Planar Circuit Design II
4.5 Miniature Planar Components
k k
k k (i) An accurate expression for the inductance of a square spiral in free space was developed by Mohan et al. [16]
L=K1𝜇oN2davg
1+K2𝜌 nH, (4.34)
whereNis the number of turns on the spiral,davgis the average side length of the spiral inμm,𝜌is the fill factor defined by
𝜌=douter−dinner
douter+dinner (4.35)
andK1andK2are factors determined by the geometry of the spiral. Mohan and colleagues [16] gave these values as K1=2.34 andK2=2.75 for a square spiral, andK1=2.25 andK2=3.55 for an octagonal spiral.
(ii) In a microstrip configuration the presence of the ground plane beneath the spiral will tend to reduce the inductance, usually by around 10%. Gupta and colleagues [2] quote an expression for a correction factor,Kg, to represent the effect of the ground plane as
Kg =0.57−0.145 ln (wf
h )
, (4.36)
where wf is the width of the spiral track andhis the thickness of the substrate. The effective inductance of the microstrip spiral is then given by
Leff =KgL, (4.37)
whereLis the free-space inductance given by Eq. (4.34).
(iii) A spiral inductor will exhibit some self-capacitance, made up of the capacitance between adjacent turns and the capacitance to the ground plane.
(iv) It follows from the last point that the spiral inductor will have a self-resonant frequency – the inductor should be operated well below this value.
(v) The inductor will also have a significant series resistance, whose value will be determined by the number of turns and the narrowness of the track. In order to make a compact component the width of the track will normally be small, leading to high ohmic loss. This high loss is a particular feature of many spiral inductors, and leads to relatively low Qvalues for this type of component.
(vi) The shape of a rectangular spiral inductor is such that there will be significant discontinuities, particular at the track corners, and appropriate compensation needs to be applied to the dimensions of the spiral.
(vii) For designs at frequencies above a few GHz, Eqs. (4.36) and (4.37) should be used only to give a prototype design, which should then be refined using electromagnetic simulation.
(viii) Spiral inductors may also be realized in a circular format. This will reduce the discontinuities, but at the cost of a slight increase in design complexity.
(ix) Spurious radiation from spiral inductors can also be a problem in practical circuits, leading to unwanted coupling between components and package resonances. Work reported by Caratelli and colleagues [17] using a full-wave finite-difference time-domain (FDTD) analysis showed how the level of radiation is affected both by the geometry of the spiral, and the materials from which it is made.
4.5.2 Loop Inductors
In circuit situations where only a small amount of inductance is required, the inductor can be formed from a simple, single loop as shown in Figure 4.19. The single loop provides a compact component with few discontinuities, although some caution is needed in the design to avoid generating excess capacitance across the entry to the loop.
k k
4.5 Miniature Planar Components 147 w
rr
Entry gap
Figure 4.19 Loop inductor.
Points to note:
(i) For a single loop inductor the free-space inductance is given approximately by:
L=12.57r (
ln [8𝜋r
w ]
−2
) nH, (4.38)
where
L=inductance
r=mean radius of loop in cm
w=width of track forming loop in cm.
Whilst this expression for the inductance of a loop is frequently quoted in the literature, it takes no account of the size of the gap at the entry to the loop. Thus, it should be regarded as very approximate, and used to establish an initial value in a practical design, which should then be refined using electromagnetic simulation.
(ii) In order to avoid unwanted capacitance across the entry to the loop, the size of the gap is normally made of the order of 5w, wherewis the width of the track.
(iii) The self-capacitance of the loop inductor tends to be less than for a spiral inductor. This capacitance is essentially due to the capacitance per unit length of the conductor. Thus, the self-resonant frequency of the loop is higher than for a spiral inductor.
Example 4.10 Determine the inductance provided by each of the following structures, and comment upon the results.
Each of the structures is formed from a flat (ribbon) conductor of width 0.8 mm of negligible thickness.
(i) A three-turn rectangular planar spiral inductor with a pitch of 1.4 mm and an inner dimension of 4 mm.
(ii) A single-loop inductor of radius 5 mm. (Neglect the effect of the entry gap.) (iii) A straight 12 mm length of conductor.
Solution
(i) . Pitch=w+s=0.8+s=1.4 ⇒ s=0.6 mm.
Referring to Figure 4.18:
douter=dinner+6w+4s= (4+6×0.8+4×0.6)mm=11.2 mm, davg=0.5× (11.2+4)mm=7.6 mm.
The fill factor for the spiral is found using Eq. (4.35):
𝜌=11.2−4
11.2+4 =0.474.
Substituting in Eq. (4.34), and using theK-factors for a square spiral:
L(spiral) =2.34(4𝜋×10−7) × 32×7.6
1+ (1.75×0.474)H=87.34 nH.
(ii) Using Eq. (4.38):
L(loop) =12.57×0.5× (
ln
(8𝜋×0.5 0.08
)
−2 )
nH
=19.21 nH.
(Continued)
k k (iii) Using Eq. (4.26):
L(straight ribbon) =0.002×1.2× [
ln
(2×1.2 0.08
)
+0.5+0.2235× (0.08
1.2 )]
μH
=9.40 nH.
Summary:
Circuit Inductance (nH)
Spiral 87.34
Loop 19.21
Straight ribbon 9.40
Comment: The spiral and loop inductors specified in this example would occupy approximately the same area of substrate, but the spiral configuration gives substantially more inductance. The theory used makes no allowance for the entry gap of the loop; if a practical gap were considered then the inductance of the loop would be rather less than the 19.21 nH that was calculated. As would be expected, the self-inductance of the straight ribbon is less than the inductance of the other configurations, but is still around 50% of that of the loop. This suggests that using straight lengths of line to provide small values of inductance may be preferable to using a loop, which would in practice have more self-capacitance and hence a more troublesome self-resonant frequency.
4.5.3 Interdigitated Capacitors
The conventional configuration of conductors forming an interdigitated capacitor is shown in Figure 4.20. The structure employs a number of fingers to couple two metallic conductors. The use of fingers effectively creates a long coupling gap between the two conductors. Normally, the widths and spacing of the fingers is the same, i.e.wf=s. This type of capacitor was originally developed for use in MMICs, but it can also be conveniently implemented in hybrid circuits by forming the coupled fingers in a microstrip conductor.
A comprehensive analysis of interdigitated capacitors was reported by Alley [18], who developed the following approxi- mate expression for the total capacitance of the structure
C= (𝜀r−1)l[(N−3)A1+A2]pF, (4.39)
where𝜀ris the dielectric constant of the substrate,lis the overlapping finger length inμm,Nis the number of fingers, and A1andA2are correction factors which are functions ofh/wf wherehis the thickness of the substrate andwf is the width of each finger. Alley [18] provided curves from whichA1andA2could be obtained for a given ratio ofh/wf. Subsequently,
wf s
l
Figure 4.20 Interdigitated capacitor.
k k
4.5 Miniature Planar Components 149
Bahl [19] used a curve-fitting technique to obtain the following closed-form equations from which values ofA1andA2can be calculated:
A1=4.409 tanh
⎡⎢
⎢⎣ 0.55
( h wf
)0.45⎤
⎥⎥
⎦
×10−6pF∕m. (4.40)
A2=9.92 tanh
⎡⎢
⎢⎣ 0.52
( h wf
)0.5⎤
⎥⎥
⎦
×10−6pF∕m. (4.41)
Points to note about interdigitated capacitors:
(i) They are normally used to provide values of capacitance up to around 1 pF. For values above 1 pF the dimensions of the components can become rather large.
(ii) The overall dimensions are normally made less than one quarter wavelength, so they can be regarded electrically as lumped components.
(iii) They suffer from a number of discontinuities and so the expression given in Eq. (4.39) for the total capacitance should be regarded as an approximation.
(iv) In a microstrip line the interdigitated capacitor can be designed for a good series match.
(v) They are precise components and are normally used where precise values of capacitance are important, such as in filters and matching networks.
(vi) TheQ-factor of an interdigitated capacitor can be increased by increasing the ratio of finger width (wf) to gap size (s).
(vii) Various CAD models for interdigitated capacitors that include the effects of losses are available in the literature, for example, Zhu and Wu [20]. These models permit precise design of low-loss interdigitated capacitors in a microstrip format.
4.5.4 Metal–Insulator–Metal Capacitor
In situations where the capacitance needed is greater than can be provided by an interdigitated structure, then a metal–insulator–metal (MIM) capacitor can be used. This type of capacitor, sometimes referred to as an overlay capacitor, can provide relatively high capacitance, but with less precision than an interdigitated capacitor. The MIM capacitor is suitable for inclusion in either monolithic or hybrid RF integrated circuits. The structure of a MIM capacitor in thick-film microstrip is shown in Figure 4.21; for clarity, the plan view of the component is shown in Figure 4.21a, and the side view in Figure 4.21b. In this structure a dielectric is printed over the open-end of the microstrip line forming the lower electrode of the capacitor. Another microstrip line is then printed over the dielectric to form the upper electrode. Normally three layers of dielectric are printed successively to avoid pin holes forming in the dielectric. If pin holes exist in the dielectric,
Dielectric overlay Bottom track
Bottom track
Top track
Top track
Substrate Ground
(a)
(b)
Capacitor
Figure 4.21 MIM capacitor: (a) plan view and (b) side view.
k k together. It should be noted that the width of the upper electrode is chosen to maintain the correct impedance match to
the microstrip line.
The approximate capacitance of a MIM capacitor can be found using simple parallel-plate capacitor theory as C= 𝜀m𝜀oA
d , (4.42)
where𝜀mis the dielectric constant of the printed dielectric between the electrodes,𝜀ois the permittivity of a vacuum,Ais the overlapping area of the electrodes, anddis the thickness of the printed dielectric.
Substituting𝜀o=8.854×10−12F/m in Eq. (4.42) gives the capacitance of the MIM capacitor as C=8.854𝜀mA
d×106 pF, (4.43)
wheredis inμm andAis in (μm)2. Points to note about MIM capacitors:
(i) MIM capacitors are useful as de-coupling capacitors, where relatively high capacitance values are needed, but without high precision.
(ii) Equation (4.42) is an approximation as it takes no account of the fringing effects at the edges of the electrodes.
(iii) Whilst MIM capacitors can conveniently be made using thick-film technology, there can be reliability problems asso- ciated with the need to print the upper electrode over the edge of the dielectric. This is known as the ‘print-over edge effect’, and relates to the thinning of the conductor as it is printed over the edge of the dielectric.
(iv) Since the resistance of the dielectric layer between the electrodes is extremely high, the only source of ohmic loss in a MIM capacitor is the resistance of the electrodes. Consequently theQof this type of capacitor can be relatively high.
(v) If the MIM capacitor is made using thick-film technology, some trimming of the structure is possible to control the capacitance value. But if precise capacitance values are required, a better solution is to use an interdigitated structure.
Example 4.11 Calculate the capacitance of each of the following structures, and comment upon the results:
(i) A microstrip interdigitated capacitor with the following specification:
Number of fingers=4 Finger width=50μm Finger spacing=50μm
Overlapping finger length=2 mm Substrate thickness=250μm Substrate relative permittivity=9.8
(ii) A MIM capacitor with the following specification:
Relative permittivity of dielectric separating the electrodes=7.8 Thickness of dielectric separating the electrodes=24μm Area of electrodes=350×2000 (μm)2
Solution (i) .
h wf =250
50 =5.
Using Eq. (4.40):
A1=4.409 tanh[0.55× (5)0.45] ×10−6pF∕m
=3.583×10−6pF∕m.
Using Eq. (4.41):
A2=9.92 tanh[0.52× (5)0.5] ×10−6pF∕m
=8.15×10−6pF∕m.
k k