Q1.2
.docx
School
University of Colorado, Boulder**We aren't endorsed by this school
Course
ECEN 5807
Subject
Electrical Engineering
Date
Dec 19, 2024
Pages
2
Uploaded by ConstableRamMaster5771
1.Question 1The questions in this practice Quiz refer to the switching circuit model of a peak current-mode controlled point-of-load buck regulator in SyncBuck_switching_CPM_CL.asc, which is provided among the course LTspice files. For the purposes of this Quiz, the parameter Va in the CPM modulator should be set to 0 0V. Perform transient simulation of the POL regulator and examine inductor current i(L1), output voltage v(out), and control voltage v(ctrl) waveforms. In the time interval between 0and10μs, what is the peak value of the inductor current obtained by transient simulation? You may use a waveform cursor to determine the value. Express the value in Amps, and round to 2 significant digits. 1 / 1 point0.57CorrectThe correct answer is (approximately) 0.58 A.2.Question 2What is the value of the control voltage v(ctrl) at the time when the inductor current is at thepeak value found in Question 1? You may use a waveform cursor to determine the value. Express the value in Volts, and round to 3 significant digits. 1 / 1 point1.04CorrectThe correct value is approximately 1.04 V.3.Question 3The CPM modulator model includes an "offset" parameter Voffset, which is preset to 1 1V. You may right-click on the model and inspect the model parameters in SpiceLine and SpiceLine2.Ideally, given the control voltage v(ctrl), and the current-sense resistance Rf, the peak inductor current should be equal to (v(ctrl)-Voffset)/Rf. When this expression is applied to v(ctrl) found in Question 2, the ideal peak inductor current value is found to be somewhat lower compared to the actual peak inductor current value found in Question 1. Why is this the case?Check the correct answer below 1 / 1 pointThe output voltage ripple is not zero.There is some tolerance in the inductance L1.The equivalent current sense resistance Rf is too large.There is a delay in the transistor M2 turn-off due to delays in the CPM comparator, the gatedriver, and the time it takes to turn the transistor off. The switching losses cannot be neglectedThe equivalent current sense resistance Rf is too low. The inductor winding resistance RL is not zero.The conduction losses cannot be neglectedThere is a delay in the transistor M1 turn-off due to delays in the CPM comparator, the gatedriver, and the time it takes to turn the transistor off. There is a dead-time of Td = 30ns between the on-times of the transistor M1 and the transistor M2.
CorrectAs discussed in the lecture, M1 turns off with some delay after the intersection between thesensed current and the control voltage.