Blog for students taking ECEN5807 Modeling and Control of Power Electronics, ECEE Department, University of Colorado at Boulder, Spring 2013
Thursday, March 31, 2011
Lecture 28 Slide 9
Tuesday, March 29, 2011
line commutated rectifier applications
In lecture 30, it was mentioned that the chapter on line commutated rectifiers may be eliminated as they don’t provide power factor correction (PFC) and thus not much used. I talked to Prof. Maksimovic about that statement and he clarified that there are indeed applications where such rectifiers are well suited for relative simplicity, low cost, or power levels. I would like to take this opportunity to mention one such application.
The application I am familiar with is where the ac voltage is not from the power line, but from an alternator (for instance, the generator in a small wind turbine system). The alternator source has no PFC requirement, per se, as it is isolated from the power grid. In such systems, the simple (line commutated type) rectifier is used to convert the variable frequency ac output of the wind turbine generator, i.e., alternator, to (variable) DC, which is then boosted and then inverted to power line AC (for connection to the power grid). The line commutated rectified DC is generally boosted using a simple conventional boost circuit. That boost circuit does provide some power factor correction action. As a result, the alternator current no longer looks like current spikes (when the rectifier diodes conduct), but actually with a nice current waveform (although, not quite sinusoidal). In my experience, the generator current lags the generator voltage by no more than, say, 15degrees, in the worst case (which is generally under light loads and also depends on the generator winding inductance and resistance). With such a design, one does has to be careful about the alternator current waveform at various operating powers. Even though there is no requirement for PFC, it is important that the alternator current waveform be not too much distorted or too much out of phase with the alternator voltage, as it results in torque ripple and acoustic effects. High current harmonic content may also lead to excessive loss in the windings.
Component differences in peak vs. avg. current mode control?
Wednesday, March 23, 2011
HW8 Prob 2
Friday, March 18, 2011
Midterm Exam Solution?
I do not see one, nor do I see anyway to get a PW for it on the CU Learn.
thanks
Mark
Friday, March 11, 2011
HW 7 Spice solution , part D link ?
thanks
Mark
Thursday, March 10, 2011
HW7 Prob2(b)
Wednesday, March 9, 2011
Post replies
Tuesday, March 8, 2011
HW7 Problem 1 Zout
Sunday, March 6, 2011
HW 7 problem 2 steady state
solving DCM buck, and boost for M
I do not understand how to solve these for M (M= output Volage/Vg in steady state). The buck-boost on page417 was easy.
Can anyone help me to understand how to solve for these?
thanks
Mark
Thursday, March 3, 2011
HW5, prob.1, determination of s^2 numerator zero contributed by L2C1
In HW5, problem 1, I cannot seem to determine the s^2 numerator zero contributed by L2,C1. For determining the first order zero contributed by C1, I assumed a samll resistance 'r' in series with L2 and the result was: v^test/i^test = r. Now, to determine the zero contributed by L2C1 together, I have set C1 as short (abnormal) and put the test source across L2 (with shorted L1 represented by a small resistance 'r') . I was expecting the resistance see would be 1/r....but I cannot get that.
Any hints would be useful.
Regards,
nitish
Wednesday, March 2, 2011
Null Loop Gain Question
Problem 11.1
are we supposed to derive a model of the switches for DCM?
The problem does not actually say to do that, however part 1 is a step in that direction.
thanks
Mark
Lecture 21 Quesion?
thanks
Mark
Tuesday, March 1, 2011
prob. 11.1(e); Lp =0; does Re become 0? which means j1 and j2 = infinity?
In prob. 11.1(e); Lp =0; Does that allow us to make Re = 0 as it has Lp in its Numerator ? Making Re=0 would make j1 and j2 = infinity as they have Re in their denominator.
Any comments would be useful.
Thanks,
Nitish
DCM waveform levels for the buck-boost converter (fig. 11.3)
In fig. 11.3, the buck-boost converter DCM waveform levels are shown in terms of instantaneous values of vg and v....but shouldn't they be written in terms of their average values ,vg> and
For example, v2(t) waveform level during interval d1Ts is shown to be vg-v. But shouldn't it be
Also, eqn. 11.1 should be written as ipk =
I would appreciate any comments.
Regards,
Nitish Agrawal
General Feedback Theorem
lecture 18, slide 34 (when expressed in terms of R, L, C....)
I did not understand the lecture 18, slide 34 summary statement, "when expressed in terms of R, L, C, and M (not D), the small signal transfer functions are he same in DCM and in CCM".
Can anyone shed light on this?
Regards,
Nitish