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I have a problem getting the state-space representation of a buck converter in DCM.

Buck-Converter

There are two switch cycles between one period.

If the switch is closed the circuit looks like this:

Switch-Closed

After applying KVL you get for the state variables:

\begin{align} \dfrac{\mathrm di_L}{\mathrm dt} &= \dfrac{1}{L}\cdot(U_{in}-u_C)\\ \dfrac{\mathrm du_C}{\mathrm dt} &= \dfrac{1}{C}\cdot(i_{L}-i_{R}) \end{align}

The same procedure can be followed when the switch is open:

Switch-Open

So the state variables are:

\begin{align} \dfrac{\mathrm di_L}{\mathrm dt} &= -\dfrac{u_C}{L}\\ \dfrac{\mathrm du_C}{\mathrm dt} &= \dfrac{1}{C}\cdot(i_{L}-i_{R}) \end{align}

So, there is a procedure for extracting the state variables.

I would like to know how to get:

\begin{align} \dfrac{di_L}{dt}=~...\\ \dfrac{du_C}{dt}=~... \end{align}

when the buck converter is working in DCM.

In State-space average Modeling of DC-DC Converters with parasitic in Discontinuous Conduction Mode (DCM) on page 20-21 the authors describe how they got to a solution, but I don't understand their method (or the way they describe it).

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    \$\begingroup\$ I don't know much about state space models, but your equation for \$\frac{di_L}{dt}\$ with the switch open is only valid for the first part of the time the switch is open. Once \$i_L\$ drops to zero (because this is DCM), you need a another equation. \$\endgroup\$
    – The Photon
    Commented Oct 28, 2019 at 18:02
  • \$\begingroup\$ yes sure, the state-space model is only valid for CCM \$\endgroup\$
    – rst_tk
    Commented Oct 28, 2019 at 18:04
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    \$\begingroup\$ Hi, I am not sure if this for homework but I would not recommend to use the state-space averaging (SSA) technique to model a CCM or DCM converter whether this is a buck or buck-boost. The SSA is quite complex and requires matrix manipulations with the need to build a small-signal model at the end. The PWM switch model is the modern way to go. Associated with the fast analytical circuit techniques, it is unbeatable in speed and ease of analysis. Furthermore, SSA fails to predict that the DCM buck is still a heavily damped second-order converter what the PWM switch did. \$\endgroup\$ Commented Oct 28, 2019 at 19:21
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    \$\begingroup\$ You can have a look here to learn about the PWM switch model: cbasso.pagesperso-orange.fr/Downloads/PPTs/… \$\endgroup\$ Commented Oct 28, 2019 at 19:22
  • \$\begingroup\$ Thanks, but this is no homework. Im working for my thesis to develop a Model of an Cascaded-Buck-Boost for Control. I wanted to implement an state-regulator. I have done a LOT of Matrix Manipulations, and everything is OK! But i want to add the DCM-Mode. But i will have a look to the doc, thank you! \$\endgroup\$
    – rst_tk
    Commented Oct 28, 2019 at 19:27

1 Answer 1

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There are three states each cycle:

  1. Switch closed, your first equation applies (less the transistor drop).
  2. Switch open, diode conducting; your second equation applies (less a diode drop, but I assume from your use of a IGBT that you consider a diode drop to be negligible).
  3. Switch open, diode not conducting; in theory the coil has neither current nor voltage (in practice it'll be ringing like a bell), and the cap voltage is only affected by the output current

Your challenge is that the time that the converter stays in the switch open, diode conducting state depends on the coil current when the switch opens -- the switch-over happens when the circuit wants it to, not from any external control.

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  • \$\begingroup\$ Ok, so the third state is the reason of DCM. But how can i get to the equations mentioned in the Literature i provided? \$\endgroup\$
    – rst_tk
    Commented Oct 28, 2019 at 18:50
  • \$\begingroup\$ That is somebody's Bachelor's thesis. I wouldn't trust it very far at all. \$\endgroup\$
    – TimWescott
    Commented Oct 28, 2019 at 19:03
  • \$\begingroup\$ yes, youre right. But there should be a mathematical decription for duc/dt and diL/dt if the converter runs in DCM? \$\endgroup\$
    – rst_tk
    Commented Oct 28, 2019 at 19:05
  • \$\begingroup\$ What is \$di/dt\$ for an open-circuit inductor? What is \$dv/dt\$ for a capacitor connected to a resistor? \$\endgroup\$
    – TimWescott
    Commented Oct 28, 2019 at 19:07
  • \$\begingroup\$ di/dt = 0 , dv/dt = -1/RC ? \$\endgroup\$
    – rst_tk
    Commented Oct 28, 2019 at 19:09

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