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Trouble Understanding What DDR266 is...
I have a problem...
I am on TigerDirect.com and I have reached a problem... The specs for the 'Mach Speed - Venom Mach V6FP - Socket A MATX Motherboard with Audio, Video, USB, AGP 4X and 10/100Mbps LAN Support' states the following: SNIP .... .... .... """ # System Memory: - 2 x 168-pin DIMM Sockets Supports PC133/PC100 SDRAM up to 2.0GB System Memory. - 2 x 184-pin DDR Module Socket Supports DDR200/DDR266 DDR SDRAM up to 2.0GB System memory. """ SNIP I understand what PC133/PC100 is, however, I DONOT understand what DDR200/DDR266 is... I was reading in other groups that DDR266 ~ to PC2100 : Is this true? If so why is it true? How would one figure this out??? Thanks in advance for reading! |
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"Snail" wrote in message
om... I have a problem... I am on TigerDirect.com and I have reached a problem... The specs for the 'Mach Speed - Venom Mach V6FP - Socket A MATX Motherboard with Audio, Video, USB, AGP 4X and 10/100Mbps LAN Support' states the following: SNIP ... ... ... """ # System Memory: - 2 x 168-pin DIMM Sockets Supports PC133/PC100 SDRAM up to 2.0GB System Memory. - 2 x 184-pin DDR Module Socket Supports DDR200/DDR266 DDR SDRAM up to 2.0GB System memory. """ SNIP I understand what PC133/PC100 is, however, I DONOT understand what DDR200/DDR266 is... I was reading in other groups that DDR266 ~ to PC2100 : Is this true? If so why is it true? How would one figure this out??? Thanks in advance for reading! In PC100 and PC133 RAM, the number tells the speed of the stick, in MHz. Therefore, the PC100 runs at 100MHz and PC133 at 133MHz (mind you, these are maximum speeds, lower speeds WILL work fine). Similarly, the 200 and 266 refered to are telling the "DDR" speed. Remember, there isn't a 266MHz bus, it is actually 133MHz double-pumped to yield the bandwidth equivelant of a 266MHz (single-pumped) bus. It's just marketing. Yes, DDR266 = PC2100. It's the same thing. You can find the equivelance in PC if you know the DDR#, by multiplying by 8. Reason is that in every clock cycle the bus transfers 64 bit, which will also be called a 64-bit wide path. Now, 64 bits is equal to 8 bytes. And your figure was already in MHz so, if you take the MHz and multiply by Bytes/Hz, you end up with MB. Look closely, and see that it is Bytes/Hz, not Bytes/MHz, because one clock cycle = 1Hz, so there are 1 million cycles in 1MHz, therefor your 8 bytes really turns into 8MB, so you could rewrite it as (266)MHz times 8MB/MHz, and you get 266 times 8, yielding your 2100MB/sec. Hope I haven't confused you, but just take the MHz and multiply by 8. *IF* they quote the actual speed, since the 266MHz RAM really runs at 133MHz, then multiply by 16, because you can transmit data on both edges of the signal. |
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