Lecture #35: data transfer
• Last lecture:
– Communications synchronous / asynchronous
– Buses
• This lecture
– Transmission lines
11/22/2004
EE 42 fall 2004 lecture 35
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Transmission Lines
• Light travels one meter in approximately 3
nanoseconds. (c=3x108 meters per
second in air or in a vacuum)
• When rise times are comparable to
propagation times, then Transmission line
effects become important.
• Transmission line effects are already
important for USB, Ethernet, Personal
computer busses.
11/22/2004
EE 42 fall 2004 lecture 35
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Long signal paths
• Signal delay
– Component delay + interconnection delay
• Signal integrity
•
•
•
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Reflections
Waveform distortion
Signal attenuation
Crosstalk
EE 42 fall 2004 lecture 35
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Gate delay
• So far, the only delay that we have considered is the
time needed to charge up the capacitance due to lines,
and due to the capacitance of the gates of the next
stage.
• This implies that if we could only push enough current,
we can make the delay as short as we like.
R
Wire
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EE 42 fall 2004 lecture 35
Gates
4
Gate delay
• If we want to speed up logic, we can
increase the drive
– reduce the pull up and pull down resistance
• or we can reduce the capacitance.
– Shorter lines, narrower lines, reduced
dielectric constant of dielectric between wires.
This is limited, however by the speed of light
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EE 42 fall 2004 lecture 35
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Light speed
• With lines that are long enough, or switching
speeds are high enough, then we can not
consider the delay to be simply that due to
charging up the capacitance. The signal will
propagate along a wire at the speed of light (in
the dielectric, which is slower than that in air or
vacuum)
• Quite a bit of current is necessary to pull up a
line that fast, for a 1 volt signal, 10-20 milliamps
are required.
11/22/2004
EE 42 fall 2004 lecture 35
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Inductance of wire
• The reason that transmission can not be sped
up past a certain limit is due to the fact that
pushing current through a wire creates a
magnetic field.
• This means that the conductors used for
communications have an inductance which the
current must be driven through.
• If changes in current are slow, or the path is
short, this can be neglected, but for high speed
communications it dominates
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EE 42 fall 2004 lecture 35
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Inductor
Inductor:
dI
V L
dt
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EE 42 fall 2004 lecture 35
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Transmission line model
• To properly account for the inductance of
the wires in a communications link, we
need to put in the distributed inductance
and capacitance all along the wires
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EE 42 fall 2004 lecture 35
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Transmission line model
• Why do we need both the signal wire and the
return path?
• Because a small loop has much more
inductance than a small loop, so the return path
needs to be kept very close to the signal wire.
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EE 42 fall 2004 lecture 35
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Differential pair
• The return path can be made by running a
wire close to the signal wire, and carrying
the opposite voltage and current,
+V
-V
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EE 42 fall 2004 lecture 35
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Ground plane
• Or the return path can be made close to
the signal wire by providing a ground
plane under all of the signal wires,
B
D
A
C
11/22/2004
EE 42 fall 2004 lecture 35
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Transmission Line
• When a signal wire is driven with that fast a rise
time compared to its length, a signal will travel
along the wire at the speed of light in the media.
• It is even possible to turn off the current, and
have the pulse that is already on the line
continue to propagate toward the destination
• So you don’t have to wait for one bit to arrive
before you send the next
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EE 42 fall 2004 lecture 35
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Transmission Line
• As the voltage pulses propagate down the
line, there is a current pulse which travels
down the line with them.
• The current is always in both conductors,
for example forward in one conductor, and
backward in the other
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EE 42 fall 2004 lecture 35
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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
11/22/2004
EE 42 fall 2004 lecture 35
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Coax
• Coax (coaxial) line is an
unbalanced transmission
line. The central wire
carries the single ended
signal, and the shield is
attached to ground, and
carries the return
current.
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EE 42 fall 2004 lecture 35
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Twisted pairs
• One wire in each pair carries +signal voltage and
current. The other wire carries the current in the
opposing direction, either by being connected to ground,
or by being driven by -signal
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EE 42 fall 2004 lecture 35
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Ribbon cable
• Ribbon cable is inexpensive to make, and
to connect. It is used by running balanced
signals (+V and –V) on adjacent wires, or
by grounding every other wire to provide
isolation and a current return path
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EE 42 fall 2004 lecture 35
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Strip line
In a PC (printed circuit) board high frequency lines
are run in between ground planes, which creates
an unbalanced transmission line called a strip
line.
The wider the metal trace, the lower the impedance
of the transmission line. Impedances are typically
50-100 ohms.
Try out an impedance calculator:
http://www1.sphere.ne.jp/i-lab/ilab/tool/s_line_e.htm
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EE 42 fall 2004 lecture 35
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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 35
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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)  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)
Where t is time, and x is distance along the line
11/22/2004
EE 42 fall 2004 lecture 35
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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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