All, I am sorry, but I can't watch this thread go on without saying something, because I am about to burst... This idea of estimating the necessary buffer strength by the total capacitance is very old, and ill fated for high speed systems. As someone correctly pointed out, it may only be good when the edge rates and trace lengths are such that the signal arrives to all receivers way before the buffer is done switching. (I am deliberately using these words here, because talking wavelengths and frequencies may be too mysterious to many). Unless you are working with such slow buffers (or such impossibly short distances), you must forget about this idea all together. In today's commonly used technology the buffer is done switching way before the signal arrives at the end of the wire. This means that the wire is not a wire, but a transmission line. If so, the buffer strength depends on the impedance of the T-line, or the termination style (none, parallel, series). The load capacitance at the end of the line has not much to do with sizing the buffer. Consider this equivalent circuit for a point to to point topology. Driver - T_line - Receiver, where the driver is a 10 Ohm Thevenin circuit, i.e. an ideal step function source with a 10 Ohm series resistor, the T_line is a 50 Ohm series resistor (yes series R, because we are talking about the condition while the signal is propagating in the line), and the receiver is a 10 pF capacitance to ground. If you look at this circuit you will see that it is a simple RC circuit. The waveform on the capacitance will be an exponentially shaped curve, and its time constant is RC to reach the 63% point (50+10)*10e-12 = 0.6ns. Now, if you remove the 10 Ohm Thevenin resistor from the circuit, i.e. turn the driver into a superconductor, the equation changes to: 50*10e-12 = 0.5ns! Was it worth it? From this you can see that the edge rate at the receiver will not depend too much on the drive strength, but more on the ZC relationship between the T-line impedance and the input capacitance. Also, the amount of time the signal needs to travel along the T-line is determined by Maxwell's wave equations, which are completely independent from the driver strength. The driver strength is only important when you start figuring out overshoot for unterminated systems, or signal swing for parallel terminated systems. In those cases the T-line and/or the parallel termination look like a resistor to the driver, and we are talking about voltage dividers. The only place the capacitance may come into effect is of you have (evenly) distributed loads (receivers) along the T-line. That will reduce the effective impedance of the T-line, but this, again is like a resistive load to the buffer with a lowered value... So, in short, forget about capacitive loading in your calculations! Arpad Muranyi Intel Corporation ================================================================== -----Original Message----- From: Bob Patel [mailto:whizplayer@xxxxxxxxx] Sent: Thursday, January 10, 2002 9:19 AM To: Ingraham, Andrew Cc: si_list Subject: [SI-LIST] Re: SV: Total load capacitance Hi! Andrew, This load capacitance estimate is for determining the drive strenght needed for an ASIC I/O. This capacitance is beiong calculated wiht the fact that I need to maintain the characteristic impedance to 50ohms and this way I ensure that the driver can source/sink the required amount of current in order to maintain the rise and fall times of the signal. Thanks Bob "Ingraham, Andrew" <Andrew.Ingraham@xxxxxxxxxx> wrote: > Hi! Anders, Even if I take a transmission line > approach the total capacitance faced by the driver > will always be there i.e. even if the edge rates are > faster and the driver is weak then the rise & fall > time will not be maintained at the receiver and if the > C is too large then the driver may not be able to > drive it above Voh. SO, either way this lumped > capacitance approach has to be taken into account to > determine the drive needed. That depends. Take the 1000 meter cable with Zo = 50 ohms, and put a 50 ohm load on the end of it. Driving it, you can't tell if it's 0 meters or 1000 meters long. It looks the same to the driver. (Ignoring discontinuities, non-ideal loads, and cable losses.) If it didn't have the 50 ohm load, the situation is different. If the cable is long enough, the initial edge rates would be determined by the ability of the driver to drive the characteristic impedance of the cable; but eventually the reflections would come back and modify the waveforms in some way that you can't quantify by simply adding up the cable capacitance. If the signal is periodic and the electrical length is an exact multiple of half the repetition rate, then the amplitude (and therefore edge rates) might be enhanced by the reflections. If it is an odd multiple of a quarter of the repetition rate, the reflection might kill the edge rates. The answer to your question also depends on what you need the capacitance estimate for, i.e., power estimation, vs. output levels or edge rates. Finally, I would think that Cout of the driver has already been accounted for so you don't need to include it, but check with the ASIC vendor to be sure. Regards, Andy ------------------------------------------------------------------ To unsubscribe from si-list: si-list-request@xxxxxxxxxxxxx with 'unsubscribe' in the Subject field or to administer your membership from a web page, go to: //www.freelists.org/webpage/si-list For help: si-list-request@xxxxxxxxxxxxx with 'help' in the Subject field List archives are viewable at: //www.freelists.org/archives/si-list or at our remote archives: http://groups.yahoo.com/group/si-list/messages Old (prior to June 6, 2001) list archives are viewable at: http://www.qsl.net/wb6tpu