EEE6323S12-HW4-sol

# (t0 is the time point where signal “a” toggles 1

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Unformatted text preview: (t0 is the time point where signal “a” toggles) 1 1 1 2 1 2 L t t t out C V D D O U T V D D O U T L O U T t t VDD L O U T O U T L DD dV E i V d t i V d t C V d t dt C V d V C V = • = • = • = = ∫ ∫ ∫ ∫ 5.2 Energy dissipated in the switching transistor 1 1 1 2 ( ) ( ) ( ) ( ) 1 2 t t t out Dis VDD DD OUT VDD DD OUT L DD OUT t t VDD L D D O U T O U T L DD dV E i V V d t i V V d t C V V d t dt C V V d V C V = • − = • − = • − = − = ∫ ∫ ∫ ∫ 5.3 Total energy supplied 2 2 2 1 1 2 2 L Total Dis C L DD L DD L DD E E E C V C V C V = + = + = Or 1 1 1 2 t t t out Total VDD DD VDD DD L DD t t VDD L D D O U T L DD dV E i V dt i V dt C V dt dt C V dV C V = • = • = • = = ∫ ∫ ∫ ∫ Case 2: 5.1 Energy stored on the capacitor C L (t0 is the time point where signal “a” toggles) Same as in Case 1: 2 2 1 DD L C V C E L = . That is, the energy stored depends on the final voltage, regardless of how the capacitor was charged. 5.2 Energy dissipated in the switching transistor The switch will dissipate one half of the energy supplied to the cap at every step, or 2 2 8 1 2 2 1 DD L DD L Diss V C V C E = Ο Π Ξ Μ Ν Λ = . Since the switch charges the cap in two steps, the total energy dissipated is twice, or 2 2 4 1 8 1 2 DD L DD L TDiss V C V C E = = . Try doing the integration and check that you get the same result....
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(t0 is the time point where signal “a” toggles 1 1 1 2...

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