Blog for students taking ECEN5817 Resonant and Soft Switching Techniques in Power Electronics, ECEE Department, University of Colorado at Boulder, Spring 2012
Saturday, February 27, 2010
Expression for PIV-eliminating IF
Do you guys get the radius as JF? Then PIV=1+JF= 1+IF/(nVg) sqrt(Ll/Cd). I am not having any luck in eliminating L and somehow expr.essing it in terms of Qr
But we could also say that at t=t3, il which is complementary to id, will instantaneously drop from Ilp to IF which corresponds to a straight line on the jl axis from Jlp to JF so then the circle will start from this point with center at -1,JF and radius of JF going clockwise. If we start with a radius of Jlp, then going clockwise because md and jl are both goin in -ve direction, we cross md axis twice and jl axis once meaning current is going to zero twice and voltage is going to zero once, which I don't think is happening.
Inductor current cannot instantaneously change, by definition; il(0+)=il(0-). So, at t=t3, the current il is greater than IF. At theta=0, the trajectory starts at m=0 (vd=0) and j=J0, which is greater than JF.
Also, I get that J0 starts with a positive slope and m(0) starts with a negative slope, so the trajectory moves CCW until it hits the m axis, at which point D1 turns off (we're ignoring losses). This is ok because the inductor current is 0 at that point.
Sean, Thanks for the clarification. But there is still one doubt, as per your explanation PIV should be -1-J0, because you are drawing your circle with a radius of J0. I cannot see how you get the expression for Mmax above.
I have my center at (-1,JF), and the initial point is on the j axis because M0=0, and it is at a point farther from the origin than JF because the current is initially larger than IF, so the radius of the circle is A=sqrt(1+(J0-JF)^2). Max voltage occurs when theta=pi, because this is farthest away from the m=0 axis. So, Mmax=-1-A.
If anyone thinks I am out of my tree, please tell me.
Sean, I might be wrong in what I am saying here but what I think is we start at 0,J0 (where J0 is the peak), then there is a straight line going left till we reach -1 (cirresponding to v/g reacning -nVg), then the circle starts from this point with a radius of J0. The reason I think we cannot start the circle at 0,J0 is because when vd=0, there is no resonant ckt yet since Cd is shorted out so that has to be represented by a straight line since the tank is not formed untill vd rises to -nVg.
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9 comments:
I get the radius as J0>JF, so PIV=Mmax*nVg, and Mmax=-1+sqrt(1+(J0-JF)^2).
But we could also say that at t=t3, il which is complementary to id, will instantaneously drop from Ilp to IF which corresponds to a straight line on the jl axis from Jlp to JF so then the circle will start from this point with center at -1,JF and radius of JF going clockwise. If we start with a radius of Jlp, then going clockwise because md and jl are both goin in -ve direction, we cross md axis twice and jl axis once meaning current is going to zero twice and voltage is going to zero once, which I don't think is happening.
Inductor current cannot instantaneously change, by definition; il(0+)=il(0-). So, at t=t3, the current il is greater than IF. At theta=0, the trajectory starts at m=0 (vd=0) and j=J0, which is greater than JF.
Also, I get that J0 starts with a positive slope and m(0) starts with a negative slope, so the trajectory moves CCW until it hits the m axis, at which point D1 turns off (we're ignoring losses). This is ok because the inductor current is 0 at that point.
Sorry, two posts ago I meant to say theta=theta0, not 0.
Sean,
Thanks for the clarification. But there is still one doubt, as per your explanation PIV should be -1-J0, because you are drawing your circle with a radius of J0. I cannot see how you get the expression for Mmax above.
I have my center at (-1,JF), and the initial point is on the j axis because M0=0, and it is at a point farther from the origin than JF because the current is initially larger than IF, so the radius of the circle is A=sqrt(1+(J0-JF)^2). Max voltage occurs when theta=pi, because this is farthest away from the m=0 axis. So, Mmax=-1-A.
If anyone thinks I am out of my tree, please tell me.
Sean,
I might be wrong in what I am saying here but what I think is we start at 0,J0 (where J0 is the peak), then there is a straight line going left till we reach -1 (cirresponding to v/g reacning -nVg), then the circle starts from this point with a radius of J0. The reason I think we cannot start the circle at 0,J0 is because when vd=0, there is no resonant ckt yet since Cd is shorted out so that has to be represented by a straight line since the tank is not formed untill vd rises to -nVg.
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