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 (RS = 10 ohms) and a risetime tr = 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.

Figure 1—A short transmission line with a capacitive load sometimes rings horribly.

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 C1 shunting the signal to ground, followed by a series inductance L, followed by another capacitor C2 shunting to ground.


Figure 2—Under ordinary conditions the pi model is accurate to about 2 1.792564e-307n the transmission-line delay T is less than 1/6th the rise or fall time.

The value of the components in the pi model are calculated from the transmission line parameters, where if td equals the raw, unloaded one-way delay of the transmission line and Z0 its characteristic impedance, then

L = td*Z0 [1]
C1 = C2 = (1/2)(td/Z0) [2]

The pi-model applies in an accurate way only to transmission lines whose delay td is short compared to the signal rise of fall time. When td 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 td is enlarged to 1/3rd the rise or fall time the accuracy deteriorates to about 20%.

In the present example td is about 1/6th the risetime tr, so the pi model applies. Note that the ratio td/tr 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 (CL = 0). Without worrying too much about math, let me just tell you that the resonant frequency fr of an unloaded line driven by a low-impedance source works out to (1/4)(1/td). Given that in this case td = (tr/6), a little algebraic re-arrangement shows that:

fr = (1/4)(1/td) = (1/4)(6/tr) = (3/2)(1/tr) [3]

The expression for fr shows that the resonant frequency of an unloaded line lies at a point three times higher than the knee frequency fknee = (1/2)(1/tr) 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).

Figure 3—Increasing the load CL pulls the resonant frequency lower. In this circuit loads greater than 12 pF resonate at frequencies below the knee frequency associated with the driver.

The figure shows for CL = 0 a towering resonance at about 1.4 GHz. This is the unloaded resonance associated with the transmission line delay. Increasing the load CL 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).

Figure 4—The resonance associated with capacitive loading manifests itself as overshoot and ringing in the time domain.

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 CL 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 CL 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

CL << C1 = (1/2)(td/Z0) [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

(CL + 2*C1) > (L/(RS*RS)) = (tdZ0/(RS*RS)) [5]

From the critical-damping expression [5] you can see the effect of increasing RS. If RS is increased to the point where it equals the characteristic impedance of the line (RS = sqrt(L/(2*C1)) then the line damps itself even with CL = 0. That's the benefit of raising RS. The higher you make RS the more natural damping you get. The lower you make RS the harder you have to work to damp the line elsewhere (i.e., by making CL 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