Ringing on a transmission Line

8/17/2019 Ringing on a transmission Line

1/33

EEE 272 High Speed Digital Design

Lab 4: Ringing on a Transmission Line

Submitted by:

Prem Bhaskara

Under the Guidance of:

Dr. Milica Markovic


2/33

Bounce Diagrams:

Draw the bounce diagram for a 100Ω transmission line terminated in a 200Ω resistance. The source is

a 10V step function, with internal resistance of 50Ω. The line is 10cm long. Dielectric inside the

transmission line is air. Draw 4 bounces. Label the time when the generator is turned on as 0. Label

each other time step at both load and generator sides.

Calculations of electrical length:

=ℎ

ℎ, =0.01

3 ∗ 1 08sec = 0.33

= 0.33

ℎ ℎ = 360

∗

=1

=

1

1= 1

= ∗ 360

=

(0.33 ∗ 360)

1= 120°

∴ = 120°

Calculation of reflection coefficient at the load and generator side:

= + +

= 10 + 10050 + 100

= 6.666

=200 − 100

200 + 100= 0.333

=50 − 100

50 + 100= −0.333


3/33

Bounce Diagram:

Schematic:


4/33

5

Fig. 1.4: Simulation Results

Thus our simulation results obtained match with our hand calculations.


5/33

6

3. Ringing on a transmission lineMake a new schematic diagram in ADS and add Transient analysis simulator. Select the start, stop and

step time after you make the circuit.

1. Re-create diagrams in Figure 8-16, page 302, as discussed below.

(a) Place a Sources-Time Domain source VtPulse with rise-time of 0.5ns and frequency of 100MHz.

Add a 10 resistor in series with the generator. This resistance simulates the internal resistance of the

generator.

(b) Place an ideal transmission line on the schematic (TLIN), and connect it to the source.

(c) Set the transmission line impedance of the line to be 50-, and the length of the transmission line to

be 20% (then 30%, adn 40%) of the rise time of the VtPulse. Remember that you don't have to set the

electrical length of the line at 100MHz frequency. It's better if you set it at f=1GHz. Show calculations.

What is the actual length of this transmission line?

Ans:

To set the length of the transmission line to 20% of the rise time,

Electrical length = 2 * 180 * TD

Case 1: TD = 20 % or rise time

here, rise time = 0.5 ns, so TD = 0.5 ns * 20 / 100 = 0.1 ns

So, the electrical length (θ) = 2 * 180 * 0.1 = 36 degree

Circuit with 20% rise-time

Fig 1 Schematic


6/33

7

Fig 1.1 Simulation


7/33

8





Fig1.2 Schematic

Fig 1.3 Simulation


8/33

9





Fig 1.4 Schematic

Fig 1.4 Schematic


9/33

10

To calculate actual length (L),

TimeDelay (TD) = L / λ

Where, λ = C / ( freq *√ )Where, C = speed of light in air = 12 inches / ns

freq = 1 GHz

εr = 1

So, L = TD * λ = TD * C / freq

For case 1: TD = 0.1 ns

So, L = 0.1 * (12 /10-9) / 1 * 109 = 0.1 * 12 = 1.2 inches


So, L = 0.15 * (12 /10-9) / 1 * 109 = 0.15 * 12 = 1.8 inches


So, L = 0.2 * (12 /10-9) / 1 * 109 = 0.2 * 12 = 2.4 inches

(d) Leave the other side of the transmission line open.

(e) Label the generator side between the resistor and the line as vin and the resistor side as vout.

(f) Simulate the above circuit and show the input and output voltage as a function of time on the same

graph. What can you conclude? What is the maximum lenght of the line before signal-integrity

problems arise?

According to the simulation result, as we increase the time delay from 20% rise time to 40 % rise time,

the ringing also increases. We can conclude that if the time delay is short compared with the rise time,

the multiple bounces will be smeared over the rise time and it will be negligible ringing. If we decreasethe time delay less than 20%, we can hardly see any ringing.

The maximum length for an unterminated line before signal integrity problem arise is roughly:

Lenmax < RT

Where, Lenmax = the maximum length for an unterminated line, in inches

RT = the rise time , in nsec

Here, RT = 0.5 ns, so Length should be less than 0.5 inches.


10/33

11

2. Repeat the above simulation if source-series termination is applied to rectify the problem. What is

the source-series termination? Explain what do you have to add to the circuit? How does that prevent

ringing?

Ans: Source-series termination : As we know, the origin of the ringing is the impedance discontinuities

at the source and the far end and the multiple reflections back and forth. To prevent this ringing, we can

add impedance on the source side, to match the impedance of the transmission line so that when the

reflected signal comes back to the source and if it sees the same impedance then it will not reflect back.

This addition of additional resistance, we call it as a source series termination as we add the additional

resistor in the series at the input of the transmission line, so we call it as source-series transmission.

In our circuit, the impedance of the transmission line is 50Ω and the impedance of the voltage source is

10Ω. To match the impedance at the input side with the impedance of the transmission line, we have to

add additional resistor of 40Ω in series with the source so that the total impedance will be 10Ω + 40Ω =

50Ω. We can see the addition of source-series termination resistor in the circuit shown below.

As we know, ringing happens only when there is a change in impedance in the transmission line. So,

when the reflected signal from far end travels back to the source and reaches the series terminating

resistance, it sees looking into the source resistance and that is just the same as transmission line; so thesignal will encounter no impedance change and there will be no reflection. Thus, source-series

termination prevents ringing. We can see the output result in the output waveform in below figure.


11/33

12

Circuit with source-series termination resistor

Fig 2 Schematic

Fig2.1 Output of the simulation


12/33

13

3. Re-create diagrams in Figure 8-20, page 308, for three widths of transmission line 25, 50! and 75 on

an electrically long line. What should the percent change of the characteristic impedance be to keep

the reflection noise below 5%? Demonstrate this with your simulations. What should the length of the

line be in terms of percent of rise time to keep the discontinuity negligible? Prove it with your

simulations.

Ans:

To keep the reflection noise to less than 5% of the voltage swing requires keeping the characteristic-

impedance change to less than 10%. The reason behind this is the impedance discontinuities contribute

to reflection noise. The more the impedance discontinuities, the more reflection noise.

We can see in our examples of three transmission lines with different characteristic impedance – 25, 50

and 75 Ω.


13/33

14

Circuit with the transmission line : 25 Ω impedance

Fig 3 Schematic

Fig 3.1 Simulation

Circuit with the transmission line : 50 Ω impedance


14/33

15

Fig 3.2 Schematic

Fig 3.3 Simulation Results


15/33

16

Circuit with the transmission line: 75 Ω impedance

Fig3.4 Schematic

Fig3.5 Simulation

We can conclude from the simulation results that reflection noise can be kept low to 5% of the voltage

swing by keeping the characteristic impedance change to less than 10%. In simulation, we can see, when

we keep characteristic impedance change from 25 Ω to 50 Ω or 75 Ω to 50 Ω, there is always the


16/33

17

reflection noise. But, when we kept the characteristic impedance change from 50 Ω to 50 Ω, there is no

reflection noise on the transmission line.

What should the length of the line be in terms of percent of rise time to keep the discontinuity

negligible? Prove it with your simulations.

To keep the discontinuity negligible, the length of the line should be such that the time delay (TD) is

shorter than 20% of the rise time.

To prove this, we have kept the length of the transmission line such that the timedelay will be 40%, 30%

and 20% of the risetime, the impedance as 25 ohm, and then we have observed the output for each

circuit.


17/33

18

Circuit with the transmission line: length as 40% of the rise-time

Fig 3.6 Schematic

Fig 3.7 Simulation


18/33

19


Fig3.8 Schematic

Fig 3.9 Simulation


19/33

20


Fig 3.10 Schematic

Fig3.11 Simulation



20/33

21

Fig3.12 Schematic

Fig3.13 SimulationFrom the simulation results, we can conclude that the length of the line should be such that the time

delay will be less than 20% of the rise time to keep the discontinuity negligible.


21/33

22

4. Re-create diagrams in Figure 8-22, page 311, for three stub lengths (0, 20%, 40% and 60% of RT.

What should be the stub’s time delay of the stub transmission line 25Ω, 50Ω and 75Ω on an

electrically long line? What is the maximum length of the stub in terms of percent of rise time that will

keep the reflection noise acceptable? Demonstrate this in your simulation, and find the actual length

of the stub in that case.

Short stub reflections

Figure 3.4.1: Schematics for Short stub reflections due to varying stub lengths (0%, 20%, 40%, and 60%

of RT)


22/33

23

Figure 3.4.2: Simulations for Short stub reflections due to varying electrical lengths

We changed the electrical length of the transmission line stub to 0, 20%, 40% and 60% of rise time and

observed reflections for different electrical lengths.

Here we are given different stub lengths we have calculated electrical length for those different cases as

below:

Case 1: TD = 0 % of rise time

Here, rise time = 0.5 ns, so TD = 0.5 ns * 0 / 100 = 0

So, the electrical length (θ) = 2 * 180 * 0= 0 degree


Here, rise time = 0.5 ns, so TD = 0.5 ns * 20 / 100 = 0.1 ns









23/33

24

From our class notes we know that,

Maximum physical length = Speed of the signal * Time Delay

We know that for FR4 speed of the signal = 6 inches/nsec.

So the Maximum physical length can be given as


Here, rise time = 0.5 ns, so TD = 0.5 ns * 0 / 100 = 0

So, the Maximum electrical length = 6inches/nsec * 0 = 0



So, the Maximum electrical length = 6inches/nsec * 0.1 = 0.6 inches







For: TD = 20 % of rise time it is close to TLIN with no discontinuity () whereas for = 60 % ofrise time, reflections are high because for higher length. Thus Shorter the length of the stub, lesser the

reflections due to discontinuity

Hence length should be shorter than the rise time of the signal as said in eqn 8.16 from the book

Maximum acceptable length for a discontinuity is _


24/33

25

What is the maximum length of the stub in terms of percent of rise time that will keep the reflection

noise acceptable? Demonstrate this in your simulation, and find the actual length of the stub in that

case.

Since for 20% of the rise time have Length of the stub as 0.6 inches, which is greater than the rise time

value o.5nsec. Thus we can approximately say that Maximum length of the stub can be kept as as high

as 15% of the rise time

TD = 15 % of rise time



Thus actual length of the stub will be 0.45inches

Since 0.45 < 0.5 so it implies that Lstubmax < Rise Time

Thus at 15% of the rise time the reflection noise will be acceptable.

Figure 3.4.3: Schematics for Short stub reflections due to stub length 15% of the Rise Time and ideal

case (To keep reflection noise acceptable)


25/33

26

Figure 3.4.4: Simulations for Short stub reflections due to stub length 15% of the rise time and ideal

case

Thus from the above graph we can see that maximum length of the stub should be 15% of the Rise time

that will keep the reflection noise acceptable.


26/33

27

5. Re-create diagrams in Figure 8-23, page 312, for a transmission line end-terminated in three

capacitance values 2pF, 5pF and 10pF. What is modeled by these capacitances? Give several

examples. Can you estimate the new rise times (at the beginning and end of TL) and the increase in

time delay for these specific cases? How do your calculations compare to your simulation results?

Reflections due to capacitive load

Figure 3.5.1: Schematics for reflections due to varying capacitive load


27/33

28

Figure 3.5.2: Simulations for reflections due to varying capacitive load

If the rise time is short compared to the charging time of the capacitor, then initially, the voltage will rise

up very fast and the impedance will be low. But as the capacitor charges, the voltage across it gets

smaller and dV/dt slows down. The exact behavior will depend on the characteristic impedance of the

line (Z0), the capacitance of the capacitor and the rise time of the signal.

Thus this capacitors in the circuit model to the “RC circuit” behavior where C is the capacitance of the

load and R is the characteristic impedance of the transmission line, Z0.

Calculation of Rise Time for 2pF, 5pF, 10pF:

We have the formula from eqn. (8-20)

RT = 2.2 * Z0 * C

RT2pF = 2.2 * 50 * 2 = 0.22 nsec

RT5pF = 2.2 * 50 * 5 = 0.55 nsec

RT10pF = 2.2 * 50 * 10 = 1.1 nsec


28/33

29

Calculation of Time Delay for 2pF, 5pF, 10pF:

We know that Total Capacitance is given by =

For 2pF

Time Delay is TD = CT ∗ Z0 = 2 ∗ 50 = 0.1

For 5pF


For 10pF


Thus from our simulation graphs it is clear that the higher the capacitor the higher the rise time the

higher the time delay. Thus our calculations show the similar results as that of the simulations.


29/33

30

6. Re-create diagrams in Figure 8-24, page 314, for a transmission line with capacitive load in the

middle of a trace for three capacitance values: 2pF, 5pF and 10pF. Calculate the maximum capacitance

for a specific rise-time and transmission line impedance. Compare the value with your simulations.

What should be done to keep capacitive discontinuities from causing excessive undershoot noise?

Reflection from Capacitive Loads in the Middle of a Trace

Fig. 3.6.1: Schematic with three different capacitance values


30/33

31

Fig. 3.6.2: Simulation results due to varying capacitive discontinuity

Maximum allowable capacitance for a rise time of .5nsec and 50Ω transmission line impedance is 2pF.Asit is clear from the simulation results capacitance above 2pF has more discontinuities.

In order to keep capacitive discontinuities from causing excessive undershoot noise ,keep the

capacitance less than four times the rise time (rise time in nsec).

For a line with characteristic impedance of 50 Ω, Maximum allowable rise time for a capacitive

discontinuity is given as = 4 ∗

Maximum capacitance for rise time of 0.5 ns, = 4 ∗ 0.5 = 2

For 0.5 nsec, maximum allowable capacitance is 2 pF. For 0.5 nsec rise time, capacitance above 2pF will

have more undershoot problems


31/33

32

7. Re-create diagrams in Figure 8-34, page 329, for a transmission line with inductive load in the

middle of a trace for three inductance values: 1nH, 5nH and 10nH. What is the maximum acceptable

inductive discontinuity for this example? Compare calculations and simulations. How can we

compensate this series inductance, if its large value is unavoidable? Propose a solution and simulate

it.

Reflections from Inductive Discontinuities

Fig. 3.7.1: Schematic with three different inductive discontinuity


32/33

33

Fig. 3.7.2: Simulation Results for reflections due to varying inductive discontinuity

Maximum allowable inductive discontinuity is given as < 0.2 ∗ 0 ∗ Lmax< .2*50*50


33/33

Compensation circuit for an inductive discontinuity

Fig. 3.7.3: Compensation Schematic

Fig. 3.7.4: Simulation results

Ringing on a transmission Line

Documents

Transcript of Ringing on a transmission Line