Question

A designer has to implement a 6- input AND gate. He has available a library containing the cell-types enumerated below together with their properties in terms of capacitance.

Inv NOR2 NOR3 NAND2 NAND3 NAND6
Cin 48fF 48fF 48fF 48fF 48fF 48 fF
Cout 85fF 101fF 117fF 105fF 132fF 200fF

Determine the average power dissipation of an implementation using a 6-input NAND ( and an inverter). The circuit operates at a supply voltage of 3V and a clock frequency of 20 MHz
Assume that all the inputs have an equal chance of being 0 or 1. You may ignore the power dissipated by the 6 input signals.

Answer 1

To determine the average power dissipation of an implementation using a 6-input NAND gate and an inverter, the power dissipation of each gate needs to be considered. The average power dissipation can be calculated using the supply voltage, average current, effective capacitance, and clock frequency. Specific formulas and calculations can be used to find the average power dissipation in this case.

Step-by-step explanation:

To determine the average power dissipation of an implementation using a 6-input NAND gate and an inverter, we need to consider the power consumed by each gate. The power dissipation of a gate can be calculated using the formula:

P = VDD * IDD * Ceff * f

Where P is the power dissipation, VDD is the supply voltage (3V in this case), IDD is the average current drawn by the gate, Ceff is the effective capacitance, and f is the clock frequency (20 MHz in this case).

For a 6-input NAND gate, the average current can be calculated as:

IDD = (INAND6 + IINV) / 2

To find the effective capacitance, we can sum up the capacitance values of the NAND6 and INV gates, considering the inputs as well:

Ceff = CNAND6 + CINV + (6 * Cin)

Using the given capacitance values, we can calculate the average power dissipation of the implementation.