Survey
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
Pass-Transistor Logic Inputs B Switch Out A Out Network B B • N transistors • No static consumption NMOS-only switch C=5V C=5V M2 A=5V A=5V B B Mn CL M1 VB does not pull up to 5V, but 5V - VTN Threshold voltage loss causes static power consumption Solution 1: Transmission Gate C A C A B B C C C=5 V A=5V B CL C=0V Resistance of Transmission Gate 30000.0 Rn (W/L)p =(W/L)n = 1.8/1.2 R (Ohm) 20000.0 Rp 10000.0 0.0 0.0 Req 1.0 2.0 3.0 Vout 4.0 5.0 Pass-Transistor Based Multiplexer S S S S VDD S A VDD M2 F S M1 B S GND In1 In2 Transmission Gate XOR B B M2 A A F M1 M3/M4 B B Delay in Transmission Gate Networks 5 5 V1 In 5 Vi Vi-1 C 0 5 C 0 Vn-1 Vi+1 C 0 Vn C C 0 (a) Req Req V1 In Req Vi C Vn-1 Vi+1 C C Req C C (b) m Req Req Req Req Req Req In C CC C C (c) CC Vn C Elmore Delay (Chapter 8) Vin R1 C1 1 R2 Ri-1 2 C2 i-1 Ci-1 Ri i Ci RN N CN Assume All internal nodes are precharged to VDD and a step voltage is applied at the input Vin N N = N N i Ri Cj = C i R j i=1 j=i i=1 j=1 Delay Optimization Transmission Gate Full Adder P VDD Ci A P A A P B VDD Ci A P Ci S Sum Generation Ci P B Setup VDD A P Co Carry Generation Ci A VDD P (2) NMOS Only Logic: Level Restoring Transistor VDD VDD Level Restorer Mr B A Mn M2 X Out M1 • Advantage: Full Swing • Disadvantage: More Complex, Larger Capacitance • Other approaches: reduced threshold NMOS Level Restoring Transistor 3.0 without without 3.0 with VB 1.0 -1.00 with 5.0 VX Vout (V) 5.0 2 t (nsec) 1.0 4 (a) Output node 6 -1.00 2 4 t (nsec) (b) Intermediate node X 6 Solution 3: Single Transistor Pass Gate with VT=0 VDD VDD 0V 5V VDD 0V Out 5V WATCH OUT FOR LEAKAGE CURRENTS Complimentary Pass Transistor Logic A A B B Pass-Transistor F Network (a) A A B B B Inverse Pass-Transistor Network B B A F B B A A B F=AB A B F=A+B F=AB AND/NAND A F=AÝ (b) A A B B F=A+B B OR/NOR A EXOR/NEXOR F=AÝ 4 Input NAND in CPL