Ćuk converter

A converter Ćuk is a Alimentation with cutting which converts a continuous tension into another continuous tension of lower or greater value but of opposite polarity. Contrary to the other types of converter, which use an inductance, a Ćuk converter uses a condenser to store energy. The Ćuk converter owes its name with its inventor, Slobodan Ćuk of the California Institute off Technology, which was the first with décire this topology in an article.

Orthography: Ćuk is sometimes spelled, in an incorrect way, Čuk or Cúk . Ć and Č is two different letters in Serbe.

Principle of operation

A Ćuk converter consists of two inductances, of two condensing, a switch (generally a Transistor) and of a Diode. The basic diagram of a Ćuk converter is represented figure 1.

The condenser C is used to transfer energy between the source from tension of entry (V_i) and that of exit (V_o). For that, it is connected alternatively to the entry or the outlet side of the converter thanks to the switch S and to the diode D (see figures 2 and 3).

Two inductances L₁ and L₂ are used respectively to convert the source of tension of entry and the source of output voltage (C_o) into power sources. Indeed a winds can be considered, over one short period, as a power source as it maintains this one constant. These conversions are necessary in order to limit the current when one connects the condenser C to a source of tension (V_o or V_i).

Commes the other converters (Boost, Buck, Buck-Boost or Flyback), the Ćuk converter can function with a continuous or discontinuous conduction while running. However, contrary to the other converters, it can also function with a discontinuous conduction in tension (the terminal voltage of the condenser is cancelled during part of the cycle of commutation)

Continuous conduction

If it is considered that the converter reached its permanent mode, the quantity of energy stored in each one of its components is the same one at the beginning and the end of a cycle of operation. In particular, the energy stored in inductance is given par :

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

Consequently, the current crossing inductance is the same one at the beginning and the end of each cycle of commutation. Evolution of the current in an inductance being related to the tension with its bornes :

$V\_L=L\backslash frac\{dI\}\{dt\}$

This relation enables us to see that the average tension at the boundaries of an inductance must be null in order to satisfy the conditions of permanent mode.

If one considers a null voltage drop at the boundaries of the diode and the condensers C and C_o sufficiently large to keep their constant tensions, the evolution of the terminal voltage of inductances devient :

• During the state passing, L₁ inductance is connected directly to the source of tension of entry. Consequently, $V\_\; \{L1\}\; =V\_i$. L₂ inductance, as for it, is connected in series with the condensers C and C_o. Consequently, $V\_\; \{L2\}\; =V\_o-V\_C$.

• During the blocked state, L₁ inductance is connected in series with _i and C (see figure 2). Consequently, $V\_\; \{L1\}\; =V\_i-V\_C$. The diode D being busy, L₂ is directly connected to the condenser of exit. Consequently, $V\_\; \{L2\}\; =V\_o$.

The converter being in a busy state of $t=0$ with $t=\; \backslash \; alpha\; T$ ($\backslash \; alpha$ being the cyclic Report/ratio) then in a blocked state of $t=\; \backslash \; T$ alpha with $t=T$. Median values of V_L1 and V_L2 écrivent :

$\backslash \; bar\; V\_\; \{L1\}\; =\; \backslash \; alpha\; \backslash \; cdot\; V\_i\; +\; \backslash \; left\; (1\; \backslash \; alpha\; \backslash \; right)\; \backslash \; cdot\; \backslash \; left\; (V\_i-V\_C\; \backslash \; right)\; =\; \backslash \; left\; (V\_i+\; (1\; \backslash \; alpha)\; \backslash \; cdot\; V\_C\; \backslash \; right)$

$\backslash \; bar\; V\_\; \{L2\}\; =\; \backslash \; alpha\; \backslash \; left\; (V\_o-V\_C\; \backslash \; right)\; +\; \backslash \; left\; (1\; \backslash \; alpha\; \backslash \; right)\; \backslash \; cdot\; V\_o=\; \backslash \; left\; (V\_o\; -\; \backslash \; alpha\; \backslash \; cdot\; V\_C\; \backslash \; right)$

As the two tensions are null in order to satisfy the conditions of permanent mode, one can deduce some by using the second équation :

$V\_C=\; \backslash \; frac\; \{V\_o\}\; \{\backslash \; alpha\}$

By replacing V_C in the equation of $\backslash \; bar\; V\_\; \{L1\}$ by its expression, one obtient :

$\backslash \; bar\; V\_\; \{L1\}\; =\; \backslash \; left\; (V\_i+\; (1\; \backslash \; alpha)\; \backslash \; cdot\; \backslash \; frac\; \{V\_o\}\; \{\backslash \; alpha\}\; \backslash \; right)\; =0$

What can rewrite   in the following way;:

$\backslash \; frac\; \{V\_o\}\; \{V\_i\}\; =\; \backslash \; frac\; \{\backslash \; alpha\}\; \{1\; \backslash \; alpha\}$

One realizes that this expression is the same one as that obtained for the Convertisseur Buck-Boost.

Discontinuous conduction

In tension

While running

References

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