To: Tenchusatsu who wrote (33046 ) 11/2/1999 6:18:00 AM From: Bilow Read Replies (2) | Respond to of 93625
Hi Tenchusatsu; Revisting the pin requirements of SDRAM versus RDRAM... Intel does not have data sheets on its web site for the new 840 parts, but they have package pin counts for the parts. This allows us some small comparison of RDRAM and SDRAM interface counts. I would like to include powers and grounds, but I just don't have the data sheets to compare to. Anyway, you wrote: "Try grafting dual DDR channels onto the north bridge. It's possible, if you don't mind the 200 extra pins or so. " This assumes that an SDRAM channel must use about 100 pins more than an RDRAM channel. I would like to examine this assumption in more detail. There are two Intel parts that connect to the Rambus channel but are not memory. Instead, the 82803 splits a Rambus channel into two Rambus channels, thereby doubling the amount of memory that can be addressed by the original channel. The 82804, on the other hand, takes a single Rambus channel, and turns it into a single SDRAM channel. Just looking at the pin counts, you would suppose that since a single RDRAM channel uses less than half the pins of a single SDRAM channel, then the 82803 would require fewer pins than the 82804. That is, both chips have an incoming Rambus channel. The 82803 has two outgoing Rambus channels, while the 82804 has only a single outgoing SDRAM channel. But the fact is that the 82803 is in a 324 pin package, which is larger than the 241 pin package that the 82804 is in. This suggests that the actual pin count ratio between the two interfaces (counting all the power, ground, termination voltages, etc., is below 2 to 1. I wish I had the specs. I know the following calculation is faulty, but I am going to include it anyway. Hopefully, when Intel releases the specs on the 840 series chips, we can understand what is going on: <<<Begin Erroneous calculation:>>> 82803 has 3 * Rambus channels, and 324 pins. So each Rambus channel uses about 108 pins. The 82804 has one Rambus channel and one SDRAM channel, with a total of 241 pins. If the Rambus channel takes 108, the total needed by the SDRAM would be 133. Therefore, the ratio of pin count between SDRAM and RDRAM channels is around 5 to 4. Therefore, the actual pincount savings for RDRAM over SDRAM is only 20% of the memory pins, or around 25 pins per channel. <<<End erroneous calculation...>>> Now some possible explanations for why SDRAM is showing fewer pins than expected, compared to RDRAM: (0) The SDRAM RIMM module actually has more pins than the number needed to drive it. That is because some of the signals are tied to VCC or GND, and others are tied together. But this doesn't apply to the RDRAM bus, which actually needs all of its pins. (1) Pin rounding. The 82804 just barely fits in a 241, while the 82803 has to get bumped to the next standard package. This is quite possible. (2) I/O bound silicon. The Rambus chip requires particularly large or wide I/O pads, and consequently, the 82803 is more I/O bound than its pin count would suggest. The part is thus forced into a larger package (that happens to have more pins) because of the resulting die size. This would be suggested by a large number of NC among the 82803 pins. Unfortunately, I haven't seen the data sheets yet. (3) The 82803 has a much larger number of power and ground pins. This could be due to its extraordinarily high output currents. The 2 outgoing channels each have something like 3 amps peak current total, while the incoming channel has something like 1.8 amps. This gives a total peak output current of 7.8 amps, and therefore ground bounce is fattening the pin count requirement. In addition, the low signal levels might increase the sensitivity to ground bounce, relative to standard CMOS signaling levels. Any more ideas? -- Carl