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Can i run DDR600 with this board? - Page 2

post #11 of 16
get new mobo and cheap faster ram. lower timing on thsoe and bam
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post #12 of 16
Quote:
Originally Posted by superk View Post
I have to agree, 200 is a bit pricey. Its nice ram, good quality, but I mean, $200??? Too rich for my blood. For 200 you could get an AM2 platform and would be able to get better, faster ram, for an extra 50 bucks or so...
Or a Core2Duo
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post #13 of 16
200 for ram, dang i could get a better set of ddr3 (1600MHZ 7 7 7 20)
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post #14 of 16
Thread Starter 
yeah i know $200 is a rip off but i guess its a matter of how bad you want it.
as i dont have nearly enough money to buy a new cpu ram and mobo i figure ill try to squeeze a little bit more out of good old 939. also i think my generic ram is on its death bed...

the story goes
- bidded up to $120 for the mushkin ram (4 x 512 DDR 550) lost the auction in the last 19 seconds to some other @ $123
-bidded $120 for OCZ gold ed DDR500 lost that too

and finally found some
G.SKILL 2x1GB DDR500 and won the auction for $100 inc postage.
haven't received it yet, but im hoping that it will increase my performance a little.

how far do you think i can overclock it safely? im not fully clued up when it comes to ram, and why the latency and timings change it different voltages? im hoping at least to 600

what do you think? thanks for suggestions all.
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post #15 of 16
Quote:
Originally Posted by Joshn View Post
why the latency and timings change it different voltages?
Could you repeat that question ...in English?

Voltage doesn't change timings, that's done in the BIOS.

If you mean "Why do I have to run higher voltage inorder to run memory faster and/or with tighter timings? Then the answer is that memory is basically a bunch of capacitors. they get charged up, they get discharged. Chargerd = a "0", discharged = a "1" in digital terms. They also loose charge (leak down) by themselves. If you spend less time charging them up, they don't get fully charged and will leak down to soon (ie data corruption, or more accurately, data gone!). So if you run higher speeds or faster timings, you have to use a higher voltage to get the capacitors fully charged so they will hold their data until the memory controller needs it again.

When buying high end memory, the memory ICs used (BH5 for spectacurly low latencies, TCCD for blazing speeds) is more of an indicator of overclockability than brand or manufacturer's specification.

Her's some TCCD at DDR665

Get a program called A64Info (I like v0.5) and you can play with all 31 latencies, instead of just the usual 4. Cas is the key to speed, Tref, MAL and Read Preamble are the keys to stability.
Edited by billbartuska - 6/16/08 at 6:33am
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post #16 of 16
Thread Starter 
Quote:
Originally Posted by billbartuska View Post
Could you repeat that question ...in English?

Voltage doesn't change timings, that's done in the BIOS.

If you mean "Why do I have to run higher voltage inorder to run memory faster and/or with tighter timings? Then the answer is that memory is basically a bunch of capacitors. they get charged up, they get discharged. Chargerd = a "0", discharged = a "1" in digital terms. They also loose charge (leak down) by themselves. If you spend less time charging them up, they don't get fully charged and will leak down to soon (ie data corruption, or more accurately, data gone!). So if you run higher speeds or faster timings, you have to use a higher voltage to get the capacitors fully charged so they will hold their data until the memory controller needs it again.

When buying high end memory, the memory ICs used (BH5 for spectacurly low latencies, TCCD for blazing speeds) is more of an indicator of overclockability than brand or manufacturer's specification.

Her's some TCCD at DDR665

Get a program called A64Info (I like v0.5) and you can play with all 31 latencies, instead of just the usual 4. Cas is the key to speed, Tref, MAL and Read Preamble are the keys to stability.
thankyou , and yes that answered my badly phrased question, i was just a little confused when the descriptions stated" x timings @ x voltage", and then "y timings at y voltage" but now i understand. Thanks!

Rep +
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