## Resonance in Short Transmission Line

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## Resonance in Short Transmission Line

by Dr. Howard Johnson

Many engineers use rules of thumb to determine when terminations will be necessary. One popular rule compares the risetime of the driver to the delay (length) of a transmission line. If the risetime is six times, or three times, or sometimes even only two times the raw, one-way unloaded line delay then you might be tempted to conclude that a termination is not necessary. The efficacy of such a rule hinges on a crucial hidden assumption: that the driver source impedance is not too much less than the line impedance. Violate that assumption and the rule fails miserably under certain conditions of loading.

For example, start with a simple linear driver having a
10-ohm output impedance (*R*_{S} = 10 ohms) and a risetime *t*_{r} = 1 ns. Connect this driver to a transmission line having a characteristic
impedance of 50 ohms and a physical length of 1 inch (176 ps raw unloaded trace
delay) (see Figure 1). The driver risetime in this case is about six times the
line delay, indicating at first glance no need for termination, but at the same
time the driver output impedance (10 ohms) lies far below the transmission line
impedance (50 ohms), raising the possibility of severe resonance. Will this
line ring or not? (Figure 1.)

The driver in Figure 1 has a 10-ohm source impedance, the line has a characteristic impedance of 50 ohms and a delay of 176 ps (~ 1 in. of FR-4), and the capacitive load CL varies from 0-200 pF.

Those of you familiar with analog circuits will find the
pi model helpful in visualizing how this circuit functions (Figure 2). The pi
model for a short section of transmission line is composed of a capacitor *C*_{1} shunting the signal to ground, followed by a series inductance *L*,
followed by another capacitor *C*_{2} shunting to ground.

The value of the components in the pi model are calculated
from the transmission line parameters, where if *t*_{d} equals the
raw, unloaded one-way delay of the transmission line and *Z*_{0} its characteristic impedance, then

*L*=

*t*

_{d}*

*Z*

_{0}[1]

*C*

_{1}=

*C*

_{2}= (1/2)(

*t*

_{d}/

*Z*

_{0}) [2]

The pi-model applies in an accurate way only to
transmission lines whose delay *t*_{d} is short compared to the
signal rise of fall time. When *t*_{d} is less than 1/6th the rise
or fall time you can expect the accuracy of the simple pi-model approximation
to be better than about 2% under ordinary conditions of loading as used in
digital applications. When *t*_{d} is enlarged to 1/3rd the rise
or fall time the accuracy deteriorates to about 20%.

In the present example *t*_{d} is about 1/6th
the risetime *t*_{r}, so the pi model applies. Note that the ratio *t*_{d}/*t*_{r} doesn't actually determine whether a
termination is needed, only whether the pi method of analysis is useful to
predict the circuit behavior.

The beauty of the pi model is how well it illustrates the
resonant properties of a transmission circuit. The pi-model in conjunction with
a capacitive load forms a circuit virtually bristling with *L*'s and *C*'s.
That should raise in your mind immediate concerns about the possibility of
terrible resonance, which is exactly what can happen.

Let's begin the analysis assuming no capacitive loading (*C*_{L} = 0). Without worrying too much about math, let me just tell you that the
resonant frequency *f*_{r} of an unloaded line driven by a
low-impedance source works out to (1/4)(1/*t*_{d}). Given that in
this case *t*_{d} = (*t*_{r}/6), a little algebraic
re-arrangement shows that:

*f*

_{r}= (1/4)(1/

*t*

_{d}) = (1/4)(6/

*t*

_{r}) = (3/2)(1/

*t*

_{r}) [3]

The expression for *f*_{r} shows that the
resonant frequency of an unloaded line lies at a point three times higher than
the knee frequency *f*_{knee} = (1/2)(1/*t*_{r})
associated with the rise and fall time of the driver. This relationship means
that the resonance, even though technically present, has no practical effect
because the resonance occurs at frequencies outside the bandwidth of the
driving signal. This relationship is also the basis for the popular myth that
lines shorter than a certain amount (roughly 1/6th or perhaps 1/3rd the rise or
fall time) never require termination. What goes wrong with that rule of thumb
is easily seen in a frequency-domain plot showing the gain of the transmission
circuit (Figure 3).

The figure shows for *C*_{L} = 0 a towering
resonance at about 1.4 GHz. This is the unloaded resonance associated with the
transmission line delay. Increasing the load *C*_{L} pulls the
resonant frequency lower. In this circuit, any load greater than 12 pF creates
a resonance below the knee frequency associated with the driver. That's the
essence of the problem. If capacitive loading decreases the resonant frequency
associated with your transmission line to a point near (or below) the knee
frequency associated with your driver you will see ringing and overshoot in the
time-domain (Figure 4).

Going back to the pi-model analogy, those of you with
analog design experience may immediately recognize that increasing the
capacitance of the load in this circuit reduces the resonant frequency. I like
very much how the pi-model circuit illustrates this effect. A few years back,
one of my more astute students noticed that as the resonant frequency
decreases, so does the Q. The Q of a resonant circuit is a measure of the
severity of the resonance: bigger values of Q imply more ringing and overshoot,
provided that the resonance falls near (or below) the knee frequency of your
driver. Figure 3 shows how increasing *C*_{L} reduces the Q. Above
200 pF the Q is reduced to the point where the resonance disappears-producing
a critically damped circuit. In the time-domain waveforms (Figure 4) you can
see that for very large values of *C*_{L} the circuit no longer
rings, but merely produces a long, sluggish response to a crisp step input.

The various behaviors described in this note suggest two ways to combat ringing on short, unterminated transmission lines.

1. Reduce the load capacitance, raising the resonant frequency well above the knee frequency of your driver. To make this work you need

*C*

_{L}<<

*C*

_{1}= (1/2)(

*t*

_{d}/

*Z*

_{0}) [4]

2. Increase the load capacitance to the point where the line reacts in a critically-damped fashion. This approach sacrifices speed for monotonicity, a good trade in some cases (especially for relatively slow clocks produced by overly speedy drivers). To make this work you need

(*C*_{L} + 2**C*_{1}) > (*L*/(*R*_{S}**R*_{S}))
= (*t*_{d}*Z*_{0}/(*R*_{S}**R*_{S})) [5]

From the critical-damping expression [5] you can see the
effect of increasing *R*_{S}. If *R*_{S} is increased
to the point where it equals the characteristic impedance of the line (*R*_{S} = sqrt(*L*/(2**C*_{1})) then the line damps itself even with *C*_{L} = 0. That's the benefit of raising *R*_{S}. The higher you make *R*_{S} the more natural damping you get. The lower you make *R*_{S} the
harder you have to work to damp the line elsewhere (i.e., by making *C*_{L} larger or by adding other termination components).

Capacitances in the middle range between these two extremes [4] and [5] cause the worst problems with ringing and overshoot.

Best Regards,

Dr. Howard Johnson