Saturday, November 23, 2013

Brenkung Adder Design

Exe sawn-offive Summary Adder Design The objective of the cipher is to cast a 16-bit common viper. The goal of the project is to belittle the endure spell keeping power within constraints, as well as minimizing the ara. The constraints are as follows: Energy/ functioning has to be slight , t trise han t0.5nJ and fare < 200ps, and use the example cell layout, which must obey each(prenominal) the design rules. Throughout the pattern there were several adder topologies with different advantages been introduced, and we finally came to the last to choose Brent-Kung adder architecture. The solid ground we chose this topology was because its execution was simpler than Radix-4 Kogge-Stone head. As a result the layout of the circuit would be less complicated. Even though it is a bit slower than the Kogg-Stone, it would talk us time doing the layout in the end. After choosing the topology, we had to come up with the logic style to implement the bl We had imple mented the logic blocks victimization transmission accesss to gain more speed. However, for simplicity in sizing the Motivations: Since we wanted the adder to operate as close as possible, we came up with some ideas to minimize the propagation delay such as: we change the Brent-Kung tree, added 3 more shipping products and cut the number of stages for the critical path from 7 to 5. is a professional essay writing service at which you can buy essays on any topics and disciplines! All custom essays are written by professional writers!
For the Brent-Kung tree in swan to get the carryout C14, the adder has to wait for the product of C7&C11, past C11&13 and finally C13&P14. On the other pile for the modified Brent- Kung tree C7 is directly producted with C1 3 at the fourth stage. This helps the C14 to! be in stock(predicate) at the fifth stage. The modified Brent-Kung tree is shown below as well as the critical path, which is highlighted in red. Loading evaluate: Rwire=Rsq*L/W=0.75*1.8mm / 4?=281.2Ohm Cin = 24.7fF(from the sum of all the access input capacitance of the very first stage.) Since the adder movement a 1.8mm long bus with 6 loads every bit distributed. Each capacitive load is equal...If you want to get a full essay, order it on our website:

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