Power Electronics: Do you think my solution is all right?

[zuzuzu]

This is a homework problem in "Power Electroincs: converters, applications and design", pg.32, Problem 2-2.

I strongly believed that my solution was all right but I could not get full credit and no one supported me. Anyone who studies power electronics, would you please spend a little time to check the problem here and give me some of your opinions on this problem? Or, Is there anyone could provide the textbook model answer to compare with mine?

 

[Miguel Gaspar]

First your scales are wrong, while voltage is from 0 to 300, your current goes from 0 to 4 so in 1.jpg are unproportional, and the power must look like 2.jpg.

 

[zuzuzu]

It was a figure with two scales, what was the problems???
i.e. two plots in one figure simutaneously

 

[Miguel Gaspar]

the problem is that it doesn't give yuo the real situation
I simulate this and it give you the answear of 2.jpg, the current graphics it is very low.

 

[VVV]

Re: Power Electronics: Do you think my solution is all right

Fundamentally there is nothing wrong with your answer.

I think the teacher just did not like your swapping the 1/2 and 1/4 factors in the equation for the approximated switching power. Swapping them is correct, but useless, it does not bring anything new. At least the quantity 14/*Vd*Io had a meaning.
In the second figure you show the steady-state current slightly lower than the voltage, yet the half value points conicide, which is incorrect.

I think that's all. You got a good mark, though.

 

[zuzuzu]

You should know that both the voltage and current are linear function of time under the assumption that pure resistive component was present. Both V and I start and stop at the same instants, as the time was half, they both reach their half values, I think this is anyway all right.

 

[VVV]

Re: Power Electronics: Do you think my solution is all right

Of course they both reach their half values at the same time, but if the scales are slightly different, which they are, judging by the difference in the steady-state values, then the half-scale values will also be different.
In other words, the voltage half-scale value should have been a little higher on the scale. But in your drawing, both were at exactly the same height. In fact you even label them Vd/2, Io/2.