dkukral@encad.Wichita.NCR.COM (Dean Kukral) (11/11/89)
Advertisement: "At high speeds interconnections take on the characteristics of transmission lines and ..." What does this mean? What is a transmission line? What makes an "interconnection" into a "transmission line?" -- Dean Kukral, NCR Peripheral Products Division, Wichita, Kansas dean.kukral@wichita.ncr.com
mark@mips.COM (Mark G. Johnson) (11/12/89)
In article <868@encad.Wichita.NCR.COM> dkukral@encad.Wichita.NCR.COM (Dean Kukral) writes: >Advertisement: "At high speeds interconnections take on the >characteristics of transmission lines and ..." > >What does this mean? What is a transmission line? What makes >an "interconnection" into a "transmission line?" Transmission lines are physical structures which have certain properties that allow the propagation of electromagnetic waves. They are discussed at length in college sophomore courses on "Electromagnetics", where they are analyzed using Maxwell's equations. The standard example of a transmission line is a coaxial cable; solving the applicable Maxwell's equations is especially easy here because the fields are contained in a small, geometrically simple (cylindrically symmetric) region. "Schaum's Outline Series" (a bunch of paperbacks that are sold at bookstores as study aids and textbook supplements) has a great, inexpensive book. It's _Electromagnetics_, by Joseph A. Edminister. They also sell another one called _Transmission Lines_. Another good source of introductory information is the "Motorola MECL System Design Handbook". It's one of Moto's databooks and they'll give it to you free. . . . . . . . . . . . . . . . . One way to think about transmission lines is in the frequency domain. A real-life cable or PCboard (trace + groundplane) or other pair of electrical conductors has some physical, nonzero length. If a signal is injected into one end of that conductor pair it takes nonzero time to propagate to the other end (velocity is <= speed of light in vacuum). If the signal contains components of high enough frequency, then the wavelength of the signal can be shorter than the length of the conductor pair. Now the conductor pair must be considered a waveguide complete with reflections, propagation velocity, etc. Note that this is only true for long conductors or very high frequency signals, or both. That's why transmission line effects are negligible in opamp circuits etc. Also note that a square wave can contain VERY high frequency components, determined essentially by the risetime and falltime of its edges. Doctor Fourier tells us that a perfect squarewave, one with zero risetime and zero falltime, has a frequency spectrum that extends to infinity. However, real square waves have finite risetimes, thank goodness. For example a 74LS04 drives from 0.2 to 3.2 volts in about 15 nanoseconds. The 15ns edge rate is low enough that transmission line effects are negligible for wires less than about 7 feet long. The newer digital circuits like the blazing fast ECL ones, have edge rates of about 1ns. This means that with ECL you have to consider your "wires" ("interconnects") to be waveguides, reflections and all, if they are more than 4 inches long. And the newest CMOS like the FACT and 74AC series devices, have fast edges too. So beware. -- -- Mark Johnson MIPS Computer Systems, 930 E. Arques, Sunnyvale, CA 94086 (408) 991-0208 mark@mips.com {or ...!decwrl!mips!mark}
henry@utzoo.uucp (Henry Spencer) (11/12/89)
In article <868@encad.Wichita.NCR.COM> dkukral@encad.Wichita.NCR.COM (Dean Kukral) writes: >Advertisement: "At high speeds interconnections take on the >characteristics of transmission lines and ..." > >What does this mean? What is a transmission line? What makes >an "interconnection" into a "transmission line?" The changeover happens when the signals being sent into the line get fast enough to change in times roughly comparable to the propagation delay of the line. Consider a pair of wires with a 5kohm resistor on one end and a switched +5V source on the other. If the source switches on relatively slowly, the circuit's behavior is very simple: 1mA of current flows. Now, suppose the propagation delay is 10ns and the risetime of the source is 1ns. How much current flows 2ns after the source switches on? The source cannot "see" the resistor, because it is 10ns away and there hasn't been time for a signal to get there and back. It turns out that the current is determined by the "characteristic impedance" of the wire pair, which in turn is determined by things like wire size, spacing, insulation, etc., plus some effect from nearby objects. The characteristic impedance is often rather less than 5kohm. So "too much" current is flowing. When the current reaches the resistor, the voltage produced across it will be more than 5V. The excess voltage propagates back toward the source. When it arrives, the source will reduce its output current due to the "back pressure", generally reducing it too far. When the reduced signal arrives at the resistor, the back pressure will be reduced, and so forth. The changes in voltage, and corresponding changes in current, bounce back and forth until everything converges on the 1mA steady state. This happens even for slow signals, but when the risetime of the signal is slow compared to the delays, the reflections are trivial disturbances in the slow rise. For fast signals, the reflections are conspicuous, and they can confuse listening circuits. For example, flip-flops that trigger on rising edges may trigger more than once if reflections are strong. The fix, in this example, is to adjust the resistor so that its resistance equals the characteristic impedance of the line. Then the right amount of current flows from the start, and there are no reflections. Unfortunately, this happy situation can be arbitrarily hard to achieve in complicated real circuits, and one generally has to settle for approximations. Warning: the above is oversimplified. The reality is even worse. -- A bit of tolerance is worth a | Henry Spencer at U of Toronto Zoology megabyte of flaming. | uunet!attcan!utzoo!henry henry@zoo.toronto.edu
henry@utzoo.uucp (Henry Spencer) (11/12/89)
In article <1989Nov12.013850.7756@utzoo.uucp> I wrote: >... current is determined by the "characteristic impedance" >of the wire pair, which in turn is determined by things like wire size, >spacing, insulation, etc., plus some effect from nearby objects... I should have been more detailed here: characteristic impedance arises from the resistance of the wires (usually small) plus their capacitance and inductance. Hence the importance of physical details and nearby objects. If these assorted variables don't change along the wire, the characteristic impedance is independent of the length of the wire. Also, another complication arises: characteristic impedance is somewhat frequency-dependent, which is lots of fun for sharp-edged digital signals that are a mishmash of harmonics. -- A bit of tolerance is worth a | Henry Spencer at U of Toronto Zoology megabyte of flaming. | uunet!attcan!utzoo!henry henry@zoo.toronto.edu
mmm@cup.portal.com (Mark Robert Thorson) (11/14/89)
A good cookbook rule for TTL is:
1) Series termination -- 10 to 100 ohms placed right near the driver. I.e.
the driver feeds the resistor, and the rest of the circuit is fed from
the other side of the resistor.
2) Parallel termination -- 220 ohms to Vcc and 330 ohms to ground, placed
at the far end from the driver. The signal line should be "serialized",
i.e. any forks or loops should be removed so that it doesn't have any
long stubs between the driver and the terminating resistors. For a line
with many loads, you may wish to put drivers in parallel. If you do,
use drivers in the same package to minimize skew caused by temperature
differences.
If you need better numbers than these, substitute values while viewing the
signals on a scope.ngc@chanel.UUCP (Chris Ng) (11/17/89)
In article <24023@cup.portal.com>, mmm@cup.portal.com (Mark Robert Thorson) writes: > A good cookbook rule for TTL is: > 1) Series termination -- 10 to 100 ohms placed right near the driver. I.e. > > 2) Parallel termination -- 220 ohms to Vcc and 330 ohms to ground, placed Parallel termination draws current constantly while series termination does not. May be it doesn't matter in your design, just a side note. -- Chris Ng UUCP: {asuvax | hrc}!gtephx!ngc
mmm@cup.portal.com (Mark Robert Thorson) (11/19/89)
ngc@chanel.UUCP (Chris Ng) says: > In article <24023@cup.portal.com>, mmm@cup.portal.com (Mark Robert Thorson) wri > tes: > > A good cookbook rule for TTL is: > > 1) Series termination -- 10 to 100 ohms placed right near the driver. I.e. > > > > 2) Parallel termination -- 220 ohms to Vcc and 330 ohms to ground, placed > > Parallel termination draws current constantly while > series termination does not. May be it doesn't matter in your design, > just a side note. Another cookbook rule: use series termination where short lines with few loads are being drived, such as a microprocessor driving a few peripheral and memory chips, or an address multiplexer driving a bank of DRAM chips. Use parallel termination where the line is long and has many loads, like a system clock line that snakes all over your board.
henry@utzoo.uucp (Henry Spencer) (11/28/89)
A couple of people have asked me to elaborate a bit on my explanation of reflections, so, on the chance that it will be of general interest... Briefly summarizing my previous posting: If you apply a voltage with a risetime of 1ns to a pair of wires 10ns long with a resistor across the other ends, how much current flows 2ns after the voltage is applied? It's determined by the characteristic impedance of the wires, which is a composite of their resistance (usually small) and their capacitance and inductance. When the signal reaches the other end, if the impedance there doesn't match the characteristic impedance of the wires, reflections result. More specifically, say we have a 6V signal, a 1kohm resistor, and wires with a c.i. of 500ohms. The c.i. means we initially get 12mA of current flowing. So when the pulse reaches the resistor, it arrives with 6V and 12mA. This does not match the resistor's impedance, however. The resistor will take 6mA at 6V, so we have an excess 6mA arriving. Electrons are conserved; they have to go *somewhere*. There are two places they can go: into the resistor, or back up the wire. This is a linear circuit, so it's valid to analyze current back into the wire as if it were independent of current coming out of the wire. The wire and the resistor are a classic current divider, so (500/(1000+500)) = 1/3 of the current will go into the resistor and the rest back up the wire. The voltage will have to rise a little to do this, specifically (2mA*1kohm) = (4mA*500ohm) = 2V. So a new signal, 4mA at 2V, starts propagating back up the wire, and the voltage at the resistor is 6+2 = 8V. The voltage source will have some non-zero input impedance, probably not the same as the wire's c.i. When the new signal hits it, the same thing happens again: the current splits, with some of it going back into the source (i.e. the source's net output current being reduced) and some of it bouncing back into the wire. If the source's impedance is less than that of the wire, some of the voltages and currents involved will be negative; this does not change the overall picture. Each time the signal hits an end of the wire, some of it carries on and some of it bounces back, with the bounced signal steadily getting smaller. The net effect, observing at any specific point on the wire, is that the voltage and current bounce up and down somewhat and gradually settle toward the final value. In theory the reflections continue forever, but in practice they eventually disappear into the background noise. -- That's not a joke, that's | Henry Spencer at U of Toronto Zoology NASA. -Nick Szabo | uunet!attcan!utzoo!henry henry@zoo.toronto.edu
phil@diablo.amd.com (Phil Ngai) (11/29/89)
In article <24187@cup.portal.com> mmm@cup.portal.com (Mark Robert Thorson) writes: |ngc@chanel.UUCP (Chris Ng) says: |> Parallel termination draws current constantly while |> series termination does not. May be it doesn't matter in your design, |> just a side note. | |Another cookbook rule: use series termination where short lines with few |loads are being drived, such as a microprocessor driving a few peripheral |and memory chips, or an address multiplexer driving a bank of DRAM chips. But don't forget about the half-step you get with series termination. -- Phil Ngai, phil@diablo.amd.com {uunet,decwrl,ucbvax}!amdcad!phil AT&T Unix System V.4: Berkeley Unix for 386 PCs!