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Homework Help: DCM analysis of Cuk converter

  1. Feb 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Assuming that a CUK converter is operating in DCM mode (see attached image), find the solution for M(D,k). D is the duty ratio of the switch, and k...

    2. Relevant equations

    [tex]\frac{V}{V_g} = M(D,k)[/tex]

    [tex]k = \frac{4L_1L_2}{(L_1+L_2)(2RT_s)}[/tex]

    3. The attempt at a solution

    The sad part is, I already have the solution guide with a semi-decent walk-through, but one part of it is really confusing me. My confusion may come from doing the problem incorrectly, so I'll go through my entire work just incase. All of our analysis has used small capacitor ripple approximation and circuit averaging.

    The first subcircuit (From time 0 <= t <= D1*Ts) has the switch closed and the diode opened. Describing the inductor voltages and capacitor currents:

    [tex]V_{L1} = V_g[/tex]
    [tex]I_{C1} = -\frac{V}{R}[/tex]
    [tex]V_{L2} = -V_{C1} - V[/tex]
    [tex]I_{C2} = I_{L2} - \frac{V}{R}[/tex]

    The second subcircuit (From time D1*Ts < t <= (D1+D2)*Ts) has the switch opened and the diode closed (saturated). Describing the inductor voltages and capacitor currents:

    [tex]V_{L1} = V_g - V_{C1}[/tex]
    [tex]I_{C1} = -I_{L1}[/tex]
    [tex]V_{L2} = -V[/tex]
    [tex]I_{C2} = I_{L2} - \frac{V}{R}[/tex]

    The third subcircuit (From time (D1+D2)*Ts < t <= Ts) has both the switch opened and the diode opened. The current through the diode has tried to go negative due to the inductor current ripples and low DC offset, thus it turned off.

    [tex]V_{L1} = 0[/tex] Is this correct??
    [tex]I_{C1} = -I_{L2} = I_{L1}[/tex]
    [tex]V_{L2} = 0[/tex] Is this correct??
    [tex]I_{C2} = I_{L2} - \frac{V}{R}[/tex]

    Performing circuit averaging on the inductor voltages and equating to zero (voltage-balance across switching period is a fundamental requirement for these converters):

    [tex]<V_{L1}> = 0 = \frac{1}{T_s}\int{V_{L1}}dt = (D1+D2)V_g - D2V_{C1} [/tex]
    [tex]<V_{L2}> = 0 = \frac{1}{T_s}\int{V_{L2}}dt = (D1+D2)(-V) - D1V_{C1} [/tex]

    Doing algebra shows:

    [tex]M(D,D2) = \frac{V}{V_g} = -\frac{D}{D2}[/tex]

    But I still have an unknown: D2, so I need to also use the capacitor charge balance equations (another fundamental requirement). To do this, I need to determine the values of the inductor currents in terms of quantities I know. I can use geometry instead of calculus due to the small capacitor ripple approximation because, as the equations show,

    1) VL1 is a DC line from t=0 to D1*Ts, the current ramps up until D1*Ts
    2) then a negative DC line from D1*Ts to (D1+D2)*Ts, so current ramps back down until it reaches 0 at (D1+D2)*Ts.
    3) a 0 value from (D1+D2)*Ts to Ts. Current remains at zero.

    IL2 follows a similar behavior, but with reversed polarities.

    So I can easily perform the averaging by geometry (triangles), I find the current peaks:

    [tex]I_{L1,peak} = D_1T_sV_g\frac{1}{L_1}[/tex]
    [tex]I_{L2,peak} = D_2T_s(-V)\frac{1}{L_2}[/tex]

    This where I would write an equation for the average, set it to zero, and start trying to eliminate D2. BUT, in the solution, it is shown that there is a DC offset in the capacitor current (see attached image).

    I don't understand where the DC offset of I3 came from, or why I should have known that.

    Could someone please explain this to me.

    Attached Files:

    • Cuk.png
      File size:
      12.6 KB
    • Sol.png
      File size:
      27 KB
  2. jcsd
  3. Feb 9, 2014 #2
    I think I see my mistake now. In my subcircuit equations, I simply put things in terms of inductor currents: IL1 & IL2.

    Each of these actually need to be broken up in an AC and a DC component. I can calculate the AC component by integrating the voltage waveforms of the inductors, but the DC components just need to get carried through the simultaneous equations as an unknown variable. It turns out that both inductors have the same DC component, because in subcircuit 3 they are the exact same current. This DC component is what he labeled as I3.

    I believe I can now solve some simultaneous equations to get all unknown values.
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