Cuk converter design

The inductor L 1 gets charged and acts as a storage element with polarity from positive to negative. Technically, there will be a small amount of stored voltage in C 1. There is no flow of current to load. Case2: When the switch is turned OFF. In addition to the stored energy dissipating from L 1 , the current will flow from the source to the inductor L 1.

The reason for the flow of current in the inductor is because the current flowing from the inductor L1 is not completely DC. The flow of the current path when switched is turned OFF is current will flow from the source to inductor L 1 to capacitor C 1 and from the diode and back to the source. Capacitor C 1 will also store the voltage in it in this case. Voltage is supplied to the load when the switch is in the OFF state.

Case3: When the switch is turned ON. But now the charged capacitor C 1 which holds the voltage dissipates its energy to the load. The voltage stored in capacitor C 1 acts as a voltage source where current flows from the capacitor C 1 to C 2 and Load. It also charges the inductor L 2 in this process where diode d is reverse biased. Case4: When the switch is turned OFF. The diode is forward biased where capacitor C 2 and L 2 dissipate the energy where the output voltage is reversed.

Since we studied the working of a Cuk converter considering the ideal conditions after these 4 conditions, the mechanism of the Cuk converter will be as follows when the switch is turned ON and OFF. Referring to Fig 2,3,4 and 5 we can conclude. When the switch is turned ON, inductor L 1 gets charged and stored energy in the capacitor C 1 supplies the energy to load. Inductor L 2 supplies energy or energizes the capacitor C 1.

C 1 acts as the voltage source for the load. When the switch is turned OFF, inductor L 1 discharges the energy and supplies power to the capacitor C 2.

And L 2 supplies the power to the load which was energized when the switch was turned ON. Related Posts:. Your email address will not be published. It requires two inductors these can be coupled or uncoupled and are typically matched in value and a coupling capacitor C5 between the input and output. The coupling or blocking capacitor receives energy from the input side of the circuit and transfers it to the output side of the circuit.

With steady-state conditions i. The LT is a multitpology switching regulator with a 3. Note that the two circuits look very similar; with the exception being the Cuk's second inductor has been replaced by a Schottky diode. The lowside switch is also used in boost, SEPIC and flyback topologies, so these devices are quite versatile. The switch node always has a positive voltage applied to it.

In a Cuk design, the feedback pin may or may not see a negative voltage some devices do not allow negative voltages anywhere on the IC, some devices have a dual mode feedback pin that accepts both positive and negative voltages. Though similar in appearance, the operation of the two circuits is quite different. For the Cuk, the simplified duty cycle assuming lossless diodes and switches is given by:.

The current flowing from the input power source is continuous in other words, current flows from the input when the power switch is closed or open. When the switch is closed, both inductors have an increasing current flow the current is ramping up, but since the current in L2 is negative the two currents ramps move in opposite directions. The current in both inductors decreases when the switch opens.

Continuous current flow combined with the LC filters results in a smoother input and output current, which in turn gives low output voltage ripple noise. The inverting charge pump is closely related to a step-up converter because it combines an inductor-based step-up regulator with an inverting charge pump. In addition, high current MOSFETs tend to come in much larger packages, so meeting the ideals of low ON resistance and low Qg might violate a space requirement spec, so the selection process has to start over.

Engineering, as ever, is a compromise. Failing that, download all the results to a spreadsheet and sort from there. So it can be argued that if the input voltage is high and the output voltage is low i.

RJK Datasheet. When the inductor voltage flies negative, it is the body diode that conducts first before the gate drive to the MOSFET activates the Drain-Source channel.

FIG 7. This is indicative of the body diode starting to conduct and indeed the negative voltage is approximately If optimum efficiency is desired, it is wise to place a Schottky diode across the bottom MOSFET, so the Schottky diode can conduct the inductor flyback voltage and not the body diode. The Schottky diode will conduct the peak current flowing through the inductor, but this current will only flow for a short period of time until the bottom MOSFET switches on.

Therefore, the current rating of the diode can be a lot less than peak inductor current. MBRS Datasheet. Output Capacitor Choice. During the charging cycle of the inductor, the output capacitor has no current flowing into it, similar to a boost converter. Therefore the load current comes purely from the output capacitor. When the inductor discharges, the output capacitor it is subjected to an inrush current.

If the capacitor has any ESR effective series resistance this will develop a ripple voltage on the output capacitor. Therefore the output ripple is made up of 2 components: the ripple caused by the output capacitor discharging when the inductor is being charged and the ripple caused by the inrush current from the inductor into the ESR of the output capacitor.

The ripple caused by the discharge of the output capacitor while the inductor is charging is dictated by. If we require a discharge ripple of 0. Note that when the inductor is charging, there is zero current flowing in the output capacitor. When the top MOSFET switches off, the bottom MOSFET current and hence the capacitor charge current jumps from 0A to the peak inductor current, so it is the peak inductor current, not the ripple current amplitude that determines this component of the output ripple compare this with the ripple in a buck converter that is determined by the ripple current amplitude, not the peak inductor current.

In our example the peak current is 1. At this point is it worth trading off ESR ripple for discharge ripple and repeating the above two calculations on several combinations of output capacitor to see the effect on the ripple.

We see that it is relatively easy to meet the spec of discharge ripple since our output capacitor is only 29uF. Repeating the above calculations several times, it would appear that three 22uF capacitors in parallel will meet our overall ripple spec. Other Points. The feedback resistor values can be calculated using this spreadsheet:. Feedback Resistor Calculator. The final circuit is shown in FIG 8.

FIG 8. The Cuk Converter. This can lead to interference being generated in the circuitry supplying the input current. This is where the Cuk Converter provides a more suitable alternative. The Cuk converter has an inductor on the input and the output, so both input and output currents have no sharp changes in current.

FIG 9 shows a Cuk Converter. FIG 9. It can be seen that the circuit about has the input configuration of a boost converter and the output configuration of a buck converter. However, unlike a boost or buck converter, the controller in a Cuk converter needs to be able to respond to a negative feedback voltage.

The current in the inductor ramps according to the equation. Thus with a fixed voltage across the inductor and a fixed inductor value, the change in current with time is constant. Thus in FIG 9 the change in current with time can be represented by.

Or , Amps per second. It does this by creating a voltage across it where the right hand side tries to fly positive to push current out of the right hand end and the left hand side flies negative. Since the left hand side of the inductor is clamped to the input voltage, the right hand side of the inductor flies positive to a voltage above Vin in order to maintain current flow. The energy from the inductor flows into capacitor C5 charging it with a positive voltage which is higher than Vin.

We will work out later exactly what voltage C5 charges to, but for the moment it is sufficient to assume it charges to a voltage higher than Vin. We will call this voltage Vcap. Since the voltage Vcap is higher than Vin, the voltage across the inductor now has the opposite polarity to before. Since the voltage across a capacitor cannot change instantaneously, an equal negative going voltage appears on the anode of diode D1 so this node transitions from 0V to —Vcap.

We now have a negative amplitude square wave voltage at the right hand node of C5 being applied to an LC filter L2 and C1. The LC filter averages out this square wave to produce a flat DC voltage whose amplitude is somewhere between 0V and —Vcap. This amplitude is dictated by the duty cycle of the square wave. The inductor charge and discharge currents are equal when the circuit is in steady state.

If the Duty Cycle DC can be represented by. To determine the duty cycle in terms of the input and output voltages, consider FIG FIG Here we can see the Drain voltage going from 0V to Vcap as yet uncalculated and the ac coupled drain voltage on the anode of the diode.

The capacitor has removed the dc offset and the diode has clamped the positive excursions to roughly 0V. Now, when the circuit is regulating there will be a flat negative dc voltage on the output. Thus, when V diode is at 0V there will be a positive voltage from V diode to V out and the inductor current in L2 will ramp in a positive direction.

When V diode is negative there will be a negative voltage from V diode to V out so the inductor current will ramp to a more negative value. Equating the values of di gives. From before we know that. Vout is the magnitude of the output voltage. This is because in the above derivation, we have ignored the slope of di — it is positive in L1 when negative in L2, so cannot strictly equate the 2 statements for DC without considering this.

Knowing that. We can work out the Duty Cycle in terms of Vout and Vin. Again, Vout is the magnitude of the output voltage. The duty cycle is set by the input and output voltages only.

The above is true as long as the current in the inductor does not fall to zero. Cuk Converter Design Procedure. Our design brief is to design a Cuk Converter with an input voltage of 10V and an output voltage of 5V that can support a load of 1A.

The switching frequency should be kHz. With a 10V input and a 5V output, we can calculate the duty cycle DC as being. With a switching frequency of kHz, this represents a period of 3. The minimum ON time of the LT is ns, so this is well within spec. This is a good trade off between small inductor size and low switching losses. The inductor on the output of a Cuk Converter is configured identically to that of a buck converter.

With the buck converter, the average inductor current is equal to the output current. On the input, the Cuk Converter has an inductor configured identically to that of a boost converter and the average inductor current in a boost converter is equal to the average input current.

With an output voltage of 5V and a load of 1A, this represents an output power of 5W. With an input voltage of 10V, this represents an average input current of mA. The change in current is therefore mA. To calculate the output inductor value, we go through the same procedure. We know that. The purists would argue that since our output current is different to the input current then keeping both inductor values the same will result in a different ripple percentage in the output inductor, so the output inductor could be sized differently to reflect this, but the resulting change in circuit performance is minimal for most applications.