Ćuk converter

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For Ide-Ural festival see Çük

The Ćuk converter (sometimes spelt in various ways as Cuk, Čuk or even Cúk) is a type of DC-DC converter that has an output voltage magnitude that is either greater than or less than the input voltage magnitude, with an opposite polarity. It uses a capacitor as its main energy-storage component, unlike most other types of converter which use an inductor. It is named after Slobodan Ćuk of the California Institute of Technology, who first presented the design in the paper referred to below.

Ćuk is pronounced as Chook. Spellings as Cuk, Čuk or Cúk are common but incorrect: C, Č and Ć are three different letters in Serbian, whereas Ú is simply not used at all.

1 Operating Principle
- 1.1 Continuous mode
- 1.2 Discontinuous mode
  - 1.2.1 Current
  - 1.2.2 Voltage
2 Related structures
- 2.1 Inductor coupling
- 2.2 Sepic
3 References

[edit] Operating Principle

Fig 1: Schematic of a Ćuk converter.

Fig 2: The two operating states of a Ćuk converter.

Fig 3: The two operating states of a Ćuk converter. In this figure, the diode and the switch are either replaced by a short circuit when they are on or by an open circuit when they are off. It can be seen that when in the Off state, the capacitor C is being charged by the input source through the inductor L₁. When in the On state, the capacitor C transfers the energy to the output capacitor through the inductance L₂.

A Ćuk converter comprises two inductors, two capacitors, a switch (usually a transistor), and a diode. Its schematic can be seen in figure 1. It is an inverting converter, so the output voltage is negative with respect to the input voltage.

The capacitor C is used to transfer energy and is connected alternately to the input and to the output of the converter via the commutation of the transistor and the diode (see figures 2 and 3).

The two inductors L₁ and L₂ are used to convert respectively the input voltage source (V_i) and the output voltage source (C_o) into current sources. Indeed, at a short time scale an inductor can be considered as a current source as it maintains a constant current. This conversion is necessary because when a capacitor is directly connected to a voltage source, there is nothing to limit the inrush current that occurs. This current results in high losses. Conversely, connecting a current source to a capacitor is fine as one element limits the current (the current source) and the other the voltage (the capacitor).

As with other converters (Buck converter, Boost converter, Buck-boost converter) the Ćuk converter can either operate in continuous or discontinuous current mode. However, unlike these converters, it can also operate in discontinuous voltage mode (i.e the voltage across the capacitor drops to zero during the commutation cycle).

[edit] Continuous mode

In steady state, the energy stored in the inductors has to remain the same at the beginning and at the end of a commutation cycle. The energy in an inductor is given by:

$E=\frac{1}{2}L\cdot I^2$

This implies that the current through the inductors has to be the same at the beginning and the end of the commutation cycle. As the evolution of the current through an inductor is related to the voltage across it:

$V_L=L\frac{dI}{dt}$

it can be seen that the average value of the inductor voltages over a commutation period have to be zero to satisfy the steady-state requirements.

If we consider that the capacitors C and C_o are large enough for the voltage ripple across them to be negligible, the inductor voltages become:

in the off-state, inductor L₁ is connected in series with V_i and C (see figure 2). Therefore $V L 1 = V i - V C$ . As the diode D is forward biased (we consider zero voltage drop), L₂ is directly connected to the output capacitor. Therefore $V L 2 = V o$

in the on-state, inductor L₁ is directly connected to the input source. Therefore $V L 1 = V i$ . Inductor L₂ is connected in series with C and the output capacitor, so $V L 2 = V o - V C$

The converter operates in on-state from t=0 to t=D.T (D is the duty cycle), and in off state from D.T to T (that is, during a period equal to (1-D).T). The average values of V_L1 and V_L2 are therefore:

$\bar V_{L1}=D \cdot V_i +\left(1-D\right)\cdot\left(V_i-V_C\right) =\left(V_i+(1-D)\cdot V_C\right)$

$\bar V_{L2}=D\left(V_o-V_C\right) + \left(1-D\right)\cdot V_o=\left(V_o - D\cdot V_C\right)$

As both average voltage have to be zero to satisfy the steady-state conditions we can write, using the last equation:

$V_C=\frac{V_o}{D}$

So the average voltage across L₁ becomes:

$\bar V_{L1}=\left(V_i+(1-D)\cdot \frac{V_o}{D}\right)=0$

Which can be written as:

$\frac{V_o}{V_i}=-\frac{D}{1-D}$

It can be seen that this relation is the same as that obtained for the Buck-boost converter.

[edit] Discontinuous mode

[edit] Current

[edit] Voltage

[edit] Related structures

[edit] Inductor coupling

[edit] Sepic

Main article: SEPIC converter

Single-Ended Primary Inductance Converter (SEPIC) is able to step-up or step-down the voltage.

[edit] References

R. D. Middlebrook and S. M. Ćuk, A General Unified Approach to Modelling Switching Converter Power Stages, Proc. IEEE Power Electronics Specialists Conference, 1976 Record