There are variations on the basic Ćuk converter. For example, the coils may share a single magnetic core, which drops the output ripple, and adds efficiency. Because the power transfer flows continuously via the capacitor, this type of switcher has minimized
EMI radiation. The Ćuk converter allows energy to flow bidirectionally by using a diode and a switch.
Operating principle 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. 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 (
Vs) and the output voltage (
Vo) 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 operate in either 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).
Continuous mode In steady state, the energy stored in each inductor 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}LI^2 . This implies that the current through each inductor 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 each inductor's voltage over a commutation period has to be zero to satisfy the steady-state requirements. (Another way to see this is to recognize that the average voltage across any inductor must be zero lest its current rise without limit.) 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
Vs and
C1 (see figure 2). Therefore 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 V_{L2}=V_o • in the
on-state, inductor
L1 is directly connected to the input source. Therefore V_{L1}=V_s. Inductor
L2 is connected in series with
C1 and the output capacitor, so V_{L2}=V_o+V_{C1} The converter operates in
on state from t=0 to t=DT (
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
VL1 and
VL2 are therefore: \bar V_{L1}=D \cdot V_s +\left(1-D\right)\cdot\left(V_s-V_{C1}\right) =V_s-(1-D)\cdot V_{C1} \bar V_{L2}=D\left(V_o+V_{C1}\right) + \left(1-D\right)\cdot V_o=V_o + D\cdot V_{C1} As both average voltage have to be zero to satisfy the steady-state conditions, using the last equation we can write: V_{C1}=-\frac{V_o}{D} So the average voltage across L1 becomes: \bar V_{L1}=V_s+(1-D)\cdot \frac{V_o}{D}=0 Which can be written as: \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.
Discontinuous mode 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: L_1min=\frac{(1-D)^2R}{2Df_s} Where f_s is the switching frequency. ==Isolated Ćuk converter==