Impedance matching - driving coax lines - Page 1
Forum section: electronics / projects

author: KX36
author_rank: Contributor
timestamp: June 23, 2014, 03:45:18 pm
content: Hello,I was just reading up on impedance matching. I've never done anything high enough frequency to worry about transmission lines before. I think I get the basic principles of matching the termination to the characteristic impedance of the line to minimise reflections and I've been able to simulate a 50 ohm transmission line (with L and C components only) with various terminators in SPICE and seen the expected effects of match vs mismatch on the transient and AC analyses.What I wanted to ask about was about the practical implication of this, specifically designing a line driver. I see line driver circuits in the datasheets of fast opamps as having a gain of 2, and then a series output resistor equal to the line impedance. The voltage is doubled in the amplifier and then halved at the output of the transmission line, making the output match the input with a delay.In SPICE, driving the line from an ideal voltage source, as long as the transmission line impedance matches the termination, I get no funny business. If the voltage source has little or no series resistance, the output matches the input with a slight delay. If I add some series resistance to the source, I get the same output but at a lower amplitude. e.g. if I have 50 ohms series resistance, a 50 ohm coax and then 50 ohm termination resistance, and put a 1Vpk pulse through it, I get a 0.5Vpk out. In the simulation, the series resistance does nothing to the waveform except reduce the amplitude.I know there is a reason for the 50 ohm series resistance on the source, but what is it?Cheers,Matt

author: KJDS
author_rank: Super Contributor
timestamp: June 23, 2014, 03:54:53 pm
content: Change you output termation from 50 to 40 ohms, which isn't a huge amount, then try the effect of having the source resistor there and its affect over frequency.

author: w2aew
author_rank: Super Contributor
timestamp: June 23, 2014, 08:18:07 pm
content: Making the driver's output impedance match the transmission line is often done to help control reflections that come back down the from the load or any impedance discontinuities along the line. With the source impedance matched to the line, the reflections coming back towards the driver get absorbed and not re-reflected down the line.

author: KX36
author_rank: Contributor
timestamp: June 23, 2014, 09:14:09 pm
content: Thanks for the responses. I played with the simulation some more. Found that in this simulation, if either the source or the load impedance matches the transmission line impedance, there are no reflections. In the simulation, as long as one matches, the other doesn't matter.So does this mean, for example, a 50 ohm coax into a 1M scope input should perform reasonably well if it has a 50 ohm source impedance and no 50 ohm terminator, but that this would be further improved by terminating it in 50 ohms at the expense of having half the amplitude at the output, or is this just a trick of the idealised simulation and in the real world 50 ohms on the source and load are necessary?

author: JackOfVA
author_rank: Unknown
timestamp: June 24, 2014, 12:51:44 am
content: Most op amps will give you problems when working into a capacitive load - overshoot or oscillation. Adding series resistance will solve this problem. (Skipping over lots of important details here, of course.)If the receiving end of the cable is terminated in a resistance equal to the cable Zo, then at the op amp end, the cable should not appear capacitive or inductive, but rather like a pure resistance of Zo. However, the real world is seldom so kind and you have to think about what happens in unexpected circumstances. Adding a series resistor at the op-amp output 'back termination' as it's often called removes sensitivity to cable mis-matches from the op amp stability prospective as well as preventing reflections from the op amp end.It also helps short circuit proof the op amp - not all op amps will tolerate a short circuit without damage and it's usually considered not an ideal design practice to have a device that destroys the output op amps if inadvertently operated into a short circuit, such as might occur with a bad connector or cable. With a back termination, the op amp always sees at least the back termination resistance as a load.

author: TerraHertz
author_rank: Super Contributor
timestamp: June 24, 2014, 04:52:09 am
content: Quote from: KX36 on June 23, 2014, 09:14:09 pmThat 'doesn't matter' is not quite correct.Line terminations are rarely perfect, so there's _usually_ some reflection from the 'destination' end. With a short line and near-zero impedance at the drive end, the reflected wave gets back to the source, reflected again (inverted, not much amplitude loss), then arrives at the destination end again, Some fraction of that gets re-reflected, and so on.The reason you can think it 'doesn't matter' is that for short transmission lines terminated at one end, reflected energy very rapidly dies away due to being attenuated greatly each round trip. You'd need a very fast scope to see the small amount of edge ringing. But it's there.With both ends of the line seeing a near-correct impedance match, reflections due to the slight mis-matches die away extremely fast.

author: KX36
author_rank: Contributor
timestamp: June 24, 2014, 09:05:44 am
content: Thanks guys, I believe you've cleared it up for me.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 24, 2014, 09:27:38 am
content: Yes, source or load termination works equally well*.*But not really. It depends, but consider what happens in source termination for instance.Voltage step: at the voltage source, the line looks like 50 ohms, so in series with the 50 ohm source terminator, the line voltage steps up to 50% (half supply). The 0-50% step propagates down the line, and a lot of nothing else happens until the wave hits the far end. When that happens, it reflects, in phase; the 50% wavefront propagates back, superimposed on the line which is now already charged to 50%. The far end flips from 0 to 100%, and as the reflection propagates back, it finishes charging the line from 50 up to 100%. When the wave terminates at the source, the source end goes to 100% supply, current flow through the source termination resistance stops, and everything is in equilibrium.BTW, note that a transmission line stores energy, as it exhibits both capacitance and inductance. You can tell which is more important by the difference in impedance: to a high impedance (like an open circuit), a line looks capacitive (voltage changes slowly). To a low impedance, inductive (current changes slowly).All well and good, right, but:- If you sense the wave anywhere along the line, you get V/2 for a duration of however far from the unterminated end you're reading at. This is smack dab in the middle of, for instance, CMOS input thresholds, so you can get pretty awful digital signals (chattering, indeterminate states). You necessarily cannot transmit pulses from the source any faster than the round trip delay, otherwise the extra transitions stack up and you only get garbage.Example: the old PCI standard used 3.3V CMOS logic, source terminated from the northbridge. The maximum clock frequency was 66MHz. Why? The maximum length, from system chip to the last PCI connector, including trace length up each expansion card if inserted, incurs a double penalty for source termination. Sure, a short length motherboard (say, micro-ATX with just a couple slots, versus an ATX with all eight) might be able to achieve more, let alone if they went to the trouble of multiple PCI bridges -- but since that's not the standard, no one's going to make a card that has to support that kind of clock rate or signaling. (PCIe, besides being fully terminated differential, is point-to-point only, and also doesn't depend on matched bus lengths.)- Real lines are not ideal. There is high frequency loss and dispersion, which causes rounding or ringing. High frequency loss means, when the wavefront arrives at the far end, it doesn't step from 0-100%, but maybe 0 to 90%, and the reflected wave only returns with ~80%, so current continues to flow from the source, charging the line gradually to 100%. (You can draw the ladder diagram showing the repeated incident and reflected waves, but because dispersion scatters a sharp wavefront into a rounded hump, it's a continuous rather than discrete process, and so it looks more like an exponential time constant than a series of steps. The staircase effect can still be observed, using a mismatched line with good frequency response.)The nice thing about source termination is, any source can disconnect (go open circuit) to free up the bus, and any driver can simply hook up with a reasonable resistance (continuing the PCI example: 74HC series logic is a pretty good match to (typically) 50-100 ohm PCB traces without needing external resistance) to drive the bus. No special arrangement is required (e.g., dual end terminations). Load termination is preferred, because no voltage is lost at the source (assuming an explicit source termination resistor would otherwise be required), all the power goes into the load, and it gets there when it needs to, without bouncing back and forth (length does not limit clock speed).Source-load termination is kind of "gold standard", in that, a signal placed on the line arbitrarily is guaranteed to be absorbed when it reaches either end -- a one-dimensional anechoic chamber.Note that, because many real sources have finite impedance by their nature, power is not necessarily lost by implementing source termination; indeed, an ideal Thevenin/Norton source delivers maximum power into a perfectly matched load. The impedance of, for example, a transducer is a very real thing; tapping on a speaker, it goes "thud" when the terminals are shorted, or "doom" when left open. In one case, the energy is reflected back (acoustic - electric - back to acoustic) out of phase and opposes the stimulus; in the other case, the energy is reflected in phase and the cone resonates like a drum (well, if it's the kind with tight suspension, it will). Same with an antenna: radio waves will simply reflect off or re-radiate (in or out of phase) if the connection is open or shorted (respectively); only when it is a good match will power be absorbed (or transmitted) usefully. Three use examples: old coax networks, RS-485 multi-master, and Bus LVDS (and related standards). All three consist of a single linear transmission line, with high impedance receivers connected at various points (they don't cause any significant reflections by just watching), and the ends -- wherever they are -- are terminated. Drivers remain open circuit until active (and must be arbitrated, so multiple drivers don't get in a fight..). In setting up such a network, youmusthave a termination resistor at the furthest ends of the cable. So the coax network, you'd start with a tee and terminator at one computer, run BNC cables from computer to computer (more tees, no terminators), back and forth throughout the office, then end with another terminator on the last machine. (By the way: like PCIe, Ethernet is differential, point-to-point only. Hence, one cable, from a switch or router, to a computer -- no chaining.) The same is required for RS-485 networks, which run at higher voltages (maybe 1-4V, into 50 ohms double terminated -- looks like 25 ohms to any bus driver -- pretty beefy, good for industrial controls). I don't know offhand what applications use Bus LVDS (maybe I do, and don't realize it), but yep, same idea again, just with smaller signal levels, differential drive (and double the drive strength of regular LVDS -- 4 instead of 2mA -- to keep the compatible 200mV signal level).Tim

author: KX36
author_rank: Contributor
timestamp: June 24, 2014, 10:42:43 am
content: Thanks for the detailed explanation. I poked my simulation some more, this time splitting the transmission line in half and probing not only the input and output but after the source terminator and in the middle of the line, run for source, load, neither and both termination and was able to see what you describe. I think putting in a second pulse in sync with the reflection also helped me to see the difference. It is also interesting to read that various termination schemes are practically used where their consequences are understood and accounted for.Thanks again,Matt

author: David Hess
author_rank: Super Contributor
timestamp: June 26, 2014, 07:47:16 pm
content: Quote from: T3sl4co1l on June 24, 2014, 09:27:38 amIt is worth noting that a current step like from a transconductance output may be used instead of a voltage step for driving the transmission line.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 27, 2014, 02:02:39 am
content: Yes, that too.In fact, LVDS transmitters (and a number of other high speed, low voltage standards) are current source outputs. Not usually very good, a few kohms I think is typical. But compared to a 50-150 ohm transmission line, that's more than enough. Note that most of these links are load terminated (good fidelity, not multi-master compatible), so the source impedance doesn't matter, as long as the voltage or current developed is sufficient.Tim

author: David Hess
author_rank: Super Contributor
timestamp: June 27, 2014, 07:27:34 pm
content: I thought I remembered seeing some oscilloscope vertical delay line circuits that worked this way by driving the delay line from the output of a transconductance amplifier without source termination but was unable to find any examples to link to. The obvious advantage is twice the gain since the amplifier is only driving the transmission line impedance without the parallel termination.