Question

Assuming for typical transistors Vdd = 0.8 V, and they operate at 300 K. Calculate:

a) If we are able to switch the transistors at Landauer’s limit, then what’s the capacitance we need to have? With this capacitance, what’s the noise voltage Vn?

b) Now let’s assume that we set Vdd = 5Vn in order to have less than 1 error in 1,000,000. In this case, what’s the capacitance do we need to have? And what about switching energy now (write energy in both kT and J)?

Answer 1

4) а) The Launder's principle states that the minimum amount of energy required to erase one bit of information is, E KT In2 And this is known as Launder's limit. Here, k is Boltzmann's constant and T is the temperature of heat sink in Kelvins. At 300 K E (1.38x1023 JK) (300 K) (0.693) -21 -2.869 x102 J Also, E CV dd 2.869 x10 C(0.8) C 4.483 x10-2 F F Therefore, the capacitance we need to have is 4.483 x10 And the noise voltage with this capacitance is, E e 2.869x 1021 J -19 1.6 x109 C -17.93125 mv b) Consider that, V5V dd - 5(17.93125) = 89.65625 mv

Therefore, the switching energy is E kT In 2 -0.693KT (0.693)(1.38x10 JIK)(300 K) -2.869 x102 J Therefore, the capacitance we need to have is, E C = 2 V dd 2.869 x10 2 y (s9.65625x10) - 3.569 x1019 F