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To: gnuman who wrote (72208)	5/8/2001 12:34:20 AM
From: Bilow	Read Replies (1) \| Respond to of 93625

Hi Gene Parrott; Re: "What's the relationship between voltage levels on the bus and a current source?" E = I * Z or E = I * Z/2, depending on which applies. Re: "And why do they need wide output compliance?" I almost typed another long explanation for this, but from our previous exchange about the subject of "middle" of the bus, my EE buddy concluded that you are ineducable on this subject. He's probably right, so I'm not going to try. -- Carl P.S. Please pardon me if I ignore some of your posts on this sort of thing, I just don't feel like putting out the wasted effort right now. Send your posts to John Walliker, Scumbria, or maybe cordob or Zeev Hed. Or feel free to assume that you already know everything about it.

To: gnuman who wrote (72208)	5/8/2001 2:28:25 AM
From: pheilman_	Read Replies (1) \| Respond to of 93625

re: And why do they need wide output compliance? Look's to me like a current source is an excellent technical solution, but I'm interested in your reasoning. I would say a current source is an excellent academic solution. It can also be lower power. Practical/pragmatic engineers know the extreme difficulty of troubleshooting a current bus. Consider that all the parts are BGA and the traces are controlled impedance, it is near impossible to instrument the bus to find out what is going on. It requires a current probe and a sensing loop. Yes, at high speeds, a current source is better, but the bus and the transactions better be dead simple or it will take a whole lot of time to get it to work. I don't think DRDRAM transactions qualify as dead simple, so it is not surprising that few, make that just one company is willing to take on the task.