However changing the clock frequency by a factor of k changes the amount of

However changing the clock frequency by a factor of k

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However, changing the clock frequency by a factor of k changes the amount of time it takes for the operation to complete by a factor of 1/k. Since Energy = Power*Time, the net scaling in energy/operation in the logic is k 2 . For the scaling above, the energy per operation becomes (1 J + 0 . 1(2) J )(18 / 20) 2 = 0 . 972 J compared to 1 . 2 J in the original. A more substantial energy efficiency improvement can be gained by pipelining the combina- tional logic. When pipelining, the critical path is reduced which allows the clock rate to be increased without changing V dd . Since this application does not require a clock rate increase, the clock frequency and V dd can both be scaled down from the new clock frequency made pos- sible by pipe-lining. The total amount of time an operation take to get through the pipeline changes to become the number of pipeline stages * the delay through each pipeline stage. This changes the energy/op scaling factor because power is scaled by k 3 but time/op is now scaled by n/k . This leads to energy/op scaling by nk 2 . Let’s divide the combination logic into 2 equal parts (from the perspective of delay) and insert a pipeline register between them. We incur the additional energy use of the pipeline register but can now reduce the combinational logic delay to 8 ns. This leads to a critical path of 10 ns. This allows clock frequency and V dd to be scaled by a factor of 1 / 2 to reach the target clock frequency of 50 MHz (20 ns period). The total energy per operation of this pipelined version is (1 J + 0 . 1(3) J )(2)(1 / 2) 2 = 0 . 65 J . This process can be tried again by splitting the combinational logic into n segments with n-1 additional regsiters. Assuming the combinational logic is evenly split, the critical path when pipelined by an integer factor n is 16 /n + 2 ns. The energy/operation is (1 + 0 . 1(1 + n ))((16 /n + 2) / 20) 2 n To find when the energy/op is minimum, we can take the first derivative of the above expression and get 0 . 027 - 0 . 704 n - 2 - 0 . 02 n The only real zero is at n 4 . 431 . The second derivative is: 0 . 002 + 1 . 408 n - 3 This is positive for n 4 . 431 identifying n 4 . 431 as a local minimum. Since n must be an integer, we will look at n = 4 and n = 5 Version: 1 - 2019-04-10 19:32:31-07:00
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EECS 151/251A Homework 7 11 For n = 4 , the critical path delay is 16 / 4 + 2 = 4 / 3 ns. The energy/operation is (1 + 0 . 1(1 + 4))((16 / 4 + 2) / 20) 2 (4) 0 . 54 J . For n = 5 , the critical path delay is 16 / 5 + 2 = 26 / 5 ns. The energy/operation is (1 + 0 . 1(1 + 5))((16 / 5 + 2) / 20) 2 (5) 0 . 5408 J . Therefore, the minimum energy/operation is achieved for this circuit when pipelined by a factor of 4 yielding an energy/op of 0 . 54 J .
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