IDEAL TRANSMISSION-LINE FUNDAMENTALS
3.2 WAVE PROPAGATION ON LOSS-FREE TRANSMISSION LINES Transmission lines are designed to guide electromagnetic waves from one point
3.2.3 Equivalent Circuit for the Loss-Free Case
where Cis normalized to length with units of farads per meter. As a result, for a specific transmission-line geometry, we can write the left-hand side of (2-30) in terms of the capacitance:
ε∂Ex
∂t →C∂v
∂t (3-15)
To derive the circuit equivalent of the right side of (2-30), recall from equation (2-79) that current is the rate of chargeflow per second:
i= dQ
dt (3-16)
Substituting Q=Cv into equation (3-16) produces the current in terms of the voltage and capacitance:
i=Cdv
dt (3-17)
which is equal to equation (3-15). Therefore, (2-30) can be rewritten in terms of circuit parameters:
∂i(z, t)
∂z = −C∂v(z, t)
∂t (3-18)
Note that (3-18) is the classic response of a capacitor from circuit theory.
Equations (3-12) and (3-18) are the loss-free forms of the telegrapher’s equations, which describe the electrical characteristics of a transmission line.
∆I ∆I ∆I I
I
V+
Vref
Signal Conductor
Reference Conductor E
E E B
B
Figure 3-8 Segment of a transmission line showing the magnetic and electricfields, the associated differential current loops, and the voltage potentials at each conductor.
as was shown in Figure 2-18. When representing the total current as a series of small loops, the adjacent vertical components of the current elements cancel each other out, leaving a net current ofI on the signal conductor and(−I) on the reference conductor. As described in equation (2-97), a current change in a loop will change the magneticflux and will thus induce a self-inductance.
Consequently,the magneticfield of a transmission line is represented by a series inductor in the circuit model. The value of the equivalent-circuit inductance is given by
Lz=zL (3-19)
where z is the length of the differential section of transmission line andL is the inductance per unit length.
Similarly, when a voltage is applied to the transmission line between the signal conductor and the reference plane as shown in Figure 3-8 (v=V+−Vref), an electricfield is established in units of volts per meter:
v(b)−v(a)= − b
a
E·dl (2-58)
The presence of an electric field implies that charges exist on the conduc- tors and consequently implies the existence of a capacitance (as described in Section 2.4.3). Therefore, the electricfield of a transmission line is represented by a shunt capacitor in the circuit model. The value of the equivalent-circuit capacitance is given by
Cz=zC (3-20)
(a)
2 Ns
1
(b) L∆z
C∆z
L∆z
C∆z
L∆z
C∆z
L∆z
C∆z
Figure 3-9 (a) Model for a differential element of a transmission line; (b) full model.
wherezis the length of a differential section of transmission line andCis the capacitance per unit length.
Figure 3-9a shows the equivalent-circuit model of a differential element of transmission line. However, to make use of the model, we need a practical methodology to represent transmission lines that are much longer thanz. Sim- ply increasingzto scale the inductance and capacitor values for longer lengths will not produce a legitimate model unless the resonance of the LC circuit is much higher than the maximum frequency to be simulated:
1 2π√
CzLz fsimulation (3-21)
If equation (3-21) is not satisfied, the equivalent circuit will behave like an LC filter circuit instead of like a transmission line.The equivalent circuit of a trans- mission line can be represented by a series inductance and a shunt capacitor only if the LC resonant frequency is significantly greater than the maximum frequency of interest.
The correct method to scale the transmission-line model for long lengths is to cascade a sufficient number of small LC segments together until the correct overall length is achieved, as shown in Figure 3-9b. EachLC segment represents a small transmission-line section of lengthz. The trick is to choose the correct value ofzto achieve adequate model accuracy. If the value ofzis too small, it will take an inordinate amount of time to perform SPICE simulations. However, if it is too large, the model will not exhibit realistic transmission-line properties.
A good “rule of thumb” is to choose z so that the delay of each segment is
approximately one-tenth the signal rise time if working in the time domain, z≤ trc
10√ εr
(3-22a) or one-tenth the wavelength that corresponds to the maximum frequency of inter- est if working in the frequency domain,
z≤λf,max
10 (3-22b)
wheretr is the signal rise or fall time,cthe speed of light in a vacuum (3×108 m/s), εr the dielectric permittivity, and λf,max=c/(fmax√
εr) the wavelength that corresponds to the highest frequency of interest in the simulation.
When using a distributedLC model for modeling transmission lines, the num- ber of segments for time-domain simulations is determined by
Ns = l
z = 10l√ εr
trc (3-23a)
For frequency-domain simulations, Ns = l
z = 10l
λf,max (3-23b)
where Ns is the minimum number of segments required to model a transmis- sion line of lengthl. Therefore, the capacitance and inductance per segment are given by
Cz= lC
Ns (3-24a)
Lz= lL
Ns (3-24b)
whereCandLare the per unit length values of the capacitance and inductance.
Example 3-1 Create a transmission-line model for the 20-cm transmission line shown in Figure 3-10a assuming the following inductance and capacitance values and a dielectric permittivity ofεr =4.5.
L=3.54×10−7 H/m C=1.41×10−10 F/m
0.498 nH
2 (b) (a)
20 cm
Signal Conductor er = 4.5
142 1
0.198 pF
0.498 nH
0.198 pF
0.498 nH
0.198 pF Reference Conductor 100 ps
100 ps
Figure 3-10 Equivalent circuit of a transmission line.
SOLUTION Since the digital waveform has rise and fall times of 100 ps (tr = tf =100 ps), equations (3-22) and (3-23) give the required values for the model:
z= 100 ps(3.0×108) 10√
4.5 =1.41×10−3m Ns= l
z = 0.2
1.41×10−3 =141.8 segments
Since it is inconvenient to build an equivalent-circuit model with 141.8 segments, Nsis rounded up to 142. The inductance and capacitance values for each segment are calculated with equations (3-24):
Cz= (0.2)(1.41×10−10)
142 =1.98×10−13 F Lz= (0.2)(3.54×10−7)
142 =4.98×10−10 H Figure 3-10b shows the equivalent circuit for this transmission line.
3.2.4 Wave Equation in Terms ofLC
The wave equations, which were used as the basis for analyzing propagating electromagnetic fields, were derived in Section 2.3.1. To analyze transmission lines in terms of circuit parameters, we rederive the wave equation from the
telegrapher’s equations (3-12) and (3-18):
∂v(z, t)
∂z = −L∂i(z, t)
∂t (3-12)
∂i(z, t)
∂z = −C∂v(z, t)
∂t (3-18)
Assuming that the digital signals can be decomposed into sinusoidal harmon- ics using the Fourier transform, the telegrapher’s equations can be expressed in time-harmonic form where the voltage and current have the formsv(t)=V0ej ωt andi(t)=I0ej ωt. Similar representations for the time-harmonicfields were dis- cussed in Section 2.3.3. Consequently, the time-harmonic forms of the telegra- pher’s equations are
dv(z)
dz = −j ωLi(z) (3-25)
di(z)
dz = −j ωCv(z) (3-26)
Taking the derivative of (3-25) with respect tozproduces d2v(z)
dz2 = −j ωLdi(z)
dz (3-27)
and substituting (3-26) into (3-27) allows us to write an equation only in terms of voltage,
d2v(z)
dz2 +ω2LCv(z)=0 (3-28)
which is the loss-free transmission-line wave equation for voltage. Equa- tion (3-28) is a second-order differential equation with the general solution given by
v(z)=v(z)+e−j zω
√LC +v(z)−ej zω
√LC (3-29)
The termv(z)+e−j zω
√LC describes the voltage propagating down the transmis- sion line in the+z-direction andv(z)−ej zω
√LC describes the voltage propagating in the−z-direction. Note the similarity of (3-29) and (2-41), which is the solu- tion equivalent to the wave equation for the electric field. In Section 2.3.4, a propagation constant for a wave traveling in an infinite medium was defined that completely describes the medium where the electromagnetic wave is propagating:
γ =α+jβ (2-42)
Comparing the solution to the electromagnetic wave equation (2-41) to equa- tion (3-29) allows us to draw direct parallels for the propagation constant of a wave propagating on a loss-free transmission line:
γ =α+jβ =0+j ω√
LC →β=ω√
LC (3-30)
Equation (3-30) is the phase constant of a transmission line. Notice that for the loss-free transmission line, the attenuation constant (α) is zero. In Chapters 5 and 6 we describe how to calculate the losses for a transmission line and subse- quently,α.