Download Lecture 36

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
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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
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EE 42 fall 2004 lecture 36
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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
Z0 
Vpulse
I pulse
• 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.
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EE 42 fall 2004 lecture 36
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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
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EE 42 fall 2004 lecture 36
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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.
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EE 42 fall 2004 lecture 36
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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.
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EE 42 fall 2004 lecture 36
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Power
• The power required to transmit signals on a
transmission line is P=IV, so:
V02
P
Z0
• 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.
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EE 42 fall 2004 lecture 36
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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.
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EE 42 fall 2004 lecture 36
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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.
R  Z0
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EE 42 fall 2004 lecture 36
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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
V (t, x)  V0 f  (t  x / v)
Where t is time, and x is distance along the line
11/24/2004
EE 42 fall 2004 lecture 36
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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
f  (t )  sin( t )
f  (t  x / v )  sin( t  x / v )
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EE 42 fall 2004 lecture 36
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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.
V (t , x )  V0 f  (t  x / v )
V0
I (t , x ) 
f  (t  x / v )
Z0
Where t is time, and x is distance along the line
11/24/2004
EE 42 fall 2004 lecture 36
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Transmission line formulas
• Since pulses can propagate either way on
the transmission line, we have:
V (t , x )  V0 f  (t  x / v )  V0 f  (t  x / v )
V0
V0
I1 ( t , x ) 
f  (t  x / v ) 
f  (t  x / v )
Z0
Z0
I 2 ( t , x )   I1 ( t , x )
Notice that the current changes direction for the
signal propagating in the reverse direction
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EE 42 fall 2004 lecture 36
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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?
R
Z0
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EE 42 fall 2004 lecture 36
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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.
f  (t  x / v) 
 f  (t  x / v)
R
Z0
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EE 42 fall 2004 lecture 36
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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
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EE 42 fall 2004 lecture 36
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Transmission line formulas
• First, lets make our coordinate system so
that x=0 occurs at the resistor.
V (t , x)  V0 f  (t  x / v)  V0 f  (t  x / v)
becomes V (t , x)  V0 f  (t )  V0 f  (t )
and
V0
V0
I1 (t , x) 
f  (t  x / v) 
f  (t  x / v)
Z0
Z0
V0
V0
becomes I1 (t , x) 
f  (t ) 
f  (t )
Z0
Z0
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EE 42 fall 2004 lecture 36
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Transmission line formulas
• And we have VR=IRR
V (t , x)  V0 f  (t )  V0 f  (t )
 V0

V0
R R I1 (t , x)  RR  f  (t ) 
f  (t )
Z0
 Z0

• And now we just need to solve for the
reflected pulse f11/24/2004
EE 42 fall 2004 lecture 36
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Transmission line formulas
Z 0 f  (t )  Z 0 f  (t )  RR f  (t )  RR f  (t )
RR f  (t )  Z 0 f  (t )  Z 0 f  (t )  RR f  (t )
f  (t )RR  Z 0   f  (t )Z 0  RR 
f  (t ) RR  Z 0 

f  (t ) RR  Z 0 
• So if the resistor has a resistance R=Z0, then there will
be no reflection.
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EE 42 fall 2004 lecture 36
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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
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EE 42 fall 2004 lecture 36
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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
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EE 42 fall 2004 lecture 36
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Branch in a transmission line
• Let’s see what happens if we just join from
one transmission line into two
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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.
Z0
Z0
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EE 42 fall 2004 lecture 36
Z0
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Reflected amplitude
f  (t ) RR  Z 0  Z 0 / 2  Z 0 
1



f  (t ) RR  Z 0  Z 0 / 2  Z 0 
3
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.
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EE 42 fall 2004 lecture 36
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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
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EE 42 fall 2004 lecture 36
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