Output drivers with open collector outputs (also called "open drain outputs" in the case of CMOS drivers) have an output circuit that can, when enabled, drive the output voltage to 0 volts. When disabled, the output circuit presents a high impedance to the output node. To establish a high output voltage when the output circuit is disabled, an open collector driver relies on an external resistance (a pull-up resistor) connected from the output node to the positive power supply voltage. It is thus safe to connect multiple open collector output drivers to the same node since they will never be driving opposite voltages onto the node. Either (a) at least one driver is driving low and so the output voltage is 0 volts, or (b) all of the drivers are disabled leaving the pull-up resistance to establish a high output voltage.Output drivers with tri-state enable can drive the output node high, or low, or can go into a high impedance state. If multiple output drivers with tri-state enable are connected together, then there must be some arbiter circuit that ensures that the drivers are never driving the same output node to opposite values.
For both kinds of output drivers it is possible to connect multiple drivers to one node or wire. In the case of output drivers with open collector outputs, a wired-AND function is obtained "for free" on the output node (this feature is exploited in the GPIB) if multiple drivers can be on at the same time. The situation is more complicated with tri-stateable drivers. Only one driver should be enabled at a time to drive the common output node. Also, a wired-AND function is not obtained when tri-stateable drivers share an output node.
The four differential equations for a lossless transmission line are:
dV(z,t)/dz = -Lo dI(z,t)/dt (Eqn 1)
dI(z,t)/dz = -Co dV(z,t)/dt (Eqn 2)
d2V(z,t)/dz2 = Lo Co d2V(z,t)/dt2 (Eqn 3)
d2I(z,t)/dz2 = Lo Co d2I(z,t)/dt2 (Eqn 4)
The assumed wave solutions are as follows:
voltage: V(z,t) = g( t - s.z ) + f( t + s.z ) where s = sqrt( Lo . Co )
current: I(z,t) = (1/Zo).g( t - s.z ) - (1/Zo).f( t + s.z )
Recall that the impedance Zo is negative for a left-moving wave.
Note that g() and f() are functions of a single variable. Thus:
dV(z,t)/dz = (-s).g'( t - s.z ) + s.f'( t + s.z )
and -Lo.dI(z,t)/dt = (-Lo/Zo).g'( t - s.z ) - (-Lo/Zo).f'( t + s.z )
But (Lo/Zo) = Lo.sqrt(Co/Lo) = sqrt( Lo Co ) = s
So -Lo.dI(z,t)/dt = (-s).g'( z - s.t ) + s.f'( z + s.t )
= dV(z,t)/dz (Eqn 1)
Similarly,
dV(z,t)/dt = g'( t - s.z ) + f'( t + s.z )
dI(z,t)/dz = -(s/Zo).g'( t - s.z ) - (s/Zo).f'( t + s.z )
And -Co.dV(z,t)/dt = (-Co).g'( t - s.z ) + (-Co).f'( t + s.z )
But s/Zo = sqrt( Co.Lo ) . sqrt( Co/Lo ) = Co
So dI(z,t)/dz = (-Co).g'( t - s.z ) + (-Co).f'( t - s.z )
= -Co . dV(z,t)/dt (Eqn 2)
From above dV(z,t)/dz = (-s).g'( t - s.z ) + s.f'( t + s.z )
so d2V(z,t)/dz2 = (s^2).g"( t - s.z ) + (s^2).f"( t + s.z )
From above dV(z,t)/dt = g'( t - s.z ) + f'( t + s.z )
so d2V(z,t)/dt2 = g"( t - s.z ) + f"( t + s.z )
But s^2 = (sqrt( Lo.Co ))^2 = Lo.Co
Thus d2V(z,t)/dz2 = Lo Co d2V(z,t)/dt2 (Eqn 3)
From above dI(z,t)/dz = (-s/Zo).g'( t - s.z ) + (-s/Zo).f'( t + s.z )
so d2I(z,t)/dz2 = (s^2/Zo).g"( t - s.z ) + (-s^2/Zo).f"( t + s.z )
From above dI(z,t)/dt = (1/Zo).g'( t - s.z ) + (1/Zo).f'( t + s.z )
so d2I(z,t)/dt2 = (1/Zo).g"( t - s.z ) + (1/Zo).f"( t + s.z )
But s^2 = sqrt( Lo.Co )^2 = Lo.Co
Thus d2I(z,t)/dz2 = Lo Co d2I(z,t)/dt2 (Eqn 4)
Zo = (60/sqrt(3.2)) ln(4.0/0.7) = 33.541 * 1.7423 = 58.46 ohms speed of propagation = c / sqrt(3.2) = 0.559 c = 16.77 cm/ns
Zo = 120/sqrt(1.0) ln( (2.2/0.5) + sqrt( (2.2/0.5)^2 - 1) ) = 120 ln( 4.4 + sqrt(18.36)) = 259.4 ohms speed of propagation = c/sqrt(1.0) = c = 30 cm/ns
effective relative permittivity of the dielectric = 3.5 Zo = 60/sqrt(3.5) ln( 4*2.3 /(0.67*(0.8*1.5 + 0.1))) = 32.071 * ln( 9.2 / 0.871 ) = 75.6 ohms speed of propagation = c/sqrt(3.5) = 0.5345 c = 16.035 cm/ns
voltage reflection coefficient = (1.1*Zo - Zo) / (1.1*Zo + Zo) = 0.1 / 2.1 = 0.04762 Therefore the reflected voltage wave will have an amplitude of 47.62 mV transmission coefficient = (2 * 1.1 * Zo)/(1.1*Zo + Zo) = 2.2 / 2.1 = 1.04762 Therefore the voltage wave propagating forward will have an amplitude of 1.04762 volts.
Recall that light travels 30cm in 1 nanosecond. Therefore this signal
will propagate 15 cm in 1 nanosecond.
t = 0- ^ ^
| |
V | I|
| |
| |
+-----------+-----------+->z +-----------+-----------+->z
0 15cm 30cm 0 15cm 30cm
t = 0+ ^ ^
| |
V | I|
1.25V*--> 0.5c 25mA*--> 0.5c
* *
*-----------+-----------+->z *-----------+-----------+->z
0 15cm 30cm 0 15cm 30cm
t = 1ns ^ ^
| |
V | I|
1.25V*************--> 0.5c 25mA*************--> 0.5c
************* *************
*************-----------+->z *************-----------+->z
0 15cm 30cm 0 15cm 30cm
t = 2-ns ^ ^
| |
V | --> 0.5c I| --> 0.5c
1.25V************************* 25mA*************************
************************* *************************
*************************->z *************************->z
0 15cm 30cm 0 15cm 30cm
t = 2+ns ^ ^
| 0.5c <-- * |
V | * I| 0.5 c <--
************************* 25mA*************************
************************* *************************
*************************->z *************************->z
0 15cm 30cm 0 15cm 30cm
t = 3ns ^ ^
2.5V| 0.5c <-- ************* |
V | ************* | 0.5c <--
************************* 25mA*************
************************* *************
*************************->z *************-----------+->z
0 15cm 30cm 0 15cm 30cm
t = 4-ns ^ ^
2.5V ************************* |
V ************************* |
************************* 25mA*
************************* *
*************************->z *-----------+-----------+->z
0 15cm 30cm 0 15cm 30cm
t = 4+ns ^ ^
2.5V ************************* |
V ************************* |
************************* |
************************* |
*************************->z +-----------+-----------+->z
0 15cm 30cm 0 15cm 30cm
t = 5ns ^ ^
2.5V ************************* |
V ************************* |
************************* |
************************* |
*************************->z +-----------+-----------+->z
0 15cm 30cm 0 15cm 30cm
Brief explanations: At just before t=0ns, there are no voltage or current
waveforms on the transmission line. Just after t=0ns, a voltage wave of
amplitude 1.25V will appear at the left end of the transmission line,
propagating to the right at a speed of 0.5c. The amplitude is equal to
1.25 V as a result of the voltage division effect of 2.5 V from the ideal
voltage source across the 50 ohm series termination resistance and the
50 ohm characteristic impedance of the transmission line. Thus the left
end of the transmission line only sees 1.25 V. As the voltage wave
travels to the right, a current wave of amplitude (1.25 V)/(50 ohm) = 25 mA
will also travel to the right. The two waves will take 2 ns to right
the right end of the transmission line.
Just after t=2ns, a voltage wave of amplitude 1.25 V appears at the
right end, which travels to the left at a speed of 0.5 c. At the same
time, a current wave of amplitude (1.25 V)/(-50 ohm) = -25 mA appears
at the right end and begins to propagate to the left. The rightward-moving
and leftward-moving voltage waves add up since they are both positive in
magnitude; the current waves subtract to 0 mA because they are equal in
magnitude but opposite in sign. The leftward-moving waves take 2 ns
to reach the left end of the transmission line.
At time t=4ns, the leftward moving waves encounter the 50 ohm series
termination impedance, and so are completely absorbed. There are no
further reflections. The line is left with a steady voltage 2.5 volts
with zero current at all positions along the line. This is what would
should expect when one applies a voltage of 2.5 V to an open circuited
cable.