11/24/2004 EE 42 fall 2004 lecture 36 1

Lecture #36: Transmission lines

• Last lecture:– Transmission lines– Balanced and unbalanced– Propagation

• This lecture:– Transmission line equations– Reflections and termination

11/24/2004 EE 42 fall 2004 lecture 36 2

Transmission Line

• Voltage

• Current in (+) signal line

• Current in (-) signal line, or current in ground

• If the signal is conducted in a pair of lines, it is called a balanced line. If the return path is through a ground, it is called unbalanced

11/24/2004 EE 42 fall 2004 lecture 36 3

Transmission line impedance

• The ratio of the voltage of propagating pulses to the current the carry is a constant, called the impedance of the line

• A typical transmission line impedance is 50-100 ohms.• When a line is being used in this fashion, it can not be

split into two, because for a given voltage, twice as much current would be needed

• So digital transmission lines do not branch, there is only one path from one end to another.

pulse

pulse0 I

VZ

11/24/2004 EE 42 fall 2004 lecture 36 4

Transmission line impedance

• The impedance of a line depends on the width of the conductors and their separation.

• The greater the separation between the conductors, the higher the impedance

• The wider the conductors, the lower the impedance

11/24/2004 EE 42 fall 2004 lecture 36 5

Practical TL impedances

• If the separation is large compared to the width of the conductors, the impedance gets larger only logarithmically, so impedances much higher than 300 ohms or so are impossible to achieve without losses getting very large

• Low impedance lines can be made with wide conductors which are close together, but are usually undesirable because of the high current required.

11/24/2004 EE 42 fall 2004 lecture 36 6

Drivers for transmission lines

• A driver for a transmission line needs to put out a current I=V0/Z0, of course, which is quite a bit of current for a logic driver.

• In general this requires several cascaded inverters, each with wider transistors that the previous ones.

11/24/2004 EE 42 fall 2004 lecture 36 7

Power

• The power required to transmit signals on a transmission line is P=IV, so:

• In order to reduce the power required, reduced voltages can be used.

• If voltages lower than logic levels are transmitted, then amplifiers (comparators) are needed at the inputs to convert the voltage levels.

0

20

Z

VP

11/24/2004 EE 42 fall 2004 lecture 36 8

Power (2)

• Notice that the power required to transmit information on a transmission line does not depend on the data rate.

• This means it is an advantage to reduce the total number of signal lines, by increasing the data rate.

• The data rate is limited by the CMOS drivers, and by the quality of the transmission line.

11/24/2004 EE 42 fall 2004 lecture 36 9

Transmission line model (from the driver’s perspective)

• Since transmitting on a long transmission line means providing a current I=V0/Z0, to the driver it looks exactly like a resistance of R=Z0.

0ZR

11/24/2004 EE 42 fall 2004 lecture 36 10

Transmission line formulas

• If a transmission line has the same cross section from one end to the other, then propagation on it can be written

)/(),( 0 vxtfVxtV

Where t is time, and x is distance along the line

11/24/2004 EE 42 fall 2004 lecture 36 11

Transmission line formulas

• What this means is that f+ can be any function of a single argument, sin(t) for example. Plugging in the argument (t-x/v) shifts the function, so that at points some distance x down the line, you see a delayed version of the function, delayed by the time x/v

)/sin()/(

)sin()(

vxtvxtf

ttf

11/24/2004 EE 42 fall 2004 lecture 36 12

Transmission line formulas

• Since the transmission line has the same cross section at each point along its length, it has a constant ratio of current to voltage, Z0, and the current pulse has the same shape.

)/(),(

)/(),(

0

0

0

vxtfZ

VxtI

vxtfVxtV

Where t is time, and x is distance along the line

11/24/2004 EE 42 fall 2004 lecture 36 13

Transmission line formulas

• Since pulses can propagate either way on the transmission line, we have:

),(),(

)/()/(),(

)/()/(),(

12

0

0

0

01

00

xtIxtI

vxtfZ

Vvxtf

Z

VxtI

vxtfVvxtfVxtV

Notice that the current changes direction for the signal propagating in the reverse direction

11/24/2004 EE 42 fall 2004 lecture 36 14

Transmission line into a resistor

• What happens at the end of a transmission line? For example an open circuit R=∞? Where does the power go?

RZ0

11/24/2004 EE 42 fall 2004 lecture 36 15

Transmission line into a resistor

• A voltage pulse coming down the transmission line and hitting the resistor can generate a reflected pulse traveling back up the transmission line.

RZ0

)/(

)/(

vxtf

vxtf

11/24/2004 EE 42 fall 2004 lecture 36 16

Transmission line formulas

• Since the voltage at the end of the transmission line is the same voltage which appears across the resistor, and since the current in each of the wires of the transmission line is the current in the resistor (don’t count it twice!) we can set up an equation to find the reflected pulse

11/24/2004 EE 42 fall 2004 lecture 36 17

Transmission line formulas

)()(),( becomes

)/()/(),(

and

)()(),( becomes

)/()/(),(

0

0

0

01

0

0

0

01

00

00

tfZ

Vtf

Z

VxtI

vxtfZ

Vvxtf

Z

VxtI

tfVtfVxtV

vxtfVvxtfVxtV

• First, lets make our coordinate system so that x=0 occurs at the resistor.

11/24/2004 EE 42 fall 2004 lecture 36 18

Transmission line formulas

)()(),( R

)()(),(

0

0

0

01R

00

tfZ

Vtf

Z

VRxtI

tfVtfVxtV

R

• And we have VR=IRR

• And now we just need to solve for the reflected pulse f-

11/24/2004 EE 42 fall 2004 lecture 36 19

Transmission line formulas

• So if the resistor has a resistance R=Z0, then there will be no reflection.

0

0

00

00

00

)(

)(

)()(

)()()()(

)()()()(

ZR

ZR

tf

tf

RZtfZRtf

tfRtfZtfZtfR

tfRtfRtfZtfZ

R

R

RR

RR

RR

11/24/2004 EE 42 fall 2004 lecture 36 20

Transmission line termination

• Since a transmission line carries pulses of voltage and current, there must be somewhere for the current to go.

• FET devices are very high impedance, so they don’t absorb that current

• All that is required to absorb the current at the end of the transmission line is a resistor which has the same resistance as the impedance of the line Rtermination=Z0

• If a termination is not provided, then a reflection will be created, propagating back up the line

11/24/2004 EE 42 fall 2004 lecture 36 21

Changes along a TL

• If a transmission line changes in impedance at a point along its length, then that will cause a reflection

• If a transmission line branches, that will cause a reflection

• Transmission lines must be build with uniform impedance along its length, from one end to another with no branches

11/24/2004 EE 42 fall 2004 lecture 36 22

Branch in a transmission line

• Let’s see what happens if we just join from one transmission line into two

11/24/2004 EE 42 fall 2004 lecture 36 23

Replace transmission lines with a resistor to model their input

• Each of the outgoing transmission lines need a current I=V/Z0, so they can each be modeled by a resistor.

• The transmission line from the left now sees a termination into a resistance R=Z0/2.

0Z 0Z0Z

11/24/2004 EE 42 fall 2004 lecture 36 24

Reflected amplitude

3

1

2/

2/

)(

)(

00

00

0

0

ZZ

ZZ

ZR

ZR

tf

tf

R

R

So there will be a reflected pulse, inverted, at 1/3 the amplitude, and each of the branches will see a pulse of 2/3 the voltage of the original.

11/24/2004 EE 42 fall 2004 lecture 36 25

Transmission line bus• To use transmission lines in a bus, each driver

along the bus must be connected very close, and they must not consume significant current from pulses going by. Each end of the bus must have terminators for each transmission line