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I'm not an EE major, but I was curious what causes overshoot and undershoot.

Where can I find a detailed explanation of the process? I heard it has to do with the parasitic capacitance, but there were no details.

Parasitic capacitance seems to have the effect of drawing more current into or out of the line, but I don't directly see how this would make a signal overshoot.

Also, how does an improperly terminated signal result in overshoot? If the signal is not properly absorbed and reflects backwards, I would expect some of the energy not to be transmitted into the receiver. Thus, if anything wouldn't this cause the voltage to undershoot because it doesn't receive as much power as it is expecting?

JYelton
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James Joshua Street
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    Complex impedance. – Ignacio Vazquez-Abrams Mar 03 '15 at 19:20
  • For the latter part: https://en.wikipedia.org/wiki/Transmission_line But the transmission line theory is not for the faint hearted. – jippie Mar 03 '15 at 19:24
  • Phase delay. The feedback signal isn't 'quick' enough so the circuit doesn't 'know' that it has already compensated for the input. – HKOB Mar 03 '15 at 19:26
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    In addition to parasitic capacitance there is also parasitic inductance. This all gets rolled up into the complex impedance Ignacio mentions. – Austin Mar 03 '15 at 23:03
  • Pretty good answers. I answered a similar question a month ago or so. You can check it out. http://electronics.stackexchange.com/questions/152030/causes-of-ringing-in-a-pcb/152040#152040 – user57037 Mar 04 '15 at 07:54

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You really only need to read about and understand the reflection coefficient. Wikepedia: http://en.wikipedia.org/wiki/Reflection_coefficient is a good simple starting point.

As you can see, it's all about impedance mismatch. You do not need standing waves or any of that nonsense - just some kind of edge going from one characteristic impedance ZS to another ZL.

If ZL is bigger than ZS you get overshoot. If it's the other way around you get undershoot. Positive or negative reflection coefficient.

Let me know if you need more to fully comprehend this.

Rolf Ostergaard
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If you have a series inductance and a shunt capacitance with very little resistance, you have a resonant circuit. If you stimulate the resonant circuit with a fast positive or negative edge, it will resonate, which includes overshoot and undershoot. Traces can act as series inductors, and digital inputs typically have some capacitance. So you have all the makings right there in the typical high-speed digital circuit. Adding a damping resistor between source and load can dramatically reduce overshoot and undershoot. Also, some digital signals have controllable output strength. Lower drive strength will reduce overshoot and undershoot.

user57037
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  • hey, where can I read about why a series inductance and shunt capacitance result in a resonant circuit? It seems intuitively true, but I would like to understand it more. – James Joshua Street Mar 04 '15 at 18:52
  • @JamesJoshuaStreet, any electrical engineering text book. Or you can look up LC circuit or LCR circuit, probably on wikipedia. The resonant frequency is determined by LC, and R primarily determines damping factor. If the circuit is heavily damped, then there will be no overshoot. – user57037 Mar 04 '15 at 20:21
  • sorry I guess I was confused. I know about LC circuits to some extent, but I thought when you mentioned a shunt capacitance that would change things. I thought to be a shunt capacitor that the capacitor would be in parallel with the RL part of the circuit, rather than in series like I vaguely remember from learning about RLC circuits in intro physics. I wasn't sure how that would effect the circuit behavior. Anyway I will try to review it sometime then. thanks – James Joshua Street Mar 04 '15 at 20:29
  • Oh, yeah. I say "shunt" because it is shunt from the IC input to GND. Sorry for the confusing terminology. It is industry standard jargon, and sometimes I forget I am even using it. – user57037 Mar 04 '15 at 20:32
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I suppose, your question (overshoot, undershoot) not only concerns "standing waves" or similar effects but also "simple" 4-poles (like amplifiers, filters), correct? In this respect, the terms "overshoot" and "undershoot" are used to describe the step response of such a device.

(a) Overshoot: If a system of (at least) second order has two complex poles (a pole pair) the step response will exhibit overshoot. It is possible to find a relation between the pole quality factor Qp and the amount of overshoot (in %). For passive circuits, this is possible only for RLC topologies (resonant effect); for active RC circuits overshoot can be observed in case of feedback (wanted or unwanted).

(b) Undershoot: This phenomenon can be observed for active circuits which have a "non-minimum phase" zero in the right half of the complex s-plane (RHP). A simple example for a system having a step response with "undershoot" is the second-order allpass.

LvW
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In a nutshell: impedance mismatches can cause standing waves or "moving" waves along a conductor. If they build up constructively you get overshoot; if they build up destructively you get undershoot. The specifics are complex and depend on the driving source, the transmission line, the length and the losses in the line.

There is an absolutely fantastic video on impedance matching which helps to demonstrate this and other concepts better than any drawing or text I've ever seen. It uses mechanical wave generators and wave guides, but the principle is exactly the same in electric circuits. It was made by AT&T in the (I think) 1950s.

It's called Similarities in Wave Behaviour and it is worth the time to watch.

akohlsmith
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  • Yes, but that's not the only cause... or rather standing waves only matter if the line is long enough, but you can still get oveshoot even without that being the case. [Mkeith's answer](http://electronics.stackexchange.com/a/152040/54580) to the similar question was a bit more comprehensive. – Fizz Nov 01 '15 at 01:58