Overshoot and ringing

[doofus32]

Today, in high speed IC, the rising time for the output is often shorter than 1 ns.

This cause ringing on the line. This ringing will cause the voltage to go over the voltage that are sendt out. In the teori the double voltage.

Will this double voltage have the possibility to damage the input/output of the IC/Other IC on the board.


Reply on another forum:

Short lines ring just as well, just at higher frequencies (than your scope may be able to see), and you CAN kill both ends of the connection by driving excessive energy into them.
The solution is a little transmission line theory. If you have a transmission line (trace on pcb) longer than 1/10 the risetime (for 1ns risetime about 2 cm, quite short), then you need to think.

OP's scenario where you get the double voltage on the output is correct, and shows a problem: Four times the required ENERGY to change the state of the wire was fed into it, and now the excess energy has to dissipate somewhere, and it does so in the receiver, transmitter, and as radiated emissions. That leads us to the art of termination.

First the problem circuit described. Suppose you have a 3.3V driver with 20 ohm output impedance, and a 100 ohm transmission line connected to it. If connected directly to the driver, the driver, when switching, sees a 120 ohm load (the impedance of the driver plus the impedance of the transmission line), and drives a 2.75V change into the wire at a current of 27.5 mA. After some time this step has propagated to the other end of the wire, and is reflected since there's nothing there to absorb the energy. It is reflected back at the transmitter, and will arrive there in a little while, but right now we have the 'outbound' current, plus the reflected current, giving us 5.5V at the output. If there's some poor device there, its protection diodes will take some of the energy. If not, then when the pulse returns to the transmitter, the transmitter sees 5.5V, and may not like that. The pulse will bounce back and forth untill SOMETHING has absorbed the energy.

There are two solutions. First, if the line was infinitely long, the pulse would never be reflected. This is what you do in end termination: Place a 100 ohm resistor (aka. infinitely long transmission line simulator) to ground, or some other voltage. Now the receiver sees the 2.75V step, and there is no reflection or ringing.

The second option is to insert 80 ohms in series with our driver. When it switches, it sees a (20+80+100) 200 ohm impedance, and drives 16.5 mA into the 200 ohm load. This means that the voltage across the input of the transmission line is half of the 3.3V, and the voltage step that propagates down the line is 1.6V. When it reaches the end of the transmission line (where there's only a high impedance receiver), it is reflected back at the transmitter, but first look at the receiver's end: The sum of the incident and reflected steps is 1.6+1.6=3.3V, so it sees a sharp 0V-3.3V transition. Back on the driver end, for two times the line propagation delay, the driver has been sourcing 1.6V across the input of the transmission line. When the reflected pulse arrives, it's also 1.6V, giving a total of 3.3V, and current stops flowing. It's now at steady state with no additional reflections or ringing.

The latter is called source termination, and is best when you have point to point connections, or a few receivers at the end of the line, since a receiver at the midpoint will see a two-step transition. The upside is that it dissipates less than half the energy of the first solution.

End termination is what is used when you send 1Gbps across a hundred meters of cheap twisted pair, or in old style 10Mbps RG58 ethernet.

I wonder, can the protection diodes handle this overshoot then? Or can the protection diodes get damages? How can I calculate how much current that will flow into those protection dioes?


Another "silly" question. The termination resistors vil develop some small amount of power. How much can the handle.

In your last example:
16.5mA x 16.5mA x 80 = 0.02W (For an very short time)

Most resistors can handle less than 0.01W, even much less. (The small array resistor that are used on busses)
Can these resistors handle the extra power developed for the short time?

I asume the value the resistor producer gives us is the RMS value, is that right.

 

[Hero999]

Most SMT resistors I've seen can handle 100mW so I think you'll be alright with a 10mW pulse.

The power levels you've calculated represent the peak power. The RMS power will depend on the duty cylce of the pulses.

 

[Analog]

I used to design bus based system boards, and we always used an active terminator vs. the passive ones you described. Check into it.

 

[poopeater]

The protection diodes place in the input stages of most high speed ICs can handle voltages MUCH higher than what you are talking about. They are typically meant as 'ESD protection diodes', so can handle orders of magnitude higher voltage than you are discussing.

I have only seen a single case in which a device was damaged by excessive overshoot/undershoot. It happened on a PCI-X input device, and only when you applied a -2V/1A bias to the input pin. This NEVER occurs in a real system, and most devices can easily handle the overshoots that occur on unterminated busses.

What you need to do:

1. Look at the datasheets for the devices you are talking about. Are there input clamps (The ESD diodes I'm talking about)
2. Look at the datasheets for the devices you are talking about and look for 'maximum voltage' specs on the input lines.