Cavity Resonators

Fabrication Processes for RF and Microwave Circuits

3.6 Characterization of Materials for RF and Microwave Circuits

3.6.1 Measurement of Dielectric Loss and Dielectric Constant

3.6.1.1 Cavity Resonators

A cavity resonator can be formed from a section of hollow metal rectangular waveguide closed at both ends, with the dimensions shown in Figure 3.17.

If an electromagnetic wave is introduced into the cavity, a standing wave pattern will be set up if the length (l) of the cavity is some multiple of half a guide wavelength, i.e.

l=p×𝜆g

2, (3.12)

wherepis an integer. In general, for transverse electric modes, the mode within a resonant cavity is denoted by TE_mnp wheremandnhave been defined in Appendix 1.C.

We can find an expression for the resonant frequency of the cavity by substituting from Eqs. (3.12) and (1.83)–(1.72) to

give (p

2l )2

+ (m

2a )2

+ (n

2b )2

= (f

c )2

. (3.13)

The resonant frequency is then

f =c√(p 2l

)2

+ (m

2a )2

+ (n

2b )2

. (3.14)

The dominant cavity mode is represented by TE₁₀₁, and from Eq. (3.14) its resonant frequency will be f_o=c

√(1 2l

)2

+ ( 1

2a )2

. (3.15)

TheQ-factor of a resonant cavity is defined in the normal manner as Q= 𝜔o×average energy stored

power dissipated , (3.16)

where𝜔o=2𝜋f_o, andf_ois the resonant frequency. The energy stored in the cavity is given by the volume integral W= 1

2𝜇o∫_v|H|²dv, (3.17)

where |H| is the peak magnetic field strength within the cavity. The power dissipated will be due to the finite conductivity of the walls which allows wall currents to flow, and leads to a power loss given by the surface integral

P= 1 2R_s

∫_s|Ht|²ds, (3.18)

whereH_tis the magnetic field tangential to the walls andR_sis the surface resistance. Substituting from Eqs. (3.17) and (3.18) into (3.16) gives the cavityQ-factor as

Q_u=𝜔o𝜇o∫_v|H|²dv

R_s∫_s|H_t|²ds . (3.19)

k k aperture in the cavity wall. Evaluation of the integrals in Eq. (3.19) is straightforward [21, 22] and leads to the following

expression for the unloadedQof a rectangular waveguide cavity supporting the dominant TE₁₀₁mode Q_u= abl(a²+l²)

𝛿s(2a³b+2l³b+a³l+l³a), (3.20) where𝛿sis the skin depth, and the other parameters are defined in Figure 3.17.

The unloadedQof a resonant cavity is often written in a more approximate format as Q_u= 2

𝛿_s ×V

A, (3.21)

whereVandAare the volume and surface area, respectively, of the cavity, but this expression is only correct for cavities that have a cubic shape. However, Eq. (3.21) is useful in that it indicates that in general the highestQvalues are obtained with simple cavity shapes that have a high ratio of volume to surface area.

Example 3.7 A copper rectangular waveguide resonant cavity supporting the TE₁₀₁mode has the following dimen- sions:a=22.86 mm,b=10.16 mm,l=20 mm. Determine the resonant frequency and the unloadedQof the cavity.

[𝜎(copper)=5.8×10⁷S/m]

Solution

To find the resonant frequency At resonance l= 𝜆g

2 ⇒ 𝜆g=2l=2×20 mm=40 mm, 𝜆c=2a=2×22.86 mm=45.72 mm.

Using Eq. (1.72): 1 𝜆²o

= 1

(45.72)² + 1

(40)² ⇒ 𝜆o=30.10 mm, 𝜆o=30.10 mm ⇒ f_o=9.97 GHz.

Resonant frequency=9.97 GHz.

To find theQof the cavity The skin depth is given by Eq. (2.8):

𝛿s= 1

√𝜋𝜇of𝜎

= 1

√𝜋×4𝜋×10⁻⁷×9.97×10⁹×5.8×10⁷ m

=0.66μm.

Using Eq. (3.20):

Q_u= 1 𝛿s

×x y, where

x=22.86×10.16×20× [22.86²+20²] ×10⁻³=4285.52,

y= [2×22.86³×10.6] + [2×20³×10.16] + [22.86³×20] + [20³×22.86]

=827 109.5619 giving

Q_u= 1

0.66×10⁻⁶ × 4285.52

827 109.56 =7850.49.

UnloadedQof cavity=7850.49.

So far we have considered unloaded cavities, i.e. cavities without any means of exciting the signal within the cavity. TheQ of the cavity is then denoted byQ_u. In order to excite a signal within a cavity, there must, in practice, be either a coupling hole

k k

3.6 Characterization of Materials for RF and Microwave Circuits 103

in one of the walls of the cavity, or a coupling loop of wire projecting into the cavity. In either case the coupling mechanism will have the effect of loading the cavity.

The unloaded cavity can be represented by a seriesRLCresonant circuit, and theQis then given by Q_u=𝜔L

R .

Coupling the cavity to the outside world will provide additional loss, but no additional reactance, since any additional reactance from the coupling will be compensated for by alteration of the internal reactance of the cavity to maintain the original resonant frequency. However, the external load resistance will be coupled into the cavity, and will add to the resis- tance (R) of the resonant circuit representing the cavity, thus lowering theQof the cavity. If the external (load) resistance coupled into the cavity isR_L, the loadedQ,Q_L, of the cavity will be given by

Q_L= 𝜔L R+R_L. Inverting each side of the last expression gives

1

Q_L = R+R_L 𝜔L = R

𝜔L+R_L 𝜔L,

i.e. 1

Q_L = 1 Q_u + 1

Q_e, (3.22)

whereQ_eis defined as the external Qof the cavity;Q_ethus represents the ratio of the same total stored energy as the unloaded cavity to only the loss coupled from the outside world.

One of the common methods of making a cavity resonator is to use a short-circuited length of rectangular waveguide, with a transverse plate containing a coupling iris, positioned at some distance from the short circuit, as shown in Figure 3.18.

The lengthlis thus the length of the resonant cavity.

In Figure 3.18, a small circular coupling iris is shown at the centre of the transverse plate. This iris behaves as a shunt inductive susceptance with a normalized value given by

b_L= −j3ab

8𝛽r³, (3.23)

where𝛽=^2𝜋∕_𝜆gand𝜆gis the guide wavelength. Thus, in the plane PP^′the equivalent circuit of the resonator can be repre- sented as a normalized susceptive inductanceb_Lin parallel with a normalized admittancey_in, which represents the input admittance of a length of waveguide terminated with a short circuit. From Eq. (1.33) we know

y_in= −jcot𝛽l. (3.24)

Therefore, for resonance we require

b_L+y_in=0, (3.25)

b

a

2r Short circuit Transverse

plate

≡ l

y_in b_L P P

b

Pʹ Pʹ

Figure 3.18 Waveguide cavity resonator and its equivalent circuit.

k k

8𝛽r³ +cot𝛽l=0. (3.26)

Since the iris presents a shunt inductance the length of the cavity must be slightly less than𝜆g/2 in order thaty_in provides a capacitance to cancel the inductance of the iris. However, the value of shunt inductance provided by a typical iris is very small, and only a very slight reduction in the cavity length is needed, as shown by Example 3.8.

Example 3.8 A 6 GHz cavity resonator having the form shown in Figure 3.18 is to be made using a rectangular waveguide having the following internal dimensions:a=40.39 mm,b=20.19 mm. If the circular coupling iris has a radius of 3 mm, determine the required length (l) of the cavity.

Solution

At 6 GHz:𝜆o=3×10⁸

6×10⁹ m=50 mm.

Using Eq. (1.72) to find the guide wavelength:

1

(50)² = 1

(2×40.39)²+ 1 𝜆²g

⇒ 𝜆_g=63.66 mm.

Then,

𝛽= 2𝜋

63.66 rad∕m=0.0987 rad∕m.

Substituting in Eq. (3.26):

3×40.39×20.19

8×0.0987×3³ +cot𝛽l=0, 114.74+cot𝛽l=0,

𝛽l=0.9972𝜋, l=0.9972𝜋×63.66

2𝜋 mm=31.74 mm.

Comment: The calculated length is only slightly less than𝜆_g/2 (=63.66/2 mm=31.83 mm), which is the resonant length of the cavity when ignoring the effect of the iris.

The cavity resonators described so far are slightly limited in their application because the resonant frequency is fixed. It is possible to achieve some limited manual tuning of resonant cavities, usually at the expense of a slight decrease in the cavityQ-factor. The two most common tuning techniques are illustrated in Figure 3.19.

In the case of screw tuning shown in Figure 3.19a, a metal post is screwed into the waveguide parallel to theE-field.

This has the effect of introducing an admittance in parallel with that of the cavity. The diameter of the post and the depth of penetration into the guide determine the value of the admittance, and thereby the amount of frequency tuning. An alternative arrangement is shown in Figure 3.19b, and this simply uses a plunger to change the resonant length of the cavity, and hence the resonant frequency. The plunger is normally driven by a micrometre screw to provide fine frequency tuning. Also, the plunger is non-contacting to avoid variations in contact resistance between the plunger and the waveguide

(a) (b)

l Tuning

screw

Adjustment

l Non-contacting

plunger Adjustment

Figure 3.19 Manual tuning of resonant cavities: (a) use of tuning screw and (b) use of non-contacting plunger.

k k

3.6 Characterization of Materials for RF and Microwave Circuits 105

walls; this is achieved by machining slots in the sides of the plunger which reflect short circuits between the face of the plunger and the sides of the waveguide.

Cavity resonators formed from rectangular waveguide are attractive because of their relative simplicity and the high Q-values that they provide. However, their use is problematic as the frequency increases into the millimetre-wave part of the spectrum because the internal dimensions of the waveguide decrease and become inconveniently small. A good solution at millimetre-wave frequencies is to make use of the properties of over-moded circular waveguide supporting the TE₀₁mode. Over-moded means that there is the potential for more than one mode to exist within the waveguide. As was seen in Appendix 1.C of Chapter 1 the TE₀₁mode in circular waveguide exhibits the property of decreasing attenuation with increasing frequency, which means that it has the potential for providing very highQvalues for cavity resonators operating at millimetre-wave frequencies. An expression for theQof a circular cavity resonator can be determined by evaluating Eq. (3.19) for the appropriate electromagnetic field distribution. Collin [22] showed that for TE_mnpmodes in a cylindrical cavity resonator theQis given by

Q=𝜆o

𝛿s

[ 1−

(m x

)2] [ x²+

(_p𝜋a

l

)2]_1.5 2𝜋

[ x²+^2a_l

(p𝜋a l

)2

+ (

1−^2a_l ) (mp𝜋a

xl

)2], (3.27)

where

p=number of guide half-wavelengths in axial length of cavity a=radius of cavity

l=length of cavity 𝛿s=skin depth

𝜆o=free-space wavelength

mandxare defined in Appendix 1.C.

Example 3.9 Determine theQat 50 GHz of a cylindrical cavity resonator supporting the TE₀₁₉mode if the cavity is made of copper and has a radius of 25 mm.

(𝜎copper=5.8×10⁷S/m) Solution

f_o=50 GHz ⇒ 𝜆o= 3×10⁸

50×10⁹ m=6 mm.

Using Eq. (1.86) to obtain𝜆c:𝜆c=2𝜋a

x =2𝜋×25

3.83 mm=41.02 mm.

Using Eq. (1.72): 1

(6)² = 1 (41.02)² + 1

𝜆²g

⇒ 𝜆g =6.07 mm.

Length of cavity at resonance:l=9×𝜆g

2 =9×6.07

2 mm=27.32 mm.

Skin depth:𝛿s= 1

√𝜋𝜇of𝜎 = 1

√𝜋×4𝜋×10⁻⁷×50×10⁹×5.8×10⁷

m=0.30μm.

Substituting in Eq. (3.27) and noting thatm=0:

Q= 6×10⁻³ 0.3×10⁻⁶×

[

(3.83)²+

(9×𝜋×25 27.32

)2]1.5

2𝜋 [

(3.83)²+2×25 27.32

(9×𝜋×25 27.32

)2]

=20×10³×17 899.51 7793.10

=45 736.82.

k k N, that can be supported simultaneously at a given frequency in a circular waveguide of radiusais given by

N=10.20 (a

𝜆o

)2

, (3.28)

where𝜆ois the free-space wavelength.

Example 3.10 Determine the number of possible modes that can exist within the cavity resonator specified in Example 3.9.

Solution Using Eq. (3.28):

N=10.2× (25

6 )2

=177.

If a waveguide is heavily over-moded, i.e. it has a large value ofN, it is difficult to avoid exciting unwanted modes. Also, when measuring theQ-value of an over-moded cavity it is difficult to identify the correct resonance curve from amongst the resonances of the unwanted modes. These problems can be overcome through the use of a mode filter. A mode filter is a section of waveguide that inhibits the propagation of all TE and TM modes, except those in theTE_0nfamily. The physical realization of the required mode filter is possible because all TE and TM modes, except those in the TE_0nfamily, have axial wall current components. Thus, the required mode filter should have anisotropic wall impedance, with low impedance in the circumferential direction to allow the low-loss propagation of the TE₀₁mode, and high axial impedance to attenuate all the unwanted modes. These requirements can be met by making the mode filter from a section of helical waveguide, whose structure is shown schematically in Figure 3.20. The normal method of manufacturing this type of filter is to start by winding enamelled-coated copper wire on a precisely dimensioned steel mandrel having the same diameter as the desired internal diameter of the waveguide. The wire is normally of the order of 40 swg (standard wire gauge), which has a diameter

∼130μm. The wire is wound in a tight helix and secured in place with resin, the outside of which is coated with absorbing material to prevent wave propagation along the outside of the helix. Finally, the structure is encased in a protective cylinder, and the mandrel withdrawn. Because a small diameter wire has been used, the circumferential impedance is very low, but the axial impedance is very high since axial currents would have to travel through the enamel surrounding the wire core. The difference between the transmission loss of the TE₀₁mode travelling through wire-wound helical waveguide and through solid copper tube of the same diameter depends on the amount of enamelling, i.e. on the ratiod/Das defined in Figure 3.20.

But it can be shown [23] that ifd/D>0.8, the difference in loss is less than 9%.

Whilst the helical mode filter will inhibit the propagation of all modes other than those in the TE_0nfamily, it does mean that the TE₀₁will only exist on its own at frequencies below the cut-off value of the TE₀₂mode, as shown in Example 3.11.

Outer cylindrical casing

Resin

Absorber

Enamel-coated wire helix

d

D Cu

Enamel coating

Cross-section of enamel-coated wire Figure 3.20 Typical structure of a helical waveguide.

k k

3.6 Characterization of Materials for RF and Microwave Circuits 107

Example 3.11 Determine the frequency range over which only the TE₀₁ mode will propagate through a helical waveguide whose internal diameter is 50 mm.

Comment: The first root of J^′₀(x) occurs when x=3.8317, and the second root when x=7.0156.

Solution Using Eq. (1.86):

Cut-off wavelength of TE₀₁mode:𝜆_c,01= 2𝜋×25

3.8317 mm=41.00 mm.

Cut-off wavelength of TE₀₂mode:𝜆c,02= 2𝜋×25

7.0156mm=22.39 mm.

Then,

f_c,01= c 𝜆c,01

=3×10⁸

41.00 Hz=7.32 GHz, f_c,02= c

𝜆c,01

= 3×10⁸

22.39 Hz=13.40 GHz.

Frequency range: (7.32–13.40) GHz.