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Minimally Allocating Always-on State Retention Storage for Supporting Power Gating Circuits

机译:最小地分配用于支持功率门控电路的始终开启状态保留存储器

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State retention storage is indispensable in power gated circuit since the circuit state should be saved somewhere before sleeping so that the circuit can continue the execution from the saved state when waking up. Thus, it is essential to allocate the always-on retention storage minimally to reduce the leakage power. Recently, it is demonstrated that a substantial saving on total retention storage can be achieved simply by allowing some flip-flops to retain 2 or 3 bits if needed, as opposed to strictly allocating every flip-flop to retain exactly 1 bit. In this work, we propose an optimal solution to the problem of allocating state retention storage, constraining the size of every flip-flop’s retention storage never to exceed l bits where l is 2 or 3. Specifically, we transform the allocation problems constraining the wakeup latency constraint l to 2 and 3 clock cycles into unate covering problems and solve them optimally with three objective options: minimizing total bits of retention storage, directly minimizing total leakage power consumed by retention storage, and minimizing total implementation area of retention storage. Through experiments with benchmark circuits, it is shown that our optimal algorithm is able to further reduce the total bits of retention storage up to 3.42%, the leakage power on retention storage up to 9.91%, and the retention storage area up to 4.46% while l is set to 3 over that produced by the conventional best-known allocation heuristic.
机译:由于电路状态应在睡眠前的某处保存,所以状态保持存储是必不可少的,因为在醒来时电路可以继续从节省状态继续执行。因此,必须尽可能地分配始终对保留存储,以减少泄漏功率。最近,证明可以简单地通过允许某些触发器来维护2或3位,而不是严格分配每个触发器来保留1位。在这项工作中,我们提出了对分配状态保留存储的问题的最佳解决方案,约束每个触发器的保留存储的大小永远不会超过L位,其中L为2或3。具体地,我们改变了约束唤醒的分配问题延迟约束L到2和3个时钟周期的对统覆盖问题,并用三个客观选项进行最佳解决:最大限度地减少保留存储的总比特,直接最小化保留存储所消耗的总漏电,并最大限度地减少保留存储的总实现区域。通过基准电路的实验,表明我们的最佳算法能够进一步将保留存储的总比特降低至3.42%,泄漏功率高达9.91%,保留存储区域高达4.46% L由传统的最着名的分配启发式产生的。

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