Blog for students taking ECEN5817 Resonant and Soft Switching Techniques in Power Electronics, ECEE Department, University of Colorado at Boulder, Spring 2012
Thursday, April 1, 2010
HW8 DCM equations derivation
Has anyone attempted the first question? Any hints on how to start?
You should easily get the first two equations just by applying sine/cosine rule. just focus on triangle which has beta angle and radius in this case is 1.
I am stuck with fourth equation. Any hints? how do I introduce angle terms in the equation?
This is where I'm stuck too. I have 2 relationships from the boundary conditions....equation 3 from (5-45) which comes from the 2 boundary condition relationships solved like in the chapter notes....equation 5 is by definition and equation 6 comes from similar procedures as in class lecture............
Given that: I'm not sure what equations to manipulate, sub into each other, to get the 4th relation.....???
I don't know if this going to help or not, I am still struggle too.
I try to get a hint from time domain plot (pg 22). Here is what I found: On time domain plot, during dicontinous period wo*t = alpha...tilta, inductor current raise from IL1 to I. The equation for this can be written as:
I - IL1 = Vg/L * (tilta - alpha)/wo then, it can be writen as: (I - IL1) * L*wo / Vg = tilta - alpha
Note: Vg/(L*wo) = Ibase, then I/Ibase = J, and IL1/Ibase = JL1. Thus, equation above can be written as: J - JL1 = tilta - alpha
The last step, that I am still struggle is how to express JL1 as function of alpha, beta, and J.
Vojkan - Thanks for the post, it was very helpful to me. When going through the JL0 boundary case, it is not clear to me how to choose the angles in DCM. I chose sigma = pi - beta. In this case JL0 = J+r2*sin(sigma) = J+r2*sin(β). How did you cancel out the r2 term on the left-hand side?
From JL0 boundary condition we can get equation A: sin(β)=sin(α)-(J+JL1)*cos(α) From Mc0 boundary condition we can get equation B: 2-cos(β)=cos(α)+(J+JL1)*sin(α)
Pb1 Equation 3: can be obtained by: sin(α)*equation A+cos(α)*equation B
Pb1 Equation 4: can be obtained by: sin(α)*equation B-cos(α)*equation A
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9 comments:
You should easily get the first two equations just by applying sine/cosine rule. just focus on triangle which has beta angle and radius in this case is 1.
I am stuck with fourth equation. Any hints?
how do I introduce angle terms in the equation?
This is where I'm stuck too. I have 2 relationships from the boundary conditions....equation 3 from (5-45) which comes from the 2 boundary condition relationships solved like in the chapter notes....equation 5 is by definition and equation 6 comes from similar procedures as in class lecture............
Given that: I'm not sure what equations to manipulate, sub into each other, to get the 4th relation.....???
Hi All,
I don't know if this going to help or not, I am still struggle too.
I try to get a hint from time domain plot (pg 22). Here is what I found:
On time domain plot, during dicontinous period wo*t = alpha...tilta, inductor current raise from IL1 to I. The equation for this can be written as:
I - IL1 = Vg/L * (tilta - alpha)/wo
then, it can be writen as:
(I - IL1) * L*wo / Vg = tilta - alpha
Note: Vg/(L*wo) = Ibase, then I/Ibase = J,
and IL1/Ibase = JL1.
Thus, equation above can be written as:
J - JL1 = tilta - alpha
The last step, that I am still struggle is how
to express JL1 as function of alpha, beta, and J.
Vojkan - Thanks for the post, it was very helpful to me. When going through the JL0 boundary case, it is not clear to me how to choose the angles in DCM. I chose sigma = pi - beta. In this case JL0 = J+r2*sin(sigma) = J+r2*sin(β). How did you cancel out the r2 term on the left-hand side?
r2 = 1 for DCM
From JL0 boundary condition we can get equation A:
sin(β)=sin(α)-(J+JL1)*cos(α)
From Mc0 boundary condition we can get equation B:
2-cos(β)=cos(α)+(J+JL1)*sin(α)
Pb1 Equation 3: can be obtained by:
sin(α)*equation A+cos(α)*equation B
Pb1 Equation 4: can be obtained by:
sin(α)*equation B-cos(α)*equation A
Determine arc radii r1 and r2:
r2=1
ζ=0
r1*sin(ξ)=J+JL1
r1*cos(ξ)=1
JL0 boundary condition:
JL0=J+sin(β)
JL0=J+r1*sin(α-ξ)
The result is eq. A: sin(β)=sin(α)-(J+JL1)*cos(α)
Mc0 boundary condition:
Mc0=1-cos(β)
Mc0=r1*cos(α-ξ)-1
The result is eq. B: 2-cos(β)=cos(α)+(J+JL1)*sin(α)
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