For the switching frequency expression I got a result that was proportional to the output current and included the conversion ratio M (=V/Vg) and the peak current, but did not have any terms for the inductance (which is part of the peak current equation) or output voltage (which is part of the conversion ratio M). The expression I have makes sense in that the frequency moves in the right direction as load, on time (t1), peak current, and conversion ratio are varied.
Anyone else getting the same thing? I ask because the problem statement asks for an expression of freqeuncy with L and V but these all got swallowed up in other terms in my result.
Blog for students taking ECEN4797/5797 Introduction to Power Electronics, ECEE Department, University of Colorado at Boulder, Fall 2011
Friday, September 30, 2011
Friday, September 23, 2011
Wednesday, September 21, 2011
HW4 buck-boost efficiency
Anyone else made it to part e) of the buck-boost problem in HW4?
I get mired in ugly algebra when trying to calculate the efficiency or M(D) of the full buck-boost (including switching loss) analytically. I can come up with an expression that only has D, D', V, Vg, and the given constants, but it's so big I'm afraid I've missed a good simplifying assumption.
I've tried making some approximations to get rid of the ugliest bits, but want to be sure I'm not missing a key piece here.
I get mired in ugly algebra when trying to calculate the efficiency or M(D) of the full buck-boost (including switching loss) analytically. I can come up with an expression that only has D, D', V, Vg, and the given constants, but it's so big I'm afraid I've missed a good simplifying assumption.
I've tried making some approximations to get rid of the ugliest bits, but want to be sure I'm not missing a key piece here.
Q2.9 from HW1
Hi can anyone please tell me why delta_IL2= VgDTs/(2L) from the HW solution.
It's a buck converter so delta_I= (Vg-V)DTs/(2L)
Can anyone please explain the difference?
HW 4 Buck Boost Diagram Error
Note that there is a mistake in the schematic diagram of the non-inverting buck-boost converter: Q4 is reversed.
The drain of Q4 must be attached to the output capacitor and the source of Q4 attached to the junction of the inductor and drain of Q3. Otherwise, when Q3 turns on the output capacitor will discharge through the body diode of Q4.
Side note: in this configuration the converter is symmetrical about the inductor and can transfer power in either direction. This is useful with a battery where the battery can be charged with a voltage source higher or lower than the battery voltage and then the converter can transfer power from the battery back to the source.
The drain of Q4 must be attached to the output capacitor and the source of Q4 attached to the junction of the inductor and drain of Q3. Otherwise, when Q3 turns on the output capacitor will discharge through the body diode of Q4.
Side note: in this configuration the converter is symmetrical about the inductor and can transfer power in either direction. This is useful with a battery where the battery can be charged with a voltage source higher or lower than the battery voltage and then the converter can transfer power from the battery back to the source.
Tuesday, September 20, 2011
Couple questions on HW4
For the Buck-Boost problem, when we solve for the buck or boost part, does that mean we just choose one to work out? Seems like "buck or boost" is two different cases, so I'm confused why the problem keeps saying we need to solve 2 cases instead of 3.
For parts b) and c) I am essentially just using the two examples the professor showed in lecture 11, except for part b) where the problem statement states that Q2 is initially on, meaning the inductor current of Q1 will be multiplied by D prime, and not D. When I go to solve it in part d) though this doesn't line up for the transformer since the other side of the transformer is of the form D*Vg. Can we just arbitrarily sketch our waveforms such that the duty cycle lines up with what we need?
Also, why don't we include the reverse recovery of the body diodes of Q1 and Q4 when the cycle is transitioning between Q1 to Q2 and Q4 to Q3?
Thanks
For parts b) and c) I am essentially just using the two examples the professor showed in lecture 11, except for part b) where the problem statement states that Q2 is initially on, meaning the inductor current of Q1 will be multiplied by D prime, and not D. When I go to solve it in part d) though this doesn't line up for the transformer since the other side of the transformer is of the form D*Vg. Can we just arbitrarily sketch our waveforms such that the duty cycle lines up with what we need?
Also, why don't we include the reverse recovery of the body diodes of Q1 and Q4 when the cycle is transitioning between Q1 to Q2 and Q4 to Q3?
Thanks
Sunday, September 18, 2011
Estimate capacitor ripple voltage
Any idea why while estimating cap. ripple voltage, we cannot use the technique for 2 poles i.e. using the charge in and out of cap to find dV.
Isn't it assumed (by definition of capacitor) that only the ripple current flows int he cap and the steady state current is taken by the load? Why does this method produce a different result like in HW1, first question (Q2.2 I think, it a buck-boost), part iii.
Thanks.
Wednesday, September 14, 2011
ECEN 5797 Study Group
Would anybody be interested in starting a ECEN 5797 distance study group?
Sunday, September 11, 2011
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