IDEAL TRANSMISSION-LINE FUNDAMENTALS
3.4 TRANSMISSION-LINE PARAMETERS FOR THE LOSS-FREE CASE Earlier, we discussed how an electromagnetic wave propagates on a transmis-
3.4.3 Transmission-Line Parameters for a Microstrip
calculated from (3-45) if the capacitance is calculated withεr =1:
L= 1
c2Cεr=1 (3-46)
wherec is the speed of light in a vacuum andCεr=1is the capacitance with the relative dielectric permittivity εr set to 1. Substituting (3-44) into (3-46) with εr =1 gives the value of theinductance per unit length for a coaxial line:
L= ln(b/a)
c22π ε0 H/m (3-47)
The solution to this partial differential equation can be found in terms of two ordinary differential equations, assuming that the potential can be represented by the product of a function for each coordinate [Jackson, 1999]:
(x, y)=X(x)Y (y) (3-49)
If equation (3-49) is substituted back into (3-48) and then divided by , the result is
1 XY
d2XY dx2 + 1
XY d2XY
dy2 = 1 X
d2X dx2 + 1
Y d2Y
dy2 =0 (3-50) where the partial derivatives are replaced by total derivatives because each term involves only one variable. This allows us to write two separate ordinary differ- ential equations:
1 X
d2X
dx2 = −β2 (3-51)
1 Y
d2Y
dy2 =β2 (3-52)
The constantβ must satisfy the conditionsβ2+(−β2)=0 to satisfy Laplace’s equation. The differential equations in (3-51) and (3-52) are solved by finding the roots of the characteristic equations:
X(x)=C1nejβx+C2ne−jβx=C1n(cosβx+jsinβx)
+C2n(cosβx−jsinβx) (3-53)
Y (y)=C1n eβy+C2n e−βy =C1n (coshβy+sinhβy)
+C2n (coshβy−sinhβy) (3-54) The potential is then calculated by substituting (3-53) and (3-54) into (3-49):
(x, y)=X(x)Y (y)=[C1n(cosβx+jsinβx)+C2n(cosβx−jsinβx)]
·[C1n (coshβy+sinhβy)+C2n (coshβy−sinhβy)] (3-55) Note that since the solution is periodic, β must be an integer. The boundary conditions that must be applied to solve (3-55) are
(x, y)=0 when x= ±d
2 (3-56)
which is the potential at the sidewalls and
(x, y)=0 when y=0,∞ (3-57)
which is the potential at the ground plane and at a point infinitely far away. This means that we must consider two different solutions: region 1, which exists from y=0 toh(in the dielectric), and region 2, which exists from y=hto infinity (in the air), where the potentials must be equal at the boundary of these regions.
Looking at X(x) in (3-55), the boundary condition in (3-56) is met if C1n= C2n andβis chosen so that the cosine term equals zero whenx= ±d/2. When C1n=C2n,X(x) is reduced to
X(x)=C1ncosβx and when β=nπ/d for oddn andx= ±d/2,
X(x)=0 yielding an appropriate form forX(x):
X(x)=C1ncos nπ d x
(3-58)
The second termY(y) in (3-55) will satisfy the boundary condition in (3-57) at y=0 ifC1n = −C2n , yielding
Y (y)=C2nsinhβy=C2n sinh nπ d y
(3-59)
At y= ∞, the boundary condition (3-57) is satisfied whenC1n =0:
Y (y)=C2n (coshβy−sinhβy)=C2n e−βy (3-60) This allows us to write equations for the potential that satisfies the boundary conditions by multiplyingX(x) andY(y) for the appropriate boundary conditions, where the product of the constants have been renamedAn andBn for clarity.
(x, y)=
∞ n=1odd
Ancos nπ d x
sinh nπ d y
when 0≤y < h (region 1)
(3-61a) ∞
n=1odd
Bncos nπ d x
e−(nπ/d)y whenh≤y <∞ (region 2) (3-61b) The potential at the signal conductor (y=h) must be continuous, so
Ancos nπ d x
sinh nπ d h
=Bncos nπ d x
e−(nπ/d)h
yielding
Ane(nπ/d)hsinh nπ d h
=Bn
which allows the equations for the potential to be written in terms ofAnalone:
(x, y)=
∞ n=1odd
Ancos nπ d x
sinh nπ d y
when 0≤y < h
(3-62a) ∞
n=1odd
Ansinh nπ d h
cos nπ d x
e−(nπ/d)(y−h) whenh≤y <∞ (3-62b) To get the electricfield between the signal conductor and the ground plane, we apply equation (2-65),Ey= −∇φ= −∂/∂y. Sinced(sinhax)/dx=acoshax andd(eax)/dx =aeax, the electricfields become
Eyn= − ∂
∂yAncos nπ d x
sinh nπ d y
= −nπ An
d cos nπ d x
cosh nπ d y
for region 1 and
Eyn= − ∂
∂yAnsinh nπ d h
cos nπ d x
e−(nπ/d)(y−h)
= nπ An
d sinh nπ d h
cos nπ d x
e−(nπ/d)(y−h)
for region 2, yielding
Ey(x, y)=
− ∞ n=1odd
nπ An
d cos nπ d x
cosh nπ d y
when 0≤y < h
(3-63a) ∞
n=1odd
nπ An
d sinh nπ d h
×cos nπ d x
e−(nπ/d)(y−h) whenh≤y <∞ (3-63b)
To calculate the coefficientAn, we must assume a charge distribution on the signal conductor. As afirst-order approximation, it can be assumed that the charge (ρ) is spread out uniformly across the signal conductor aty=h. Consequently, there will be no variation ofρ withx on the strip, but it will be zero off the strip.
ρ(x)=
1 when|x|< w/2
0 when|x|> w/2 (3-64) From equation (3-4), the electricfields above and below the signal conductor are equated to the charge density:
ρ(x)=(ε0Ey1)−(ε0εrEy2)= ∞ n=1odd
ε0nπ An
d cos nπ d x
×
sinh nπ d h
+εrcosh nπ d h
(3-65)
If we temporarily assign all terms in (3-65) that are not a function ofx (except An) to a constant term k as in (3-66), it is easy to see that it takes the form similar to the Fourier series shown in (3-68) whena0 andbm are zero:
ρ(x)= ∞ n=1odd
Ankcos nπ d x
(3-66)
k=ε0nπ d
sinh nπ d h
+εrcosh nπ d h
(3-67)
f (x)= 1 2a0+
∞ n=1
amcos mπ d x
+ ∞ n=1
bmcos mπ d x
(3-68)
This allows the use of Fourier series techniques to solve for the coefficient,An: w/2
−w/2ρ(x)cos mπ d x
dx=
d/2
−d/2Ankcos nπ d x
cos mπ d x
dx Setting m=n, we arrive at
w/2
−w/2ρ(x)cos nπ d x
dx =
d/2
−d/2Ankcos2 nπ d x
dx (3-69)
After substituting (3-67) fork, integration of both sides of (3-69) yields 2d
nπsinnπ w
2d = Ankd 2
An= 4 sin(nπ w/2d)
ε0[(nπ )2/d]
sinh((nπ/d)h)+εrcosh((nπ/d)h) (3-70) The voltage of the signal conductor can now be calculated with respect to the ground plane atx=0 using (3-1) and (3-63a):
v= − b
a
E·dl = − h
0 Ey(x =0, y) dy = ∞ n=1odd
Ansinh nπ d h
(3-71)
The total charge is calculated with Q=
w/2
−w/2ρ(x)dx=w (3-72) So the capacitance per unit length is calculated from (3-71) and (3-72) using (2-76):
C= Q
v = w
∞
n=1odd
Ansinh[(nπ/d)h] (3-73)
and the inductance per unit length is calculated using (3-46) byfirst calculating the capacitance with (3-73) with a relative dielectric permittivity of unity (εr =1):
L= 1
c2Cεr=1 (3-46)
With (3-73) and (3-46), we can calculate several useful quantities, such as the phase constant (3-30), the characteristic impedance (3-33), and the effective rel- ative dielectric permittivity (εeff). To understand how to calculate εeff, we can examine the formulation of a parallel-plate capacitor, as derived in Example 2-3:
Cεr,eff
Cεr=1 = ε0εr,effA/d
ε0A/d =εr,eff (3-74)
Consequently, the effective permittivity is determined by calculating the capac- itance of the microstrip with (3-73) using the correct value of the dielectric constant and dividing by the capacitance calculated with (3-73) when εr =1.
Since the dimensions that determine the capacitance remain identical, all that remains is the effective dielectric permittivity.
The accuracy of (3-73) and (3-74) is reasonable, but two approximations made during the derivation degrade the accuracy. First, we assumed that the signal conductor is infinitely thin, which is not realistic because microstrip transmission lines manufactured on printed circuit boards often have conductor thicknesses of dimensions similar to the dielectric height. There is little benefit of deriving a finite thickness formula here because the most useful method would employ numerical solutions that are not covered in this book. Second, the charge distri- bution on the signal conductor is not uniform. The charge distribution near sharp edges is derived in Section 3.4.4 and applied to this formulation.