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Oscillator Resonator Design Tutorial J P Silver

E-mail: john@rfic.co.uk

1 ABSTRACT This paper discusses the design of various types of resonator that form the heart of any oscillator de-sign.

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 defi-nitions 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 per-formance 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 resona-tors.

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 two-element 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 cir-cuit.

Q = LR

ω..2

Q = 2.R.Lω

Figure 1 Schematic of a two element, lumped resona-tor, 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 trans-mission while the parallel network allows signal trans-mission. The main problem with such a simple resona-tor 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 inter-digital 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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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 ele-ment 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 quarter-wave coaxial resonator is formed by short-ing the centre conductor of a coaxial line to its shield at one end, leaving the other end open-circuited. 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:

1-12-r

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 reso-nator is made from semi-rigid coaxial cable that con-tains a dielectric of PTFE, which has a relative permit-tivity 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 coax-ial cable from the conductivity of the sheath and the loss tangent of the dielectric will set the Q of the reso-nator. Most coaxial cables especially semi-rigid 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 rigid-semi

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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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 - 1.56E-8Ω.m - Gold - 2.04E-8Ω.m - Silver - 1.63 E-8Ω.m - Nylon 3.0 109-1011Ω.m 0.012@3GHz PTFE 2-2.1 1E-16 0.0004@10GHz

Polythene HD

2.25 >1014Ω.m 0.0004@10GHz

PVC flexi 4.5 109-1012Ω.m

Table 1 Design data for a range of materials com-monly 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 quarter-wave coaxial resonator is formed, by plating a piece of dielectric material with a high relative per-mittivity using a highly conductive metal. A cylindrical hole is formed along the axis of a cylin-der 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 resona-tor.

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 short-circuited coaxial line elements simulate high-Q, temperature stable ‘ideal’

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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 ca-pacitance. The resonator is usually used below the self-resonant 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’ induc-tor the resonator must be used below the self-resonant 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 silver-plated ceramic resona-tor 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= 1-4 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 self-resonant frequency of 1.241GHz.

3.10 DIELECTRIC RESONATOR [7] At lower frequencies the length of W/d ratio of a coax-ial 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 low-loss metal. In addition they are mounted on planer circuits as shown below (figure 35) and are coupled to a trans-mission 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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a

b

Figure 8 Schematic diagram of a dielectric reso-nator showing the key dimensions.

The most common resonant mode in dielectric resona-tors 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 ap-proximate estimation of the resonant frequency in TE01δ mode of an isolated dielectric resonator, the fol-lowing 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 dielec-tric resonator for a frequency of ~ 7GHz. A manufac-turer of dielectric resonators – Transtech can supply two relative permittivities of 30 and 38. The Trans-Tech D8733-0305-137 puck was selected with the following parameters, εr = 30, Diameter = 7.75mm, Height = 3.48mm, the resonant frequency can be esti-mated 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 free-space. 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 ac-count 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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( ) ( )[ ]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 micro-strip 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 micro-strip line and the corresponding circuit diagram. The resistor L simulates the coupling of the L-C 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

QQ

+1Q

Q

5000 of Q unloaded a withResonator a use weIf

L

UUL

β

ββ

Trans-Tech have a CAD package [15] to calculate various design parameters using their dielectric resona-tors. We can use the CAD package to calculate a plot of the coupling coefficient β vs distance from the cen-tre of the micro-strip 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 dis-tance 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 mi-cro-strip line centre.

3.13 TRANSMISSION LINE RESONATOR [9] Over a narrow bandwidth L-C lumped components can be realised using short-circuit and open-circuit trans-mission lines. If we analyse a transmission line termi-nated in a load ZL we can define the transformed im-pedance 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 imped-ance looking into a load ZL via a length of transmis-sion line. If we now have the case where the transmis-sion 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 divid-ing 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 design-ing 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 sub-strate 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 open-circuit stub we can calculate the electrical length re-quired for an inductance of 0.7nH. Therefore a open-circuit 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 em-pirical 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 equa-tions that are used to calculate the dimensions of the micro-strip lines and characteristic impedance [8]. The first equation describes the effective relative permittiv-ity which, differs from the specified value due the width of the micro-strip 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 micro-strip/substrate thickness) for a given characteristic impedance and effective relative permitivity:

ro

rr

r

ro

2Z377 = B where

0.517-0.293+1)-Ln(B2

1+1)-Ln(2B-1-B2

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 INTER-DIGITAL MICRO-STRIP CAPACITORS [11]

Normally resonators need to be lightly coupled in or-der to maintain a high Q, this can be done by using a filter arrangement or by using very small value capaci-tors. Normal chip capacitors can go as low as 0.1pF, but for smaller capacitance it is convenient to use transmission line inter-digital capacitors.

Sheet

12 of 20

Literature on the subject is very scarce so a basic de-sign formula was used to get the initial dimensions and the final dimensions were optimised during RF simula-tions. The basic formula for the inter-digital 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 inter-digital ca-pacitor the model was analysed in Libra RF CAD with a finger width and gaps of 0.1mm and number of fin-gers 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 micro-strip inter-digital ca-pacitor 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 micro-strip band-pass filter. The basic principle involves using open circuit trans-mission 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 resona-tors can be end coupled or parallel coupled using the gaps between them as the low value coupling capaci-tors. It is also possible to use inter-digital 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 ca-pacitance 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 Tun-ing diode type MA46H071. The data sheet gives the following parameters for the diode:- C = 0.9-1.1pF @ 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 fre-quencies 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 re-verse-biased 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 di-ode. 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 ex-pect 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 trans-formed in order not to load the resonant circuit, lower-ing 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 verti-cal 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 8.313E-05 83.13327512 0.4741742 0.4741126 6.16E-05 61.59596914 0.4249156 0.4248679 4.769E-05 47.68836416 0.3861476 0.3861094 3.815E-05 38.1479118 0.3547394 0.3547081 3.13E-05 31.29731720 0.32871 0.3286838 2.62E-05 26.199086

Table 2 Calculated data of the capacitance varia-tion with temperature for the MACom varactor diode.

3.22 TEMPERATURE COMPENSATION A popular method of temperature compensation in-volves the use of a forward bias diode. The voltage drop of a forward biased diode decreases as the tem-perature 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 tem-perature 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 al-low calculation of the unloaded circuit Q. The follow-

Sheet

15 of 20

ing example (shown in Figure 20) shows a simple L-C tuned circuit but with losses added.

L ~ 2.5uH

Q =100 @ 100MHz

C = 1pF

Q = 200 @ 100MHz

RIND=163KΩ

RC=318KΩ

Figure 20 Simple L-C circuit with component losses added

The equivalent parallel loss resistance for each com-ponent 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 resis-tive 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 fre-quency 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 50-ohm 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 L-C resonant circuit loaded, with 50-ohm source and load impedances.

The addition of the source and load impedances will degrade the loaded Q of the circuit as they will effec-tively 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 L-C resonant circuit reduced to one re-sistive loss component.

The loaded Q of the circuit of Figure 22 is:-

0.0159 2.5E*100E*2

24.99 Rp Q 6-6o

===πω L

This dramatic decrease in Q will give the simple L-C 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 imped-ance 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 paral-lel equivalent resistance of:

18pFC2 and 1.055pFC1 have could We

pF1C2C1C2*C1 and C2*18 C1 Therfore

18 1-50

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 L-C circuit with a capacitor tapped im-pedance 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 L-C 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 fre-quencies by the use of a microstrip gap or a inter-digital capacitor (as described in section 3.16).

4.4 INSERTION LOSS OF RESONATOR The insertion loss of a resonator is important in oscilla-tor 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 com-plexity of the oscillator. The insertion loss of the reso-nator is a function of loaded and unloaded Q ie:-

Q unloaded Q and Q loaded Q where

QQ

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 simpli-fied 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 equiva-lent 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 capaci-tor 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 resona-tor 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 50-ohm 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 combi-nation of the equivalent parallel resistance of the reso-nant 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

9-9 ==

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 50-ohm source and load impedances. The Q was graphi-cally 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 trans-former or by the use of coupling capacitors. For this example we shall consider the use of coupling capaci-tors on the varactor circuit. Figure 30 shows the im-plementation 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

9-9

=

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 inter-digital 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 23.710-120log-

QQ-120log- (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 50-ohm source, load impedances and coupling ca-pacitors. 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 accept-able insertion loss of ~4dB.

4.7 REFERENCES [1] Oscillator Design and Simulation, Randall W Rhea, 1995 ,Noble Publishing, ISBN 1-884932-30-4, chap 4. [2] RF Design Guide, Peter Vizmuller, 1995, Artech-house, ISBN 0-89006-754-6, p237. [3] Trans-Tech 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 0-471-58060-0, p 403. [5] Dielectric Resonators, D Kajfez & P Guillon, 1990, Vector fields, ISBN 0-930071-04-2. [6] Microwave Engineering, David Pozar, 1993, Addi-son Wesley, ISBN 0-201-50418-9, p354-358. [7] Card V3 Dielectric resonator design software by Scillasoft Consultants for Trans-Tech, www.alphaind.com. [8] Micro-strip coupling model by Patrick Champagne, “Better coupling model of DR to micro-strip ensures repeatability”, Microwaves & RF Sept 1987, p113-118. [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 ,Prentice-Hall , ISBN 0-13-588534-5, p30-32. [11] Microwave Field Effect Transistors, Raymond S Pengelly, 1994, Noble Publishing, ISBN 1-884932-50-9, p473-476. [12] Motorola Semiconductor Application Note – Tun-ing diode design technique – AN847/D, 1994. [13] Radio Frequency Design – Wes Hayward, 1994, The American Radio Relay League, ISBN 0-87259-492-0, p54-59.

Sheet

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[14] Oscillator Design and Simulation, Randall W Rhea, 1995, Noble Publishing, ISBN 1-884932-30-4, p 35. [15] RF Circuit Design, Chris Bowick, 1997, Butter-worth & Heinemann, ISBN 0750699469.