The twoinput NAND gate shown below is driven by two periodic

The two-input NAND gate shown below is driven by two periodic signals. Assume the shown capacitor represents the only capacitance in the circuit and the power supply level is V_dd. Determine the dynamic power consumption of the circuit.

Solution

Dynamic Power = a * C * V²_dd * f

a = Switching probability of NAND Gate

a = 1 - P_A * P_B

Lets assume P_A = 1 than P_B = 0.5

a = 1 - 0.5 = 0.5

P = 0.5 * C * V²_dd * f