be used:
⎟⎠⎞
⎜⎝⎛ += 45.3La.
.a34 F
(mm)GHz
rε
The above equation is accurate to about 2% in the range 0.5 < a/L < 2 and 30 < εr < 50 The approximate Q factor of the resonator is directly related to the dielectric loss ie tanδ.
( )ro εεωσδ
δ .. = tan
tan1 Q unloaded =
3.11 DESIGN EXAMPLE OF A DIELECTRIC RESONATOR
The following section describes the design of a dielectric resonator for a frequency of ~ 7GHz. A manufacturer of dielectric resonators – Transtech can supply two relative permittivities of 30 and 38. The TransTech D87330305137 puck was selected with the following parameters, εr = 30, Diameter = 7.75mm, Height = 3.48mm, the resonant frequency can be estimated using:
7.313GHz = 45.33.479
3.8735.30.8735.3
34
45.3La.
.a34 F
(mm)GHz
⎟⎠⎞
⎜⎝⎛ +
=⎟⎠⎞
⎜⎝⎛ +=
rε
This calculated figure assumes that the resonator is in freespace. If the resonator is mounted on a substrate in a cavity then this will significantly alter the resonant frequency. A more accurate model to take into account cavity and substrate is the Itoh and Rudokas model [7] which, is shown below in Figure 9:
L2
L
L1
a
er6
er1
er2
er4
shield
shield
Region 2
Region 1
Region 4
Region 6
Figure 9 Itoh & Rudokas model of a dielectric resonator inside a metallic shielded cavity
This model can be simplified to the numerical solution of a pair of transcendental equations:
( ) ( )
⎟⎟⎠
⎞⎜⎜⎝
⎛+
−−=
00
01
0120146
2o0
GHz)((mm)o
y291.0y
2.43+12.4048
y+2.4048=ak
2.4048 be to taken is xxak y
L height the calculate to entered isfrequency initial An
.a.150 = ak
ρ
εε
π
rr
f
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8 of 20
( ) ( )[ ]2221
1111
216
20
220
212
120
211
L.cothtanL.cothtan1 =
L Length Resonator
k.k =
: is 6 and 4 regions to common constant npropagatio The
.kk
.kk
: are 2 and 1 regions in constants nattenuatio The
1 ααβ
εβ
εα
εα
β
α
βα
ρ
ρ
ρ
−− +
−
−=
−=
r
r
r
3.12 COUPLING OF RESONATOR TO MICROSTRIP LINE [8]
For analysis of the resonator coupled to a microstrip line, the transformation shown in the Figure 10 below is used. β (coupling coefficient) is used to provide an equivalent series resistance for the resonator:
d
R
≡ L
C
Figure 10 Dielectric resonator coupled to a microstrip line and the corresponding circuit diagram. The resistor L simulates the coupling of the LC resonant circuit of the dielectric resonator.
Calculation of loaded Q:
( )
ββ
β
π
= 1QQ
+1Q
Q
*Zo*2 = R
21 = LC
L
UUL
2
−⎟⎟⎠
⎞⎜⎜⎝
⎛=
f
With the above equations it is possible to design VCO for a given Q for example if we want a minimum Q of 1000:
Ω
−−⎟⎟⎠
⎞⎜⎜⎝
⎛=
4K = 4*50*2 *Zo*2 = R
of resistor series a withresonator the replace can weCAD a on analysing For
4 = 110005000 = 1
+1Q
Q
5000 of Q unloaded a withResonator a use weIf
L
UUL
β
ββ
TransTech have a CAD package [15] to calculate various design parameters using their dielectric resonators. We can use the CAD package to calculate a plot of the coupling coefficient β vs distance from the centre of the microstrip line to the centre of the DRO puck. The plot of the analysis is shown below in Figure 11.
5
10
15
20
25
30
35
40
45
5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5
Coupling Coefficient
B
D (mm) Center to Center
Figure 11 Plot of coupling coefficient (β) with distance from the centre of the puck to the centre of the microstrip line in mm
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Therefore, in our example, the puck would be placed at a distance of 7.15mm from the puck centre to the microstrip line centre.
3.13 TRANSMISSION LINE RESONATOR [9] Over a narrow bandwidth LC lumped components can be realised using shortcircuit and opencircuit transmission lines. If we analyse a transmission line terminated in a load ZL we can define the transformed impedance in terms of the characteristic line impedance and the electrical length of the transmission line. The diagram below (Figure 12) shows a transmission line loaded with ZL.
ZLT.L ZoZ(in) →
l l=0 Figure 12 Transmission line loaded with load ZL
[ ][ ]
( ) ( )( ) ( )
ljeelee
eeZoeeZleeZoeeZlZoinZ
eZoZleZoZl
eZoZleZoZlZoIVinZ
ZoZZoZ
VV
evevZo
evevIVinZ
ljlj
ljlj
ljljljlj
ljljljlj
ljlj
ljlj
L
L
ljlj
ljlj
.sin2)(.cos2)(
)()()()(.)(
..
...)(
12
21121)(
..
..
....
....
..
..
..
..
β
βββ
ββ
ββββ
ββββ
ββ
ββ
ββ
ββ
=−
=+
⎥⎦
⎤⎢⎣
⎡++−−++
=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−−+
−++==∴
+−
==
−
+==
−
−
−−
−−
−+
−+
−+
−+
⎥⎦
⎤⎢⎣
⎡+−+−++
=∴
⎥⎦
⎤⎢⎣
⎡++
=∴
−−
−−
ljljljlj
ljljljlj
eZoeZleZoeZleZoeZleZoeZlZoinZ
lZoljZlljZolZlZoinZ
....
....
....
.....)(
.cos2..sin2.
.sin2..cos2..)(
ββββ
ββββ
ββββ
⎥⎦
⎤⎢⎣
⎡++
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
++
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+
+
lZlZolZoZlZoinZ
lljZl
lljZo
ZoZlZo
llZo
lljZo
lljZl
llZl
Zo
.tan.
.tan..)(
.cos.sin.
.cos.sin.
.
.cos2.cos2.
.cos2.sin2.
.cos2.sin2.
.cos2.cos2.
.
.l2cosby through divide
ββ
ββ
ββ
ββ
ββ
ββ
ββ
β
This equation is the general expression for the impedance looking into a load ZL via a length of transmission line. If we now have the case where the transmission line is terminated with a short circuit we find the general expression simplifies ie let ZL = 0 then
Z in Zo Zl Zo lZo Zl l
( ) . . tan .. tan .
tan .
=++
⎡
⎣⎢
⎤
⎦⎥
ββ
β = jZ ( Short circuit)o l
We can now plot the impedance (Figure 13) of the shorted length of transmission line vs electrical length and we get the following graph, which shows how the transmission line equates to lumped capacitance and inductance with resonance’s in between. In general Z(in) = R(in) + jX(in) For S/CCT R(in) = 0 ; X(in) = Zotanβ.L Zotanβ.L is purely reactive varies between  ∞ & + ∞ as L varies
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10 of 20
l=0π3π/2 π/22π
4fo 3fo 2fo fo 0
θ = β.L
← f
0λg/4λg/23λg/4λg← λg
3 3 34 4
2 2
1 1
X = Z
= 2
= . = .v
= v
o
g
tan .β
β πλ
ϑ β ω
ω
l
ll
l⎛⎝⎜
⎞⎠⎟
Figure 13 Plot of impedance against length of a short circuited transmission line. The plot shows how the reactance of the transmission line varies between inductive and capacitive reactances with resonant frequency regions in between.
Each region of figure 40 is now described: (1) If θ between 0 & π/2 tanβ.L is positive ∴X is +ve ⇒ j(ω.L)  INDUCTIVE. (2) If π/2 < θ < π tanβ.L is ve ∴ X is ve ⇒ j(1/ω.C)  CAPACITIVE. (3) If θ ≈ 0, π , 2π  X  goes to a minimum ie:  X 
θ
≅L .C
(4) If θ ≅ π/2 , 3π/2  X  goes to a maximum:  X 
θ
≅
L //C
Similarly, for a transmission line terminated by an open circuit we can repeat the analysis, but we dividing through by ZL. Note Zo/ZL tends to zero ie:
ZLT.L Zo
ZL = ∞
V=Maxat O/cct
Z(in) →
[ ] circuit) Open ( .tan
1jZ =
.tan.
.tan.
.
ie Zby bottom & top divide.tan..tan..)(
o
L
⎟⎟⎠
⎞⎜⎜⎝
⎛
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+
+
⎥⎦
⎤⎢⎣
⎡++
=
lβ
β
β
ββ
ZllZl
ZlZo
ZllZo
ZlZl
Zo
lZlZolZoZlZoinZ
Again we can plot the impedance against electrical length of the transmission line (Figure 14) to see the equivalent lumped reactance and resonance points. In general Z(in) = R(in) + jX(in) For O/CCT R(in) = ∞ ; X(in) = Zocotβ.L Zocotβ.L is purely reactive varies between  ∞ & + ∞ as L
l=0π3π/2 π/22π
4fo 3fo 2fo fo 0
θ = β.L
← f
0λg/4λg/23λg/4λg← λg
3 34 4
2
1
2
1
X = Z
= 2
= . = .v
= v
o
g
cot .β
β πλ
ϑ β ω
ω
l
ll
l⎛⎝⎜
⎞⎠⎟
4
Figure 14 Plot of impedance against length of a open circuited transmission line. The plot shows how the
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reactance of the transmission line varies between inductive and capacitive reactance’s with resonant frequency regions in between.
The previous graphs show that we can realise lumped components from transmission lines eg
3.14 DESIGN EXAMPLE OF INDUCTOR USING A TRANSMISSION LINE
The following section describes the process of designing a transmission line to have a specific inductance of 0.7nH at a frequency of 8.8GHz. The transmission line is to be etched on RT duroid substrate material, which has a relative permittivity of 2.94 and a substrate thickness of 0.25mm.
ll
for Solve 2 = where.tan
1j.Zo = Zin
0.466pF = C C =
8.8GHz at 0.7nH of inductance of 38.8 = Reactance
g
f21 2
λπβ
β
π
⎟⎟⎠
⎞⎜⎜⎝
⎛
∴
Ω
⎟⎠
⎞⎜⎝
⎛
L
Using the transmission line equation for an opencircuit stub we can calculate the electrical length required for an inductance of 0.7nH. Therefore a opencircuit stub of length 3.1mm will have an inductance of 0.7nH at 8.8GHz. As the equations show the resulting impedance is a function of the characteristic of the line and generally we use a narrow high impedance line ~ 100Ω for an inductive impedance and a wide length of line ~ 20Ω, for a capacitive impedance. For completeness the empirical equations for calculating line widths are given in the next section:
3.1mm = 293
38.950arctan
= XZarctan
=
293 = 0214.02 =
21.4mm or 0.0214m = 53.2
3E8/8.8E9 =
therefore 2.94 is used be to material the ofty permittivi Relative
o
eg
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
=
β
πβ
ελλ
l
ff
air
3.15 CALCULATION OF EFFECTIVE RELATIVE PERMITTIVITY [10]
The following section describes the empirical equations that are used to calculate the dimensions of the microstrip lines and characteristic impedance [8]. The first equation describes the effective relative permittivity which, differs from the specified value due the width of the microstrip track.
( )( )
( ) )1.18/(17.18
1432.0/
)52/(/491+1 = a and
39.00.564 = b where
.1012
12
1 =
34
24
053.0
r
r
.rr
hWLnhW
hWhWLn
wh ba
eff
+⎟⎠⎞
⎜⎝⎛+⎥
⎦
⎤⎢⎣
⎡
++
⎟⎠⎞
⎜⎝⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛+−
⎟⎠⎞
⎜⎝⎛ +
−+
+ −
εε
εεε
Calculation of W/h (width of microstrip/substrate thickness) for a given characteristic impedance and effective relative permitivity:
ro
rr
r
ro
2Z377 = B where
0.5170.293+1)Ln(B2
1+1)Ln(2B1B2
hW
2  44 Z For
επ
εεε
π
ε
⎭⎬⎫
⎩⎨⎧
⎥⎦
⎤⎢⎣
⎡−=
≤
⎟⎟⎠
⎞⎜⎜⎝
⎛+
+−
++
−=
≥
rr
rr
2
ro
12.0226.0.11
60.
21 = n where
28
hW
2  44 Z For
εεεε
ε
Zo
ee
n
n
3.16 INTERDIGITAL MICROSTRIP CAPACITORS [11]
Normally resonators need to be lightly coupled in order to maintain a high Q, this can be done by using a filter arrangement or by using very small value capacitors. Normal chip capacitors can go as low as 0.1pF, but for smaller capacitance it is convenient to use transmission line interdigital capacitors.
Sheet
12 of 20
Literature on the subject is very scarce so a basic design formula was used to get the initial dimensions and the final dimensions were optimised during RF simulations. The basic formula for the interdigital capacitor is given by:
fingers long 600um =
cm06.01)0.83(2
0.05 = L)1(N*0.83
C
:be willfingers the of length the thenfingers 2 are there that assume weif
and capacitor 0.05pF a want weif example For10um of widthfinger a and
5um of spacing finger a assumes formula This
pF in eCapacitanc C cm in fingers of Length = L fingers of Number = N Where
L).1(N 0.83 = C
F
F
F
==−
=
−
To further aid in the evaluation of a interdigital capacitor the model was analysed in Libra RF CAD with a finger width and gaps of 0.1mm and number of fingers 2,3 & 4. The graph (Figure 15) shows the relationship between capacitance and finger length.
0
0.5
1
1.5
2
2.5
3
0 0.5 1 1.5 2 2.5 3Finger Length mm
Cap
acita
nce
pF
Figure 15 Graph of a microstrip interdigital capacitor vs capacitance. The plots were calculated by analysis on HP/Eesof libra.
Transmission lines may be used as single resonators capacitively coupled to the active device, but also they may be configured as a microstrip bandpass filter. The basic principle involves using open circuit transmission lines of electrical length 180 degrees, which is equivalent to a ‘tuned circuit’ parallel resonator. What tends to differ in the topographies are the ways in which the resonators are coupled together. The resonators can be end coupled or parallel coupled using the gaps between them as the low value coupling capacitors. It is also possible to use interdigital capacitors to generate coupling capacitors less than 1pF
3.17 VARACTORS [12] Voltage variable capacitors or tuning diodes are best described as diode capacitors employing the junction capacitance of a reverse biased PN junction. The capacitance of these devices varies inversely with the applied reverse bias voltage. The general equation for calculating the capacitance of the varactor is :
exponent eCapacitanc = and 0.7V)(~ potential contact junction=
voltage, applied =V e;capacitanc diodeC where
)(
D
γφ
φ γ
=
+=
VCC D
J
3.18 DESIGN EXAMPLE OF A VARACTOR DIODE
The following section describes how information from a data sheet can be used to predict the capacitance of the varactor diode for a given reverse bias. For this example the varactor diode selected is a Macom Tuning diode type MA46H071. The data sheet gives the following parameters for the diode: C = 0.91.1pF @ 4V;cap ratio Cto/Ct20 = 5.5;Gamma=0.75;Q @ 50MHz=4500
75.0
12
0.7512
JD
)7.0(E19.3 =
bias given a for ecapacitanc a calculate to therefore,
3.19pF 0.7)+(41E =
).(C = C give to rearrange )(
+=
=
++
=
−
VC
VV
CC
J
DJ
γγ φ
φ
Sheet
13 of 20
This is obviously the ideal case as it does not take into account the case parasitics
3.19 TUNING RATIOS The tuning or capacitance ratio, TR, denotes the ratio of capacitance obtained with two values of applied bias voltage. This ratio is given by the following:
γ
φφ⎥⎦
⎤⎢⎣
⎡+2
1
1J
2J
V+V =
)V(C)V(C = TR
where CJ(V1) = junction capacitance at V1;CJ(V2) = junction capacitance at V2 (V1>V2).
3.20 CIRCUIT Q The Q of the varactor can be very important, because the varactor usually directly forms the tuned circuit and the overall Q is dominated by the worst Q factor. The Q of tuning diode capacitors falls off at high frequencies because of the series bulk resistance of the silicon used in the diode. The Q also falls off at low frequencies because of the back resistance of the reversebiased diode. The equivalent circuit of a tuning diode is often shown in the form given below in Figure 16. Rp
Cj
Rs Ls Ls’
Cc
Figure 16 Equivalent circuit of a typical varctor diode together with case and lead parasitic components.
Where Rp = Parallel resistance /back resistance of the diode. Rs = Bulk resistance of the silicon in the diode. Ls’ = External lead inductance. Ls = Internal lead inductance. Cc = Case Capacitance. Normally the lead inductance and case capacitance can be ignored, which results in a simplified circuit shown in Figure 17.
Rp
Cj
Rs
Figure 17 Simplified model of a typical varactor diode with parasitic reactance removed.
The resulting Q for the above circuit is given by :
ΩΩ
=
9
22
2
30x10 = Rp & 1 = Rs Typically
Rs.RpC)2(+Rp+Rs
C.Rp2f
fQπ
π
Therefore for a MA/COM MA46H071 we would expect the following Q’s at different frequencies as shown in the table below:
f(GHz) Q 0.05 3500 2 88 6 30
The degradation of Q at microwave frequencies means that the varactor, has to be lightly coupled, or Q transformed in order not to load the resonant circuit, lowering the loaded Q with the resultant degradation in phase noise performance. The following graph (Figure 18) of the varactor diode frequency response shows that at low frequencies the Q is dominated by the parallel term ie Qp = 2πf.Rp.C and at high frequencies by the series term Qs = 1/(2πfRs.C).
Sheet
14 of 20
0.1
1
10
100
1000
10000
100000
1 100 10000 1000000 100000000 1E+10
Frequency (Hz)
Figure 18 Plot of Q against frequency. The vertical scale is Q and the horizontal scale is frequency in Hz.
3.21 TEMPERATURE VARIATION The two mechanisms for the variation of capacitance over temperature are (i) contact potential and (ii) case capacitance. The contact potential will vary at 2.2mV/°C thus for the MACom diode we would expect the following temperature drifts as shown in Table 2.
V Cj Cj+1 decC Diff ppm/degC1 2.1426636 2.1405863 0.0020773 2077.2922 1.5144951 1.5135702 0.0009249 924.865384 0.9993492 0.9989986 0.0003507 350.691746 0.7660103 0.7658217 0.0001886 188.590138 0.6297254 0.629606 0.0001194 119.40427
10 0.5392037 0.5391205 8.313E05 83.13327512 0.4741742 0.4741126 6.16E05 61.59596914 0.4249156 0.4248679 4.769E05 47.68836416 0.3861476 0.3861094 3.815E05 38.1479118 0.3547394 0.3547081 3.13E05 31.29731720 0.32871 0.3286838 2.62E05 26.199086
Table 2 Calculated data of the capacitance variation with temperature for the MACom varactor diode.
3.22 TEMPERATURE COMPENSATION A popular method of temperature compensation involves the use of a forward bias diode. The voltage drop of a forward biased diode decreases as the temperature rises, therefore applying a changing voltage to the tuning diode. For the circuit to be effective the compensating diode must be thermally coupled to the varactor to be corrected. Figure 19 shows a method for temperature compensating a varactor diode.
R
Compensating
Vin
Varactor
Figure 19 Schematic circuit diagram, for temperature compensation, of a varactor diode
Normally, however the varactor is part of a feedback loop, which controls the frequency of oscillation eg in a PLL system. In this case, the temperature effects are generally accounted for in the loop so that external compensation is not required.
4 LOADED & UNLOADED Q [13,14,15] 4.1 UNLOADED Q
The earlier section described how the Q of a tuning diode varies over frequency and can be quite low (~ 30) at microwave frequencies. This will obviously have an effect on the loaded Q of a circuit where the individual components may have higher Q’s in the hundred’s. We therefore need to estimate the loaded Q of a resonator, with a varactor connected, in order to calculate the phase noise performance of the oscillator. It is useful to be able to simplify the equivalent Q of a circuit, so the effect of the varactor Q can be evaluated. Some basic definitions of Q in the series and parallel form are:
LR = Q External
LR = R.C =
Q circuit Parallel Unloaded
RL = Q External
R.C1 =
RL =
Q circuit Series Unloaded
L
L
ooo
o
o
o
ωωω
ωω
ω
We can take the specified Q values for inductors and capacitors from the data sheets and calculate the equivalent series or parallel resistance that distinguish the component from an ‘ideal’ component to one with a finite Q. Once the resistance has been calculated, the circuit can be simplified down to a single component or a series/parallel combination of two circuits, to allow calculation of the unloaded circuit Q. The follow
Sheet
15 of 20
ing example (shown in Figure 20) shows a simple LC tuned circuit but with losses added.
L ~ 2.5uH
Q =100 @ 100MHz
C = 1pF
Q = 200 @ 100MHz
RIND=163KΩ
RC=318KΩ
Figure 20 Simple LC circuit with component losses added
The equivalent parallel loss resistance for each component was calculated as follows
67 E108*E1*E100*2 L.
R Q
and R.C. circuit of Q Unloaded
K108 K318K163
318K*163K
//RR resistance equivalent Parallel
318K E1*E100*2
200R and K163
E5.2*E100*100*2R
C.QR and L..R
3126
o
o
CPLP
126CP
66LP
oCPoPL
===
=∴
=+
=
=
Ω==Ω=
=
==
−
−
−
πω
ω
π
π
ωωQ
A useful transformation from series equivalent resistive loss (Rs) to parallel equivalent resistive loss (Rp) is given as –
Xp Xs and Rs*)(Q Rp
10 Q For
Rs*)1(Q Rp
10 Q For
2
2
≈≈
>
+=
<
These transformations are only valid at one frequency, as they involve the component reactance, which is frequency dependant.
4.2 LOADED Q The loaded Q of a resonant circuit is dependent on three main factors:
(1) The source impedance (Rs). (2) The load impedance (RL). (3) The component Q.
The circuit used in the example of section 3.5.1 is to be loaded in a 50ohm system as shown in Figure 21.
L ~ 2.5uH
Q =100 @ 100MHz
C = 1pF
Q = 200 @ 100MHz
RRES=108KΩ
RL=50Ω
Rs = 50Ω
Figure 21 Simple LC resonant circuit loaded, with 50ohm source and load impedances.
The addition of the source and load impedances will degrade the loaded Q of the circuit as they will effectively be in parallel with the high impedance resonant circuit as shown below in Figure 22.
L ~ 2.5uH
Q =100 @ 100MHz
C = 1pF
Q = 200 @ 100MHz
RRES=108KΩ
RL=50Ω
Rs = 50Ω
=
Requ = 24.99Ω
Figure 22 LC resonant circuit reduced to one resistive loss component.
The loaded Q of the circuit of Figure 22 is:
0.0159 2.5E*100E*2
24.99 Rp Q 66o
===πω L
This dramatic decrease in Q will give the simple LC network a 3dB bandwidth of:
!! GHz6 0.0159
100MHz f ff
o
==Δ∴Δ
=Q
Sheet
16 of 20
To improve the loaded Q, given a restraining source and load impedance, we could alter the value of Xp. This however, results in either very high inductors, or very low capacitors. If we are restrained from altering the value of Xp we can either use a tapped L or C transformer or coupling L or C.
4.3 Q TRANSFORMATION The circuits shown in Figure 23 show the two methods of transforming the Q of a circuit, by the use of impedance transformers.
Rs RL
2
C2C11Rs Rs' ⎟
⎠⎞
⎜⎝⎛ +=
Tapped C circuit
Rs RL
2
n1nRs Rs' ⎟⎠⎞
⎜⎝⎛=
Tapped L circuit
n1 n
Figure 23 Impedance transformation circuits (Tapped L & C). These circuits can be used to increase the effective source & or load impedances in order to improve the loaded Q of a circuit.
If we require a Q of 10 then this will equate to a parallel equivalent resistance of:
18pFC2 and 1.055pFC1 have could We
pF1C2C1C2*C1 and C2*18 C1 Therfore
18 150
18K C2C1 1
RsRs'
C2C11Rs Rs'
r,transforme tapped capacitor a using 18K to impedance source our transform to need weTherefore
18.37K Rs for solve 081Rs
108K*Rs 15707
108K RL nscalculatio previous From RLRsRL*Rs 15707
15.7K E6.2*100E*2*10 Q.Xp Rp
2
66
==
=+
=
=⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛∴⎟
⎠⎞
⎜⎝⎛ +=
Ω
Ω=Ω+Ω
=
Ω=+
=
Ω=== −
K
π
The final circuit designed to give a Q of 10 is shown in Figure 24.
Rs = 50Ω
R L = 108KΩ
1pF ~ 1.055pFpF181.055pF*18pF C1//C2
16K3 1.055
18150 Rs'
C2C11Rs Rs'
2
2
+=
Ω=⎟⎠⎞
⎜⎝⎛ +=
⎟⎠⎞
⎜⎝⎛ +=
L = 2.5uH
C2=18pF
C1= 1.055pF
Figure 24 LC circuit with a capacitor tapped impedance transformer, to give a loaded Q of 10, when loaded with a source impedance of 50 ohms.
Sheet
17 of 20
Equally we could use a coupling capacitor between the source impedance and resonant circuit such that the resistance will equal 16KΩ.
0.1pF E16*100E*2
1 C
16K~ 50 16K
100MHz at reactance capacitor coupling Required
36coupling ==∴
ΩΩΩ=
π
The addition of a coupling capacitor to the circuit is shown in Figure 25.
Rs = 50Ω
RL = 108KΩ
L = 2.5uH
C=1pF
Cc=0.1pF
Figure 25 Addition of a coupling capacitor to the simple LC to increase the loaded Q to ~10
The required coupling capacitor is very small at 0.1pF and is probably impracticable at 100MHz. However this size of capacitor can be realised at microwave frequencies by the use of a microstrip gap or a interdigital capacitor (as described in section 3.16).
4.4 INSERTION LOSS OF RESONATOR The insertion loss of a resonator is important in oscillator design as there needs to be enough loop gain to allow oscillation. A high insertion loss resonator may require two stages of amplification around the loop that will add to the size, power consumption and complexity of the oscillator. The insertion loss of the resonator is a function of loaded and unloaded Q ie:
Q unloaded Q and Q loaded Q where
1 20log (dB) loss Insertion
UL
U
L
==
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
4.5 DESIGN EXAMPLE FOR A VARACTOR CONTROLLED RESONATOR
Consider the varactor resonator shown below in Figure 26. The capacitor combination can be simplified to a single capacitor that then forms a parallel resonant circuit with the inductor. In this example, we assume a source impedance of 50ohms.
Cdiode ~ 1pF Q =30 @ 2GHz
L ~ 7.6nH
Q =150 @ 1GHz C ~5pF
Q = 100 @ 5GHzRIND=7163Ω
RC=0.06Ω
Rcdiode= 2.65Ω
Figure 26 Schematic circuit diagram of a varactor controlled resonator for use at 2GHz. The equivalent loss resistances have been calculated using the equations of section 3.5.1
This circuit of Figure 26 can be simplified to that shown in Figure 27. The loss resistances of the capacitor arm can be added and converted to a parallel loss resistance that can be added to the loss of the inductor. The equivalent capacitor now equals 0.833pF ie 1pF // 5pF.
L ~ 7.6nH
Q =150 @ 1GHz
RIND=7163Ω
Rcdiode= 2.71Ω(series)
3343Ω (parallel)
Q of capcitor+diode
~35
Figure 27 Simplified varactor controlled resonator for use at 2GHz
Sheet
18 of 20
( )
( ) Ω===
=+
==−
331971.2*35 *Q R
35 0.062.65
E833.0*2E*.21
(series) RsXs Q
:loss parallel tolosscapacitor series of Conversion
22P
129
sR
π
Now we can calculate the equivalent loss resistance and the unloaded Q of the circuit:
23.7 E6.7*2E*2
2268
XpRp circuit the of Q Unloaded
2268 3319 // 7163 is circuit resonant the across resistance loss Equivalent
99 ==
=
Ω=ΩΩ
−π
We can see that the low Q of the inductor is going to dominate the unloaded Q of the parallel circuit. Now, if we load the circuit with 50ohm source and load impedances, (as shown in Figure 28) we can calculate the loaded Q of the circuit.
Cdiode ~ 0.833pF
L ~ 7.6nH
RRES=2268Ω RL = 50Ω RS = 50Ω
Figure 28 Resonant varactor circuit loaded with 50ohm source and load impedances.
The loaded Q of the circuit will be the parallel combination of the equivalent parallel resistance of the resonant circuit with the source and load impedances ie
0.26 E6.7*2E*2
24.73 XpRp of Q loaded a give willThis
24.73 Rp 2268
1501
501
Rp1
99 ==
Ω=∴++=
π The circuit was analysed on the CAD to confirm the Q calculations and is shown in Figure 29.
0.2 2.2 4.2 6.2 8Frequency (GHz)
Graph 1
10
8
6
4
2
0 DB(S[2,1]) *Varactor
Figure 29 Varactor resonator circuit loaded, with 50ohm source and load impedances. The Q was graphically measured at ~ 0.28.
The loaded Q is lower than the unloaded Q due to the damping effect of the low value source impedance. An oscillator with a resonant circuit with a Q of 0.24 will be very unsatisfactory, so a means of increasing the loaded Q is required. We cannot do much about the tuned circuit, but we can modify the source and load impedances either by the used of a C/L tapped transformer or by the use of coupling capacitors. For this example we shall consider the use of coupling capacitors on the varactor circuit. Figure 30 shows the implementation of coupling capacitors.
Cdiode ~ 0.833pF
L ~ 7.6nH
RRES=2268Ω
RL = 50Ω
RS = 50ΩCoupling
C Coupling
C
Figure 30 Varactor tuned circuit, with coupling capacitors, added between 50 ohm source and load impedances.
If we decide that we require a loaded Q of say 10, then we can calculate the value of the source resistors, that when placed in parallel with the tuned circuit, will give the required value of Q ie
Sheet
19 of 20
0.4pF (198) * 2E * 2
1 = capacitor series of Value
198 Rp 2268
111Rp1
95.5=10*7.6E*2E*2 = Rp .
10 of Q a give to resistance parallel Total
9
99
=
Ω=∴++=
Ω∴=
=
π
π
RLRs
RpQLX
This value of series coupling capacitor is very small but can be realised at microwave frequencies by the use of a interdigital microstrip capacitor. The coupling capacitors were added to the CAD model and analysed to confirm a Q of ~ 10, the plot is shown in Figure 31. Predicted insertion loss:
4.76dB 23.710120log
QQ120log (dB) loss
U
L =⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
1.5 2 2.5Frequency (GHz)
Loaded Q
10
8
6
4
2
0DB(S[2,1]) *Varactor
Figure 31 Varactor resonator circuit loaded, with 50ohm source, load impedances and coupling capacitors. The Q was graphically measured at ~ 10, with a resonator insertion loss of ~ 3.85dB.
4.6 CONCLUSION/SUMMARY This paper described the design of various types of resonator suitable in oscillator designs. For each type the design equations for frequency and unloaded Q were given that are required in order for an oscillator to meet a given phase noise specification. As most oscillators require electronic tuning the various design aspects of varactor diodes was given including the effects of temperature and a method of temperature compensation. Various methods of Q transformation were given using the tapped ‘C’ and tapped ‘L’ methods together with worked examples.
Finally a lumped element resonator with varactor diode was given as an example in designing a resonator to give a loaded Q of 10 and at the same time an acceptable insertion loss of ~4dB.
4.7 REFERENCES [1] Oscillator Design and Simulation, Randall W Rhea, 1995 ,Noble Publishing, ISBN 1884932304, chap 4. [2] RF Design Guide, Peter Vizmuller, 1995, Artechhouse, ISBN 0890067546, p237. [3] TransTech Application Note 1008,1010 & 1015, from www.alphaind.com. [4] Microwave Circuit Design – Using Linear and Nonlinear Techniques ,George D Vendelin,Anthony M Pavio and Ulrich L Rohde, 1990. Wiley – Interscience ISBN 0471580600, p 403. [5] Dielectric Resonators, D Kajfez & P Guillon, 1990, Vector fields, ISBN 0930071042. [6] Microwave Engineering, David Pozar, 1993, Addison Wesley, ISBN 0201504189, p354358. [7] Card V3 Dielectric resonator design software by Scillasoft Consultants for TransTech, www.alphaind.com. [8] Microstrip coupling model by Patrick Champagne, “Better coupling model of DR to microstrip ensures repeatability”, Microwaves & RF Sept 1987, p113118. [9] MSc Solid State Physics Course Notes for Unit P503 – Transmission line theory ,1999, Dr D Nixon. [10] Microstrip Circuit Analysis , David H Schrader,1995 ,PrenticeHall , ISBN 0135885345, p3032. [11] Microwave Field Effect Transistors, Raymond S Pengelly, 1994, Noble Publishing, ISBN 1884932509, p473476. [12] Motorola Semiconductor Application Note – Tuning diode design technique – AN847/D, 1994. [13] Radio Frequency Design – Wes Hayward, 1994, The American Radio Relay League, ISBN 0872594920, p5459.
Sheet
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[14] Oscillator Design and Simulation, Randall W Rhea, 1995, Noble Publishing, ISBN 1884932304, p 35. [15] RF Circuit Design, Chris Bowick, 1997, Butterworth & Heinemann, ISBN 0750699469.