April 10, 1997

RLC circuits match transmission lines to loads

Boris Dvoskin, Worcester, MA

In many systems, transmission lines drive reactive (mostly capacitive) loads. To preserve signal integrity, you should properly terminate the lines. However, the input impedance of CMOS logic ICs, for example, comprises a high resistance in parallel with capacitance. A mismatch at the line’s end causes reflections and pulse-shaped distortion. You can use simple RLC circuits to compensate for the mismatch. In RF designs, the goal in matching impedances is to transfer power with maximum efficiency. This goal effectively precludes the use of resistive components in the matching network. In digital circuitry, however, efficient power transmission is not the primary concern, so using resistors is appropriate.

Assume the load is purely capacitive, as in Figure 1. To satisfy termination rules, you need to construct a circuit that includes the load and, at the same time, exhibits constant resistive impedance in the widest possible frequency range. It’s a given that the capacitor has no resistive component and that its impedance varies from almost infinity at low frequencies to almost zero at high frequencies. Because your goal is to design a circuit with constant, resistive impedance, you should shunt the RC branch in Figure 1 with another branch that exhibits inverse frequency dependence of impedance. This branch can be a series-RL network, which has impedance R at zero frequency and a high impedance at high frequencies.

The following expression gives the impedance of the RLC network in Figure 1a:

If you impose the condition L/R=RC, the numerator and denominator cancel and the impedance is R at any frequency.

If you select R as the transmission line’s characteristic impedance, you achieve perfect matching and obtain reflection- and distortion-free signals at the end of the line. Don’t forget, however, that the load is a capacitor, so the signal at the load differs from the transmitted signal. The –3-dB cutoff frequency is f_C=1/2uppercase piRC. Consider a typical case, in which C=10 pF and the line impedance is 50ohms. The cutoff frequency is 318 MHz. Figure 2a shows the results of a Spice simulation for the network.

The bandwidth limitation causes some rounding of fast pulse edges. With the 318-MHz bandwidth, the rounding effect becomes significant for pulse rise times shorter than 1/f_C, or approximately 3 nsec. Figure 2b shows a Spice simulation of the pulse shape for a 1-nsec signal rise time. To increase the bandwidth, you could use a lower value for R, but you’d then need a lower impedance transmission line. Widening pc-board traces lowers the line impedance, for example. To further increase bandwidth, you can use high-frequency correction techniques similar to those found in wideband amplifiers. These methods rely on resonance of the reactive components at the high end of the signal spectrum.

The circuit in Figure 1b adds an inductor to the RC branch and a capacitor to the RL branch. Now, consider the input admittance, Y=1/Z. The admittance is Y=Y₁+Y₂, where Y₁ and Y₂ are the admittances of the circuit’s right and left halves, respectively. The following expressions give the two admittances:

and

It follows that

(1)

Now, if you set L/R=2RC in Equation 1, the terms in square brackets in the numerator and denominator cancel and the input admittance is purely resistive and equal to 1/R. You’ve thus constructed another RLC network that matches the transmission line at all frequencies. To calculate the frequency response between points A and B, use the expression for the transfer function (magnitude only):

(2)

where uppercase tau=L/R–time constant and lowercase omega_RES= 2uppercase pif_RES=natural resonant frequency (in radians) of the LC tank. You can see that the transfer function is unity at dc and that the denominator has a minimum near-resonance frequency. (The term in square brackets equals zero.) The transfer function has a maximum near this frequency. However, because the numerator is also frequency-dependent, peaking occurs at slightly different frequencies. The exact calculation of the peak frequency is laborious—it entails taking the derivative of Equation 2, equating it to zero, then finding the roots of the obtained equation. The solution for the peak frequency is

f_PEAK=0.78f_RES=0.78/(2uppercase pi÷LC).

The peak response occurs at a frequency slightly lower than the resonant frequency of L and C. You can now find the amount of peaking by substituting lowercase omega_PEAK=2uppercase pif_PEAK into Equation 2. Calculations show that the peaking is approximately 2.1 dB. Now, consider the bandwidth. In the presence of peaking, you should use the transfer-function value at low frequencies as a reference level, then find the frequency, f₂, at which the magnitude drops by 3 dB. Equating Equation 2 to 0.707 (–3 dB) and solving for f₂ produces f₂=2.05f_RES=(2.05)/(2uppercase pi÷LC).

If you recall the matching condition L/R=2RC and set L=2R2C, the cutoff frequency for the circuit in Figure 1b is f₂=(2.05)/(2uppercase pi÷2RC). If you compare this expression with f_C=1/2uppercase piRC, you see that the circuit in Figure 1b has a greater bandwidth than that of Figure 1a by a factor of 2.05/÷2, or 45%. The bandwidth is approximately 460 MHz. Figure 2a shows the peaking and bandwidth improvement. Figure 2b shows the transient response for a pulse with 1-nsec rise time. Note that the pulse edges are faster, though at the expense of incurring overshoots. These overshoots, however, appear only at the load terminals and not at the line terminals.
(DI #2008)

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