K/K# being single ended but 180 degrees out of phase

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Anonymous
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 This EETimes article www.eetimes.com/design/memory-design/4395832/Interfacing-QDR-II--Synchronous-SRAM-with-high-speed-FP... by Reshi Ravindranand and Ajay Bharadwaj says K and K# are not differential signals, but are single ended signals that are 180 degrees out of phase. 

   

I thought those were differential signals?

   

Thanks,

   

-Travis

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1 Solution
PriteshM_61
Employee
Employee
25 solutions authored 10 solutions authored 5 solutions authored

Hi Travis,

   


   

Please note that K and K\ clocks are not true differential clock signals. There is no differential receiver in the SRAM. The QDR uses the rising edges of K and K\ to latch input signals. Both clocks are single ended signals. Although they are not true differential, it is advised to keep K and K\ 180 degrees out of phase with respect to one another. This effect produces the ability for the QDR to perform a double data rate with one clock cycle.

   


   

Thanks,

   

Prit

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1 Reply
PriteshM_61
Employee
Employee
25 solutions authored 10 solutions authored 5 solutions authored

Hi Travis,

   


   

Please note that K and K\ clocks are not true differential clock signals. There is no differential receiver in the SRAM. The QDR uses the rising edges of K and K\ to latch input signals. Both clocks are single ended signals. Although they are not true differential, it is advised to keep K and K\ 180 degrees out of phase with respect to one another. This effect produces the ability for the QDR to perform a double data rate with one clock cycle.

   


   

Thanks,

   

Prit

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