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(elec551)[2008](s)mid~PPSpider^_10333.pdf
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(1) Design a buck converter with the following specifications: 10 marks
Switching frequency fs = 1MHz
Input voltage Vg changes from 2.4V to 4.2V
Output voltage Vo = 1.8V
Load current Io changes from 100mA to 500mA









(1a) Determine the value of the inductor if the converter always works in 3 marks
CCM.

Sol:






For the boundary case, the output current is minimum and the input voltage
is maximum,




(1b) Determine the value of the inductor if the converter always works in 3 marks
DCM.

Sol:


For the boundary case, the output current is maximum and the input voltage
is minimum,











(1c) Determine the value of the capacitor for the case of (1a) such that the 4 marks
output voltage ripple is 20mV when Vg = 2.4V and Io = 300mA.

Sol:









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(2) Consider an ideal non-inverting buck-boost converter working in 10 marks
hysteretic current mode. Note that T = t1 + t2 + t3.









(2a) If the peak inductor current Ipk is set to be 50mV and the averaged 4 marks
load current is 10mA, compute the switching frequency fs = 1/T of
the converter.

Sol:






or


(2b) Compute the output voltage ripple. 3 marks

Sol:








(2c) If the load current is decreased to 6mA, will the switching frequency 3 marks
increase or decrease? And will the output voltage ripple increase or
decrease? Explain your reason (by using inductor current waveform,
for example).

Sol: t1 and t2 are remain unchanged. t3 will increase as it take more time to
discharge the charge store in the capacitor. Thus, the period will increase
and the frequency is decrease.



The shallow area is the charge store in the capacitor, as the Io decrease,
the shallow area will increase. Thus, the output voltage ripple will increase.

(3) Consider a second order boost converter that works in CCM. 10 marks









(3a) For = 0, derive the state equations for the two states, i.e., 2


= Aix + BiVg with x =
and i = 1, 2

Sol: State 1,
,


State 2,

,










Io

t




K: Io

(3b) Employ state space averaging to solve for the output voltage Vc 2 marks
in terms of Vg and relevant parameters.

Sol: Apply SSA:

,


In steady State,









Thus,




(3c) For 0, derive the state equations for the two states, i.e., 3 marks


= Aix + BiVg with x =
and i = 1, 2

Sol: State 1,
,


State 2,

,















(3d) Employ state space averaging to solve for the output voltage Vc 3 marks
in terms of Vg and relevant parameters.

Sol: Apply SSA:

,


In stea