Lecture 28 Slide 9

My question is regarding the CPM Buck converter example on slide 9 of lecture 28. If Ma = M2 = M1 = 1A/us, how do we arrive at .11/A for Fm?

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?

I am trying to figure out the difference in components between a peak and an average current mode control (ACMC) setup. There doesn't appear any difference between the block diagram of fig. 12.12 (which is in the peak current mode control section) and the block diagram presented for ACMC. Both have the voltage loop and the current loop as well as the comparator to compare iL and ic. I am guessing, then, the differences are: 1) The current sensor in peak control will be sensing the instantaneous inductor current, while the one in average control will sense the average inductor current. 2) The comparator comparing iL and ic in peak control operates in the open loop condition (no feedback to reduce its gain) and hence puts out either "1" or "0", whereas the one in ACMC operates with a finite gain (closed loop form). 3) Finally, the current command (variously called i^c*Rs and i^c*Rc in class) would need to reflect the peak and average inductor currents respectively. Would these be the only differences between a peak control and ACMC setup? Nitish

HW8 Prob 2

At the risk of coming across as a simpleton, how do you go about calculating the steady state duty cycle for differing loads in part a? Every time I try a different approach I come up with the same duty cycle for all loads which cannot be the case.

Midterm Exam Solution?

Will there be a midterm exam solution posted?

I do not see one, nor do I see anyway to get a PW for it on the CU Learn.

thanks
Mark

HW 7 Spice solution , part D link ?

the link for the HW 7 Spice sims netlist for part D,. points to the same as part C. Is this a mistake?

thanks
Mark

HW7 Prob2(b)

In deriving the loop gain T for this converter, I'm trying to understand what the impact R1 has. Can anyone please explain what it's supposed to do? Does it affect the sensor gain H(s)? Thanks.