IDEAL TRANSMISSION-LINE FUNDAMENTALS
3.6 TIME-DOMAIN REFLECTOMETRY
3.6.1 Measuring the Characteristic Impedance and Delay of a Transmission Line
Consequently, for a digital pulse, reactive components such as inductors and capacitors will low-pass-filter the waveform, resulting in increased rise and fall times. The exception to this statement is when specificfilters are constructed using reactive components to equalize a channel, which is described is Chapter 12.
from the voltage by rearranging equation (3-102), ≡ vr
vi = ZDUT−Z0
ZDUT+Z0 (3-113)
ZDUT=Z0vi+vr
vi−vr
whereviis the incident voltage transmitted andvr is the voltage reflected from the DUT. Figure 3-43 depicts the TDR waveforms of a 60-transmission line with a length that corresponds to a delay of 250 ps. The impedance of the transmission line under test is calculated from thefirst voltage step that occurs at 500 ps using equation (3-113):
ZDUT=50·1+0.091
1−0.091 =60
We can also calculate the delay of the transmission line. Since a TDR is essentially a reflected voltage at the driver, the reflection of the DUT remains for the time it takes the signal to propagate to the end of the DUT and return to the source from the open circuit. Consequently, the duration of the reflection will correspond to twice the delay of the transmission line under test, as shown in Figure 3-43.
00 0.5 1 1.5 2
0.5 1 1.5 2 2.5
Time, ns
Voltage at A
0.90.2 0.4 0.6 0.8 1
0.95 1 1.05 1.1
Time, ns
Voltage at A
vs
Rs = 50 Ω
Zo = 50 Ω 0-2 V
A ZDUT = 60 Ω
DUT td= 250 ps
2td= 500 ps vi= 1.0 V
vi= 0.091 V
Figure 3-43 TDR waveform measuring a 60- transmission line with a length that corresponds to a delay of 250 ps.
vs
Rs = 50 Ω
Zo = 50 Ω
0-2 V A ZDUT
DUT 2 inches
0.90 0.2 0.4 0.6 0.8 1 1.2 1.4
1 1.1 1.2 1.3 1.4
Time, ns
Voltage at A
Figure 3-44 TDR profile for Example 3-7.
Example 3-7 Calculate the equivalent L and C per unit length of the 2-in.
transmission line measured with TDR profile shown in Figure 3-44.
SOLUTION The impedance is calculated from the step voltage at 1.2 V, which corresponds to the reflections from the transmission line under test.
vr =1.2−1.0=0.2 V vi = 2Z0
Z0+Rs =1 V ZDUT=Z0vi +vr
vi −vr =50· 1+0.2
1−0.2 =75 The delay is calculated from the duration of the step:
τd= 12(1.1 ns−0.5 ns)=300 ps
The distributed inductance and capacitance values are calculated by solving the impedance and delay equations simultaneously:
Z0= L
C =75 τd=l√
LC =300 ps
The inductance per unit length is calculated by normalizing the delay to line length of the transmission line under test and multiplying by the impedance:
τd
2 in.=√
LC =150 ps/in.
L= L
C
√LC =75(150×10−12)=11.25 nH/in.
The capacitance is calculated by dividing the normalized delay by the impedance:
C=
√LC
√L/C = 150×10−12
75 =2.0 pF/in.
3.6.2 Measuring Inductance and Capacitance of Reactive Structures In Section 3.5.7 we discussed reflections from reactive loads. In this section we show how to estimate the value of the capacitance or the inductance from a measured TDR profile. The analysis presented here is somewhat idealized because it assumes an ideal step for a source. In reality, ideal steps are impossible to generate. However, modern TDR measurement equipment allows the generation of very fast rise times that vary anywhere between 9 and 35 ps, which are fast enough to approximate a step function for a variety of applications.
Inductive Structures In a TDR measurement, narrow spikes such as those depicted in Figure 3-41 are indicative of an inductive component such as a bond wire, a connector pin, or a package lead frame. Assuming that the input step has a sufficiently fast rise time, the value of the inductor can be estimated by measuring the area under the inductive spike, as shown in Figure 3-45. The area is calculated by integrating equation (3-110b) after subtracting the dc offset and assuming thatRt =Z0:
vss=vs
Rt
Rt+Rs = 50
50+50 = vs 2 Aind=
∞
2τd
vs
2
1+e−
(t−2τd )(Z0+Rt ) L
−vs
2dt
= vs
2 ∞
0 e−(2Zo/L)tdt= vsL
4Z0 (3-114)
L= 4Z0Aind
vs
Therefore, through careful measurement of the area under the inductive spike, we can achieve a good approximation of the inductance. The accuracy is maximized when the rise time is very fast compared to the duration of the inductive spike.
0.6 0.58 0.56 0.54 0.52 0.5 0.48 0.46 0.60.44
0.8 1 2 1.8 1.6 1.4 1.2
Time, ns
Voltage at A
Area = Aind
vs
Rs = 50 Ω
Z0 = 50 Ω 0-2 V
A
L
Rt = 50 Ω Z0 = 50 Ω
Figure 3-45 The area under the reflection can be used to estimate the inductance.
0 0.2
0.8 0.7
0.6 0.5
0.4 0.3
0.4 0.6 0.8 1 1.2
Time, ns
Voltage at A
Area = Acap vs
Rs = 50 Ω
Z0 = 50 Ω 0-2 V
A
C Rt = 50 Ω Z0 = 50 Ω
Figure 3-46 The area under the reflection can be used to estimate the capacitance.
For long rise times or very small inductance values, the area under the curve will be masked by a rise time similar to the reflections in Figure 3-37.
Capacitive Structures In a TDR measurement, narrow dips such as that depicted in Figure 3-46 are indicative of an capacitive component such as a probe pad or a via pad. Assuming that the input step has a sufficiently fast rise time, the value
of the capacitance can be estimated by measuring the area under the capacitive dip. The area is calculated by integrating equation (3-108) after subtracting the dc offset, assuming thatRt =Z0 whereτ =Z0C/2
vcapacitor= vs
2(1−e−[2(t−2τd)/ZoC]) Acap=
∞
2τd
vs 2 −vs
2(1−e−[2(t−2τd)/Z0C])dt
= vs
2 ∞
0 e−2t/Z0Cdt= vsCZ0
4 (3-115)
C= 4Acap vsZ0
Example 3-8 Calculate the value of the shunt capacitance for the waveform in Figure 3-47 assuming the circuit shown in Figure 3-46.
SOLUTION Afine grid was overlaid on the waveform in Figure 3-47 so that the area under the curve can be estimated. The total number of squares under the capacitive dip is about 60. The area per square is
Asquare=(0.04)(0.01×10−9)=4×10−13 The total area is therefore
Atot=60Asquare=2.4×10−11
0 0.2
0.8 0.75 0.7 0.65 0.6 0.55 0.5 0.45 0.4 0.35 0.3 0.4 0.6 0.8 1 1.2
Time, ns
Voltage at A
Figure 3-47 TDR waveform for Example 3-8.
1.4 1.2 1 0.8 0.6 0.4 0 0.2
0.00 0.50 2.00 1.50 1.00
Time, ns
Voltage at A
vs
Rs = 50 Ω
Z0 = 50 Ω 0-2 V
A
L = 2 nH
C= 1 pF td = 250 ps
Z0 = 30 Ω td = 250 ps
Figure 3-48 TDR profile showing the relationship between the waveform and the indi- vidual circuit components.
yielding an estimated capacitance of C= 4Acap
vsZ0 = 4(2.4×10−11)
2(50) =0.96×10−12 F
which is very reasonable because the waveform for this example was simulated using a capacitance value of 1.0 pF. More accuracy could be obtained by using afiner grid.