Ćuk converter

A non-isolated 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 main advantage of this converter is the continuous currents at the input and output of the converter.  The main disadvantage is the high current stress on the switch.[4]

Fig. 1. Cuk converter circuit diagram.

The capacitor C₁ is used to transfer energy. It 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_s) and the output voltage source (V_o) into current sources. 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 if the capacitor were connected directly to the voltage source, the current would be limited only by the parasitic resistance, resulting in high energy loss. Charging a capacitor with a current source (the inductor) prevents resistive current limiting and its associated energy loss.

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 (the voltage across the capacitor drops to zero during the commutation cycle).

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

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:

${\displaystyle E={\frac {1}{2}}LI^{2}}$

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:

${\displaystyle V_{L}=L{\frac {dI}{dt}}}$

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 C₁ and C₂ 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_s and C₁ (see figure 2). Therefore ${\textstyle V_{L1}=V_{s}-V_{C1}}$. As the diode D is forward biased (we consider zero voltage drop), L₂ is directly connected to the output capacitor. Therefore ${\displaystyle V_{L2}=V_{o}}$
• in the on-state, inductor L₁ is directly connected to the input source. Therefore ${\textstyle V_{L1}=V_{s}}$. Inductor L₂ is connected in series with C and the output capacitor, so ${\displaystyle V_{L2}=V_{o}+V_{C}}$

Fig. 3.: The two operating states of a non-isolated 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.

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

${\displaystyle {\bar {V}}_{L1}=D\cdot V_{s}+\left(1-D\right)\cdot \left(V_{s}-V_{C}\right)=\left(V_{s}-(1-D)\cdot V_{C}\right)}$

${\displaystyle {\bar {V}}_{L2}=D\left(V_{o}+V_{C}\right)+\left(1-D\right)\cdot V_{o}=\left(V_{o}+D\cdot V_{C}\right)}$

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

${\displaystyle V_{C}=-{\frac {V_{o}}{D}}}$

So the average voltage across L₁ becomes:

${\displaystyle {\bar {V}}_{L1}=\left(V_{s}+(1-D)\cdot {\frac {V_{o}}{D}}\right)=0}$

Which can be written as:

${\displaystyle {\frac {V_{o}}{V_{s}}}=-{\frac {D}{1-D}}}$

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

Like all DC/DC converters, Ćuk converters rely on the ability of the inductors in the circuit to provide continuous current, in much the same way a capacitor in a rectifier filter provides continuous voltage. If this inductor is too small or below the "critical inductance", then the inductor current slope will be discontinuous where the current goes to zero. This state of operation is usually not studied in much depth as it is generally not used beyond a demonstrating of why the minimum inductance is crucial, although it may occur when maintaining a standby voltage at a much lower current than the converter was designed for.

The minimum inductance is given by:

${\displaystyle L_{1}min={\frac {(1-D)^{2}R}{2Df_{s}}}}$

Where ${\displaystyle f_{s}}$ is the switching frequency.

Instead of using two discrete inductor components, many designers implement a coupled inductor Ćuk converter, using a single magnetic component that includes both inductors on the same core. The transformer action between the inductors inside that component gives a coupled inductor Ćuk converter with lower output ripple than a Ćuk converter using two independent discrete inductor components.[6]

A zeta converter is a non-isolated, non-inverting, buck-boost power supply topology.

A SEPIC converter is able to step-up or step-down the voltage.