Project 1

Aim : To design and develop the converters from given table in MATLAB-Simulink.

Software Used : MATLAB 2020a

From the table I have chosen to design the projects 5, 12 and 13 which are SEPIC in CCM, Buck-Boost Converter in CCM and Cuk converter in CCM respectively.

Now I will start designing the chosen converters in their respected modes of operation.

 

SEPIC in CCM :

The single-ended primary-inductor converter (SEPIC) is a type of DC to DC converter that allows the electrical potential (voltage)  at its output to be greater than, less than, or equal to that at its input.

 

• The output of the SEPIC is controlled by the duty cycle of the control switch (S1).
• A SEPIC is essentially a boost converter followed by an inverted buck  boost converter , therefore it is similar to a traditional buck boost converter, but has advantages of having noninverted output (the output has the same voltage polarity as the input), using a series capacitor to couple energy from the input to the output (and thus can respond more gracefully to a short-circuit output), and being capable of true shutdown. 
• When the switch S1 (MOSFET) is turned off enough, the output (V₀) drops to zero volts, following a fairly heavy  transient dump of charge.
• SEPICs are useful in applications in which a battery voltage can be above and below that of the regulator's intended output. For example, a single Lithium ion typically discharges from 4.2 volts to 3 volts; if other components require 3.3 volts, then the SEPIC would be effective.

Working/Operation :

Now we will see how does the SEPIC work in Continuous Conduction Mode (CCM) and also we will design the Simulink model for the mentioned converter :

• The diagram for a basic SEPIC is shown in above. As with other SMPS (specifically DC/DC converters), the SEPIC exchanges energy between the capacitors and inductors in order to convert from one voltage to another.
• The amount of energy exchanged is controlled by switch S1(MOSFET), which is typically a transistor such as a MOSFET.
• MOSFETs offer much higher input impedance and lower voltage drop than BJTs, and do not require biasing resistors as MOSFET switching is controlled by differences in voltage rather than a current, as with BJTs.

SEPIC in CCM:

• SEPIC is said to be in continuous-conduction mode ("continuous mode") if the currents through inductors L1 and L2 never fall to zero during an operating cycle.
• During a SEPIC's steady - state operation, the average voltage across capacitor C1 (V_C1) is equal to the input voltage (V_in). Because capacitor C1 blocks direct current (DC), the average current through it (I_C1) is zero, making inductor L2 the only source of DC load current.
• Therefore, the average current through inductor L2 (I_L2) is the same as the average load current and hence independent of the input voltage.

Hence we can say that,

• Because the average voltage of V_C1 is equal to V_IN, V_L1 = −V_L2. For this reason, the two inductors can be wound on the same core, which begins to resemble a Flyback converter the most basic of the transformer-isolated SMPS topologies.
• Since the voltages are the same in magnitude, their effects on the mutual inductance will be zero, assuming the polarity of the windings is correct. Also, since the voltages are the same in magnitude, the ripple currents from the two inductors will be equal in magnitude.
• The average currents can be summed as follows (average capacitor currents must be zero).

During ON condition - ( when MOSFET is ON)

The circuit looks like this -

• When switch S1 is turned on, current I_L1 increases and the current I_L2 goes more negative.
• The energy to increase the current I_L1 comes from the input source. Since S1 is a short while closed, and the instantaneous voltage V_L1 is approximately V_IN, the voltage V_L2 is approximately −V_C1.
• Therefore, D1 is opened and the capacitor C1 supplies the energy to increase the magnitude of the current in I_L2 and thus increase the energy stored in L2.
• I_L is supplied by C2. The easiest way to visualize this is to consider the bias voltages of the circuit in a d.c. state, then close S1.

During OFF condition - ( when MOSFET is OFF)

The circuit looks like this -

• When switch S1 is turned off, the current I_C1 becomes the same as the current I_L1, since inductors do not allow instantaneous changes in current.
• The current I_L2 will continue in the negative direction, in fact it never reverses direction.
• It can be seen from the diagram that a negative I_L2 will add to the current I_L1 to increase the current delivered to the load. Using KCL , it can be shown that I_D1 = I_C1 - I_L2. It can then be concluded, that while S1 is off, power is delivered to the load from both L2 and L1.
• C1, however is being charged by L1 during this off cycle (as C2 by L1 and L2), and will in turn recharge L2 during the following on cycle.
• Because the potential (voltage) across capacitor C1 may reverse direction every cycle, a non-polarized capacitor should be used. However, a polarized tantalum or electrolytic capacitormay be used in some cases, because the potential (voltage) across capacitor C1 will not change unless the switch is closed long enough for a half cycle of resonance with inductor L2, and by this time the current in inductor L1 could be quite large.
• The capacitor C_IN has no effect on the ideal circuit's analysis, but is required in actual regulator circuits to reduce the effects of parasitic inductance and internal resistance of the power supply.
• The boost/buck capabilities of the SEPIC are possible because of capacitor C1 and inductor L2. Inductor L1 and switch S1 create a standard boost converter , which generates a voltage (V_S1) that is higher than V_IN, whose magnitude is determined by the duty cycle of the switch S1.
• Since the average voltage across C1 is V_IN, the output voltage (V_O) is V_S1 - V_IN. If V_S1 is less than double V_IN, then the output voltage will be less than the input voltage.
• If V_S1 is greater than double V_IN, then the output voltage will be greater than the input voltage.

Now I will implement this model in Matlab - Simulink environment as shown below -

The model looks like this -

 Given,

Input Voltage = Vi = 30 to 60 V,   Vo = 45 V  and P = 1 KW.

I have chosen the input voltage value as, Vi = 55 V

Increasing the switching frequency of a power converter will bring a number of benefits like passive component size can be reduced as the requirements for stored energy decrease. So I will choose switching frequency as ,

Fs = 50 KHz

lets find the required parameters and also substitute these values in the respected blocks.

(i) Duty cycle ratio:

D = Vo/(Vo + Vi)

D = 45/ (45 + 55)

D = 0.45 ≈  0.46

 

(ii) R value:

R = (Vo)2 / P

R = (45*45) / 1000

R = 2.025  Ω  ≈  3 Ω

 

(iii) Inductor Currents (IL1 & IL2) :

IL1 = P / Vi = 1000/55 = 18.18 A

IL1 = 18.18 A

IL2 = P / Vo  = 1000/ 45 = 22.22 A

IL2 = 22.22 A

Now we will find the ripple currents for IL1 & IL2

To achieve a good compromise between inductor and capacitor size a ripple current value of 10% - 30% of maximum inductor current should be chosen.

So, assume 10% ripple and ,

ΔIL1 = (18.18)*(0.1) = 1.818 A

ΔIL2 = (22.22)*(0.1) = 2.222 A

 

(iii) L1 & L2 Values:

L1 = (Vi*D) / (Fs * ΔIL1)

L1 = (55 * 0.46) / (50e3 * 1.818)

L1 = 0.272 mH

 

L2 = (Vi*D) / (Fs * ΔIL2)

L2 = (55*0.46)/ (50e3 * 2.222)

L2 = 0.202 mH

 

(iv) C1 & C2 values:

We know that, according to the circuit -

VC1 = Vi = 55 V

As we have assumed 10% of ripple for better performance,

ΔVC1 = 0.1 * 55 = 5.5 V

ΔVC1 = 5.5 V

Now to find C1,

C1 = (Vo * D) / (R * Fs * ΔVC1)

C1 = (45 * 0.46) / (3 * 50e3 * 5.5)

C1 = 24.54 µF

 

To find output capacitor C2 ,

C2 = Cout = D / (R * Fs * ΔVo/Vo)

Assume ΔVo/Vo = 1%

C2 = 0.46 / (3 * 50e3 * 0.01)

C2 = Cout = 0.3 mF

Now we have calculated all the circuit parameters and put these values in the circuit and observe the output and waveforms.

Vin = 55 V                            IL1 = 18.18 A

L1 = 0.272 mH                      IL2 = 22.22 A

L2 = 0.202 mH                      Fs = 50 kHz

C1 = 24.54 µF                       R = 3 ohms

C2 = Cout = 0.3 mF               D = 0.46

 

Input DC source :

Input voltage of 55 V is put in the block.

 

Pulse Block -

• Here i have substituted values for period and pulse width.
  
• Later pulse width value is altered in order to achieve desired output.

 

L1 & L2 blocks :

       

C1 & C2 blocks :

      

 

Diode block :

MOSFET block :

 

R block :

Since calculated resistance value is 2.025 ohms , but in order to achieve desired output I put 3 ohms.

Now we will run the model for the specified simulation time and we observe the following results.

Input DC waveform :

• The Input voltage is pure DC waveform.
• It is the value of 55 V.

Pulse Waveform:

On Zooming,

• This has duty ratio of 46 % and has a switching frequency of 50kHz.
• This pulse signal drives the MOSFET and MOSFET performs the switching operations.

 

MOSFET Voltage & Current Waveforms :

On zooming , MOSFET voltage would look like this,

• When the MoSFET is turned ON , voltage reaches the maximun voltage and the current starts increasing in the circuit.
  
• The current is nothing but an inductor current.
• The MOSFET voltage reaches maximum upto 100 V as we can see in the above waveform.
• When the MOSFET is tuned OFF, the volatge comes to zero, similarly inductor current would be zero.

On zooming, MOSFET current would look like this,

• It shows that, when voltage is maximum the current will be ,aximum.
  
• But when MOSFET is turned off, current falls to zero in the MOSFET.
• It has the peak value of 30 A as from the waveform.

 

Inductor currents IL1 & IL2:

IL1 Waveform :

• Since, the SEPIC is in the CCM the current does not reach to zero as we can see.
• It fluctuates inly between the specified values.

On Zooming,

• When switch S1 is turned on, current I_L1 increases as shown above.
• Inductor current IL1 is in continuous mode and it has the value nearer to 18 A.
• And since the ripple in the IL1 is about 1.18 A which we can see in the above waveform.
• The difference between the upper and lower peaks is approximately 1.18 A.

IL2 waveform :

• This is how the IL2 looks like.
• Therefore, D1 is opened and the capacitor C1 supplies the energy to increase the magnitude of the current in I_L2 and thus increase the energy stored in L2.
• I_L is supplied by C2. The easiest way to visualize this is to consider the bias voltages of the circuit in a d.c. state, then close S1.

On Zooming,

• Initially, the current was slight high.
• Then after some time it comes to steady state.
• later it starts decreasing and comes to steady state.

 

Diode Voltage & Current waveforms :

• This is how the diode current and voltage waveforms look like.
  
• The diode current is same as current flows in the circuit, whereas voltage is forward biased voltage.

On zooming, the diode current looks like-

• Initially diode current was at low values.
• as the time increases, it reaches peak value ad again comes to lower values.
• Later it goes to steady state same as of inductor currents, which we can see from theworking of the SEPIC as mentioned above.

On zooming diode voltage would look like this -

f

• Since, the diode operates only at forward voltage of 0.7 V we can see at 0.7 V the diode starts conducting.
• During the OFF operation of switch , the voltage reaches to negative voltage which is of maximum (negative peak).

Now we will see the capacitor voltage VC1 & VC2.

VC1 wafeform :

On zooming,

• we have assumed the 10% ripple in the capacitor design.
• So we got 5.5 V of ripple voltage which we can see here.
• As VC1 is nothing but input voltage and it fluctuates between 55 to neraly 60 V because of the assumed ripple.
• It varies between 55 V to 60 V which can be clearly seen above.

 

VC2 waveform (output capacitor):

On zooming,

• This is how the output capacitor volatge looks like.
• The voltage is same as what exactly we wanted to achieve.
• Since we assumed 1% ripple at the output side, so it only fluctuates between 44.5 to 45 V.
  

 According to the given data we should achieve an output of 45 V, so this is what we excatly got at the output side. And also display block shows the same value.

And also the output voltage waveform looks like this-

• Here we can clearly see that from 44.64 to 45.05 , the voltage is varying which is pretty fine.
• Because of assumed ripple, we did not get much variations in the output side. And also desired output is achieved.
• An also the SEPIC converter operation is achieved.

 

 

Buck - Boost Converter in CCM :

The buck–boost converter 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.

• In the inverting topology, the output voltage is of the opposite polarity than the input.
• This is a SMPS with a similar circuit topology to the boost converter and the buck converter.
• The output voltage is adjustable based on the duty cycle of the switching transistor.
• One possible drawback of this converter is that the switch does not have a terminal at ground; this complicates the driving circuitry.
• However, this drawback is of no consequence if the power supply is isolated from the load circuit (if, for example, the supply is a battery) because the supply and diode polarity can simply be reversed. When they can be reversed, the switch can be on either the ground side or the supply side.

 

Working / Operation of Buck-Boost Converter :

• The working operation of the DC to DC converter is the inductor in the input resistance has the unexpected variation in the input current.
• If the switch is ON then the inductor feed the energy from the input and it stores the energy of magnetic energy.
• If the switch is closed it discharges the energy.
• The output circuit of the capacitor is assumed as high sufficient than the time constant of an RC circuit is high on the output stage. The huge time constant is compared with the switching period and make sure that the steady state is a constant output voltage Vo(t) = Vo(constant) and present at the load terminal.

There are two different types of working principles in the buck boost converter.

• Buck converter.
• Boost converter.

Buck Converter Working :

The following diagram shows the working operation of the buck converter. In the buck converter first transistor is turned ON and second transistor is switched OFF due to high square wave frequency. If the gate terminal of the first transistor is more than the current pass through the magnetic field, charging C, and it supplies the load. The D1 is the schoktty diode and it is turned OFF due to the positive voltage to the cathode.

 

 

• The inductor L is the initial source of current. If the first transistor is OFF by using the control unit then the current flow in the buck operation.
• The magnetic field of the inductor is collapsed and the back e.m.f is generated collapsing field turn around the polarity of the voltage across the inductor. The current flows in the diode D, the load and the diode will be turned ON.
• The discharge of the inductor L decreases with the help of the current.
• During the first transistor is in one state the charge of the accumulator in the capacitor. The current flows through the load and during the off period keeping Vout reasonably. Hence it keeps the minimum ripple amplitude and Vout closes to the value of Vs

Boost Converter Working :

In this converter the first transistor is switched ON continually and for the second transistor the square wave of high frequency is applied to the gate terminal. The second transistor is in conducting when the on state and the input current flow from the inductor L through the second transistor.

The negative terminal charging up the magnetic field around the inductor. The D2 diode cannot conduct because the anode is on the potential ground by highly conducting the second transistor.

By charging the capacitor C the load is applied to the entire circuit in the ON State and it can construct earlier oscillator cycles. During the ON period the capacitor C can discharge regularly and the amount of high ripple frequency on the output voltage. The approximate potential difference is given by the equation below.

Vs + VL (both input and load voltage)

• During the OFF period of second transistor the inductor L is charged and the capacitor C is discharged. The inductor L can produce the back e.m.f and the values are depending up on the rate of change of current of the second transistor switch.
• The amount of inductance the coil can occupy. Hence the back e.m.f can produce any different voltage through a wide range and determined by the design of the circuit. Hence the polarity of voltage across the inductor L has reversed now.
• The input voltage gives the output voltage and atleast equal to or higher than the input voltage. The diode D2 is in forward biased and the current applied to the load current and it recharges the capacitors to VS + VL and it is ready for the second transistor.

When a buck converter is combined with a boost converter, the output voltage is typically of the same polarity of the input, and can be lower or higher than the input. Such a non-inverting buck-boost converter may use a single inductor which is used for both the buck inductor mode and the boost inductor mode, using switches instead of diodes,sometimes called a "four-switch buck-boost converter",it may use multiple inductors but only a single switch as in the SEPIC and Cuk topologies.

Buck Boost Converter in CCM mode -

If the current through the inductor L never falls to zero during a commutation cycle, the converter is said to operate in continuous mode. The current and voltage waveforms in an ideal converter can be seen in the below figure.

From t = 0 {\displaystyle t=0} to t = D T {\displaystyle t=DT} , the converter is in On-State, so the switch S is closed. The rate of change in the inductor current (I_L) is therefore given by,

At the end of the On-state, the increase of I_L is therefore:

Δ I L On = ∫ 0 D T d ⁡ I L = ∫ 0 D T V i L d ⁡ t = V i D T L {\displaystyle \Delta I_{{\text{L}}_{\text{On}}}=\int _{0}^{D\,T}\operatorname {d} I_{\text{L}}=\int _{0}^{D\,T}{\frac {V_{i}}{L}}\,\operatorname {d} t={\frac {V_{i}\,D\,T}{L}}}

D is the duty cycle. It represents the fraction of the commutation period T during which the switch is On. Therefore D ranges between 0 (S is never on) and 1 (S is always on).

During the Off-state, the switch S is open, so the inductor current flows through the load. If we assume zero voltage drop in the diode, and a capacitor large enough for its voltage to remain constant, the evolution of I_L is:

d ⁡ I L d ⁡ t = V o L {\displaystyle {\frac {\operatorname {d} I_{\text{L}}}{\operatorname {d} t}}={\frac {V_{o}}{L}}}

Therefore, the variation of I_L during the Off-period is -

Δ I L Off = ∫ 0 ( 1 − D ) T d ⁡ I L = ∫ 0 ( 1 − D ) T V o d ⁡ t L = V o ( 1 − D ) T L {\displaystyle \Delta I_{{\text{L}}_{\text{Off}}}=\int _{0}^{\left(1-D\right)T}\operatorname {d} I_{\text{L}}=\int _{0}^{\left(1-D\right)T}{\frac {V_{o}\,\operatorname {d} t}{L}}={\frac {V_{o}\left(1-D\right)T}{L}}}

As we consider that the converter operates in steady-state conditions, the amount of energy stored in each of its components has to be the same at the beginning and at the end of a commutation cycle. As the energy in an inductor is given by -

E = 1 2 L I L 2 {\displaystyle E={\frac {1}{2}}L\,I_{\text{L}}^{2}}

it is obvious that the value of I_L at the end of the Off state must be the same with the value of I_L at the beginning of the On-state, i.e. the sum of the variations of I_L during the on and the off states must be zero.

Δ I L On + Δ I L Off = 0 {\displaystyle \Delta I_{{\text{L}}_{\text{On}}}+\Delta I_{{\text{L}}_{\text{Off}}}=0}

Substituting Δ I L On {\displaystyle \Delta I_{{\text{L}}_{\text{On}}}} and Δ I L Off {\displaystyle \Delta I_{{\text{L}}_{\text{Off}}}} by their expressions yields:

Δ I L On + Δ I L Off = V i D T L + V o ( 1 − D ) T L = 0 {\displaystyle \Delta I_{{\text{L}}_{\text{On}}}+\Delta I_{{\text{L}}_{\text{Off}}}={\frac {V_{i}\,D\,T}{L}}+{\frac {V_{o}\left(1-D\right)T}{L}}=0}

This can be written as-

V o V i = − D 1 − D {\displaystyle {\frac {V_{o}}{V_{i}}}=-{\frac {D}{1-D}}}

From the above expression it can be seen that the polarity of the output voltage is always negative (because the duty cycle goes from 0 to 1), and that its absolute value increases with D, theoretically up to minus infinity when D approaches 1.

Apart from the polarity, this converter is either step-up (a boost converter) or step-down (a buck converter). Thus it is named a buck–boost converter.

 

Now we will do the calculations as per the given data :

Vi = 300 to 600 V,  Vo = -450 V , P = 2KW

Assume the switching frequency as Fs = 20kHz

Now I will take the input voltage Vin = 500 V , and we will do required calculations.

(i) Duty cycle ratio :

Vo/Vin = (-D) /( 1 - D)

(-450)/ (500) = -D / ( 1 - D)

0.9 = D / (1 -D)

0.9 - 0.9D = D

0.9 = D + 0.9D

D = 0.9/1.9

D = 0.4737

D = 47.37 %

 

(ii) Resistance Value :

R = (Vo)2 / P

R = (450*450) / 2000

R = 101.25 Ω

 

(iii) Inductor currents :

IL = Vo/R(1 - D)

IL = (-450) / [(101.25 )(1-0.4737)]

IL = -8.45 A

Assume,  10% ripple to achieve a good compromise between inductor and capacitor size a ripple current value of 10% - 30% of maximum inductor current should be chosen.

ΔIL = (0.1 ) (-8.45)

ΔIL = - 0.845 A

 

(iv) Inductance value :

L = (Vo) (1 - D) / (Fs* ΔIL)

L =  (-450)(1-0.4737) / (20e3 *(-0.845))

L = 14.01 mH

 

(v) Output Capacitor :

Cout = D / (R * Fs * ΔVo/vo)

Assume, ΔVo/vo = 1%

Cout = 0.4737 / (101.25 * 20e3 * 0.01)

Cout = 23.4 µF

 

Now we will design the Buck Boost converter circuit in CCM mode

The Circuit diagram looks like this -

Vin = 500 V       R = 101.25 Ω  

L = 14.01 mH     D = 0.4737      

Cout = 23.4 µF

After calculating all these values now substitute them in the above model as shown,

DC source :

We have to substitute the 500 V dc value into the block as shown.

Pulse Generator:

• Here I have substituted the period = 1/Fs
  
• Fs = 20KHz.
• And the pulse width as D = 47.31%

L block:

C Block:

 

R Block:

 

Diode Block:

 

MOSFET Block:

 

Now we will run the simulation and observe the simulation results.

Input DC waveform:

• The input voltage is pure DC.
  
• And the value is 500 V as shown in the waveform.

Pulse Waveform:

• This is how the pulse waves look like.
  
• Almost ON and OFF times are almost equal and this signal drives the gate of the MOSFET.

 

MOSFET Voltage & Current waveforms.

On zooming, MOSFET current looks like this:

 

• This is how the MOSFET current looks like.
• It is high when the MOSFET is turned ON and Zero during OFF period.
• And it has the peak value of 10 A approximately.

On Zooming, MOSFET voltage looks like-

• This is how it looks with almost the double amplitude of input voltage.
• The peak value reaches 1000 V which is double of 500 V(Vi).
• And comes to zero during OFF period.

Diode Current & Voltage waveforms.

 

Diode Current:

On zooming,

• Initially diode current is high because of switching operations.
• Later it comes to 8.5 A same as inductor current.
• Because current is same in entire circuit and current is zero when the switch is off or  when diode is not conducting.

Diode Voltage:

• The diode operates at 0.7 V forward voltage.
• When it is reverse biased the peak value reaches to almost double of input but in negative polarity as shown.

On zooming,

• This s how the voltage looks like.
• The negative voltage is avaliable during the reverse biased.

 

Inductor current:

• This is how the inductor current looks like.
• Initially, it has high value of 20A.
• Later, it comes to  5 A but does not go to zero, since it's in CCM mode.
• Further it achives the steady state of value nearly 9 A.

On zooming, it looks like:

• Here, because of 10% ripple it varies between 9 to 9.6 A as shown above.
• But this current only flows in the entire circuit in order to perform the smoother operation.

Now we will see the output voltage waveform.

• Here we exactly got the output of -450V which is inverted.
• Now we will see the waveform related to output voltage.

• This is the inverted voltage of -450V.
• The output is achived by the buck boost converter and now we will see the zoomed waveform.

On zooming,

• Because of ripples considered it varies between the -450V to -444V as shown.
• The output is fluctuating DC but we stepped it down to -450V(inverted) .
• And also it is not more fluctuating as shown here.
• And the desired output is achieved using Bck - Boost converter.

 

Cuk Converter in CCM :

The Cuk Converter is a type of buck boost converter with zero ripple current Cuk converter can be seen as a combination ofBoost and Buck converter having one switching device and a mutual capacitor, to couple the energy.

Working / Operation :

• A non-isolated Cuk converter comprises two inductors and two capacitors a switch (usually a transistor), and a diode.. Its schematic can be seen in figure. 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.

• 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, Boost and Buck-Boost) the Cuk 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).
• The topology of a Cuk converter involves a low-side switch, diode, inductors, and a coupling capacitor between the input and output terminals. In the conventional topology, the inductors are typically uncoupled. Few variants of the basic topology incorporate coupled inductors instead that are typically matched in value. In some other cases, the coupled inductors are integrated with a transformer that additionally offers enhanced step-up, volume reduction, and isolation features.
• It is evident that the converter has two inductor currents and two capacitor voltages that need to be determined. The strategy employed for circuit analysis can be summed up as the application of a volt-second balance for each inductor voltage and charge balance for each capacitor current .
• Simplification is carried out using small ripple approximation and the resulting four equations are solved to obtain the values for the four unknowns.
• From figure, it can be observed that when the switch is conducting, the diode is off and the capacitor C1 releases energy to the output.

 

Cuk Converter in CCM:

 

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:

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 d I d t {\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 C 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 V L 1 = V s − V C 1 {\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 V L 2 = V o {\displaystyle V_{L2}=V_{o}}
• in the ON state , inductor L₁ is directly connected to the input source.
• Therefore V L 1 = V s {\textstyle V_{L1}=V_{s}} . Inductor L₂ is connected in series with C and the output capacitor,
• so V L 2 = V o + V C {\displaystyle V_{L2}=V_{o}+V_{C}}

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

V ¯ L 1 = D ⋅ V s + ( 1 − D ) ⋅ ( V s + V C ) = ( V s + ( 1 − D ) ⋅ V C ) {\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)}

V ¯ L 2 = D ( V o + V C ) + ( 1 − D ) ⋅ V o = ( V o + D ⋅ V C ) {\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:

V C = − V o D {\displaystyle V_{C}=-{\frac {V_{o}}{D}}}

So the average voltage across L₁ becomes:

V ¯ L 1 = ( V s − ( 1 − D ) ⋅ V o D ) = 0 {\displaystyle {\bar {V}}_{L1}=\left(V_{s}-(1-D)\cdot {\frac {V_{o}}{D}}\right)=0}

Which can be written as:

V o V s = D 1 − D {\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.

 

Calculations:

Now we will design this circuit in the simulink and achieve the desired output.

Given,

Vi = 300 to 600V ,  Vo = -450 V,  P = 5KW;

Assume the higher switching frequency for smoother operation so take Fs= 100KHz

Now I will take Vin = 500 V;

(i) Duty cycle ratio:

Vout/Vin = (-D)/(1-D)

(-450/500) = (-D)/(1-D)

0.9 = D/(1-D)

0.9 - 0.9D = D

D = 0.4737

D = 47.37%

 

(ii) Resistnce value:

R = (Vo)2/P

R = (450*450) / 5000

R = 40.5 Ω

 

(iii) Inductor currents:

IL1 = Ii = P/vi

IL1 = 5000/500 = 10A

IL2 = Io = P/Vo

IL2 = 5000/(-450)

IL2 = -11.11 A

Assume 10% ripples,

ΔIL1 = (0.1)(10) = 1A

ΔIL2 = (0.1)(-11.11) = -1.11A

 

(iv) L1 & L2 values :

L1 = Vin * D / (Fs *ΔIL1)

L1 = (500 * 0.4737) / (100e3*1)

L1 = 2.36 mH

 

L2 = Vin * D / (Fs *ΔIL2)

L2 = (500 * 0.4737) / (100e3 * 1.111)

L2 = 2.131 mH

 

(v) C1 & C2 values:

 

VC1 = Vin - Vo

Vc1 = (500 - (-450))

VC1 = 950V

Assume, 10% ripple,

ΔVc1 = (0.1)(950)

ΔVc1 = 95 V

 

C1 = Vo * D / (R*Fs*ΔVc1)

C1 = (450 * 0.4737) / (40.5 * 100e3 * 95)

C1 = 0.55µF

 

C2 = Cout = (1-D)/(8 * L2 * Fs2 * ΔVo/Vo)

C2 = (1-0.4737) / (8*2.13e-3*100*100e6*0.01)

C2 = 0.308 µF

 

The circuit diagram would look like this :

The block parameters involved in this diagram would look like :

DC voltage source:

I have substitued the value of 500V into this block.

 

Pulse Generator:

Here I have chosen 100KHz frequenct to get a proper operation.

L1 & L2 :

 

 

C1 & C2 :

  

 

R:

 

Diode and MOSFET Blocks :

    

 

After substituting we will run the model for spectified simlation time and observe the waveforms:

Input Voltage waveform:

• The input voltage is a pure DC and its a straight line of 500 V.

 

Pulse waveform:

• The pulse signal almost has equal turn on and turn off time.
• This drives the MoSFET.

MOSFET current & Voltage waveform:

On zooming,

MOSFET Current:

• This is how the MOSFET current waveform looks like.
• During ON condition it reaches peak and during OFf condition it comes to zero.

On zooming,

MOSFET voltage:

• The voltage gets almost doubled during initial switching.
• Then in off condition again it comes back to zero and it happens through out the simulation.

 

Diode voltage & Current:

On zooming,

Diode current:

• The input current flowsin the diode 
• .During switching condition current value goes to the maximum nd comes to minimum.

On zooming,

Diode Voltage:

• Diode conducts in forward baising mode.
• During reverse biasing, voltage reaches to the negative maximum voltage as we can observe from here.

 

Inductor currents IL1 &  IL2:

IL1 :

On zooming,

• As we can see IL1 is same as calculated value and it is in CCM.
• It varies between 10.5 A to 11. 5 a because of the assumed ripple current of 1 A.
• This is the input current as mentioned before.

IL2 :

On Zooming,

•  This IL2 is basically an output current(negative).
• It has a ripple of -1.111 A as we can see here.
• This is the negative current which we can varified with the calculated value, and it varies from -11.6 to -10.8  A.

 

Capacitor voltages VC1 & VC2 :

VC1 :

On zooming,

• As calculated, VC1 is the difference between Vin & Vo.
• So it is 950 V. And due to ripple of 95 V the fluctuations are present in the waveform.

 

VC2 : (output voltage)

On zooming,

• As we can see the VC2 is exactly the required output voltage.
• It is varying continuosly and charging and dischrging through out the smulation.

 

Output Voltage:

• The required output is achieved.
• Output of -450V is displayed here.

On zooming,

• This is how the output voltage looks like.
• It is -450 V amplitude with considerable ripples.
• It varies from -450V to -452 V as we can see in the above figure.
• The desired output is achieved from the Cuk converter hence the operation is achieved successfully.

 

Conclusion: The desired SEPIC, Buck - Boost and Cuk converters are designes as per the given requirements. The supported images, output waveforms are attached. And the simulation files are attached as .slx files. And the desired outputs are successfully achieved.