5.7 Cascaded lines (Page 2/2)

Introduction to physical Page 2 / 2

We can now start to make our bounce diagram. We propagate a +5V wave and a -5V wave (separated by 100ns) downtowards the junction. Since the line is 40m long, and the waves move at $2108 m s$ , it takes 200ns for them to get to the junction. There, a -1.25V wave is reflected back towards thesource, and a +3.75V wave is transmitted into the second transmission line .

Reflection and transmission at the "t"

Since the load for the second line is

50 Ω

, and the characteristic impedance,

Z_{0 ​ 2}

for the second line is

75 Ω

, we will have a reflection coefficient,

Γ_{V_{2}} R_{L ​ 2} Z_{0 ​ 2} R_{L ​ 2} Z_{0 ​ 2} 50755075 -0.2

Thus a -0.75V signal is reflected off of the second load .

What is the magnitude of the voltage which is developed across the second load?

3 Volts!

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Reflection of transmitted pulse

What happens to the 0.75V pulse when it gets to the "T"? Well there is another mismatch here, with a reflection coefficient

Γ_{V_{2 ​ 1}}

given by

Γ_{V_{2 ​ 1}} 25752575 -0.5

(The

50 Ω

resistor and the

50 Ω

transmission line look like a

25 Ω

termination to the

75 Ω

line) and a transmission coefficient

T_{V_{2 ​ 1}} 1 Γ_{V_{2 ​ 1}} 0.5

and so we add to the bounce diagram .

When the reflected load pulse hits the junction

We could keep going, but the voltage reflected off of the second load will only be 75mV now, and solet's call it a day.

There are a couple of other interesting applications of bounce diagrams and the transient behavior oftransmission lines that we might look at before we move on to other things. The first is called the Charged Line Problem . Here it is:

The "charged line" problem

We have a transmission line with characteristic impedance

Z_{0}

and phase velocity

v_{p}

. It is

L

long, and for some time has been connected to a battery of potential

V_{g}

. At time

t 0

, the switch S, is thrown, which removes the battery from the circuit, and connects the line to a load resistor

R_{L}

. The question is: what does the voltage across the load resistor,

V_{L}

, look like as a function of time? This is almost like what we have done before, but not quite.

Initial conditions

In the first place, we now have non-zero initial conditions. For

t 0

we will have both voltages and current on the line. In order to match boundary conditions, we must do more than haveone voltage and one current, because the voltage on the line must be

V_{g}

, while the current flowing down the line must be 0. So, we will put in both a

V^{+}

and a

V^{-}

and their corresponding currents. Note that

x

is going to the left this time. Let's forget about the switch and the load resistor for a minute and just look at theline and battery. We have two equations we must satisfy

V_{0}^{+} V_{0}^{-} V_{g}

and

I_{0}^{+} I_{0}^{-} 0

We can use the impedance relationship to change to:

V_{0}^{+} Z_{0} V_{0}^{-} Z_{0} 0

I hope most of you can then see by inspection that we must have

V_{0}^{+} V_{0}^{-} V_{g} 2

OK, the switch S is thrown at

t 0

. Now the end of the line looks like this .

After the resistor is connected

We have anticipated the fact that we are going to need another voltage and current wave if we are going to be able to matchboundary conditions when the load resistor is connected, and have added a

V_{1}^{+}

and a

V_{1}^{-}

to the line. These are new voltage and current waves which originate at the load resistor position in orderto satisfy the new boundary conditions there. Now we do KVL and KCL again.

V_{0}^{+} V_{0}^{-} V_{1}^{+} V_{L}

and

V_{0}^{+} Z_{0} V_{0}^{-} Z_{0} V_{1}^{+} Z_{0} V_{L} R_{L}

We have already made the impedance substitution for the current equation in . We know what the sum and difference of

V_{0}^{+}

and

V_{0}^{-}

are, so let's substitute in.

V_{g} V_{1}^{+} V_{L}

and

V_{1}^{+} Z_{0} V_{L} R_{L}

From this we get

V_{L} R_{L} Z_{0} V_{1}^{+}

which we substitute back into

V_{g} V_{1}^{+} R_{L} Z_{0} V_{1}^{+}

which we can solve for

V_{1}^{+}

V_{1}^{+} V_{g} 1 R_{L} Z_{0} Z_{0} R_{L} Z_{0} V_{g}

The voltage on the load is given by and is clearly just:

V_{L} V_{g} Z_{0} R_{L} Z_{0} V_{g}

and in particular, when

R_{L}

is chosen to be

Z_{0}

(which is usually done when this circuit is used), we have

V_{L} V_{g} 2

Now what do we do? We build a bounce diagram! Let us stay with the assumption that

R_{L} Z_{0}

, in which case the reflection coefficient at the resistor end is 0. At the open circuit end of the transmissionline

Γ

is +1. So we have this .

Bounce diagram for the charged line problem

Note that for this bounce diagram, we have added an additional voltage,

V_{g}

, on the baseline, to indicate that there is an initial voltage on the line, before the switch is thrown, and

t

starts on the bounce diagram.

If we concentrate on the voltage across the load, we add $V_{g}$ and $V_{g} 2$ and find that the voltage across the load resistor rises to $V_{g} 2$ at time $t 0$ . The $V_{g} 2$ voltage wave travels down the line, hits the open circuit, reflects back, and when it gets to the load resistor,brings the voltage across the load resistor back down to zero. We have made a pulse generator!

Voltage across the load resistor

V t

across

R_{L}

In today's digital age, this might seem like a strange way to go about creating a pulse. Imagine however, if you needed a pulsewith a very large potential (100s of thousands or even millions of volts) for say, a particle accelerator. It is unlikely that aMOSFET will ever be built which is up to the task! In fact, in a field of study called pulsed power electronics just such circuits are used all the time. Sometimes they are built withreal transmission lines, sometimes they are built from discrete inductors and capacitors, hooked together just as in the distributed parameter model . Such circuits are called pulse forming networks or PFNs for short.

Finally, just because it affords us a good opportunity to review how we got to where we are right now,let's consider the problem of a non-resistive load on the end of a line. Suppose the line is terminated with a capacitor! Forsimplicity, let's let $R_{s} Z_{0}$ , so when S is closed a wave $V_{1}^{+} V_{g} 2$ heads down the line . Let's think about what happens when it hits the capacitor. We know weneed to generate a reflected signal $V_{1}^{-}$ , so let's go ahead and put this in the figure , along with its companion current wave.

Transient problem with capacitive load

Initial pulse hits the load

The capacitor is initially uncharged, and we know we can not instantaneously change the voltage across a capacitor (at leastwithout an infinite current!) and so the initial voltage across the capacitor should be zero, making

V_{1}^{-} 0 V_{1}^{+}

, if we make time

t 0

be when the initial wave just gets to the capacitor. So, at

t 0

Γ_{V} 0 -1

. Note that we are making

Γ

a function of time now, as it will change depending upon the charge state of the capacitor.

The current into the capacitor, $I_{C}$ is just $I_{1}^{+} I_{1}^{-} t$ .

I_{C} 0 I_{1}^{+} I_{1}^{-} 0 V_{g} Z_{0}

since

I_{1}^{+} V_{1}^{+} Z_{0} V_{g} 2 Z_{0}

and

I_{1}^{-} 0 V_{1} Z_{0} V_{g} 2 Z_{0}

How will the current into the capacitor

I_{C} t

behave? We have to remember the capacitor equation:

I_{C} t C t V_{C} t C t V_{1}^{+} V_{1}^{-} t C t V_{1}^{-} t

since

V_{1}^{+}

is a constant and hence has a zero time derivative. Well, we also know that

I_{C} t I_{1}^{+} I_{1}^{-} t V_{1}^{+} Z_{0} V_{1}^{-} t Z_{0}

So we equate and and we get

C t V_{1}^{-} t V_{1}^{+} Z_{0} V_{1}^{-} t Z_{0}

t V_{1}^{-} t 1 Z_{0} C V_{1}^{-} t 1 C V_{1}^{+} Z_{0}

which gets us back to another Diff-E-Q!

The homogeneous solution is easy. We have

t V_{1}^{-} t 1 Z_{0} C V_{1}^{-} t 0

for which the solution is obviously

V_{1 ​ homo}^{-} t V_{0} t Z_{0} C

After a long time, the derivative of the homogeneous solution is zero, and so the particular solution (the constant part) is thesolution to

1 Z_{0} C V_{1 ​ part}^{-} 1 C V_{1}^{+} Z_{0}

V_{1 ​ part}^{-} V_{1}^{+}

The complete solution is the sum of the two:

V_{1}^{-} t V_{1 ​ homo}^{-} t V_{1 ​ part}^{-} V_{0} t Z_{0} C V_{1}^{+}

Now all we need to do is find

V_{0}

, the initial condition. We know, however, that

V_{1}^{-} 0 V_{1}^{+}

, so that makes

V_{0} -2 V_{1}^{+}

! So we have:

V_{1}^{-} t -2 V_{1}^{+} t Z_{0} C V_{1}^{+} V_{1}^{+} 1 2 t Z_{0} C

Since

V_{1}^{+} V_{g} 2

we can plot

V_{1}^{-} t

as a function of time from which we can make a plot of

Γ_{V} t

Reflected voltage as a function of time

The capacitor starts off looking like a short circuit, and charges up to look like an open circuit, which makes perfectsense. Can you figure out what the shape would be of a pulse reflected off of the capacitor, given that the time constant

Z_{0} C

was short compared to the width of the pulse?

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Source: OpenStax, Introduction to physical electronics. OpenStax CNX. Sep 17, 2007 Download for free at http://cnx.org/content/col10114/1.4

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