Project 1 - Loss calculation for a DC/DC converter-MATLAB

Aim) The main aim is Design the Simulink model of BOOST converter and determine the Losses of Converter and efficiency of the Converter.

Answer)

BOOST-CONVERTER:-

        A boost converter (step-up converter) is a DC-DC power converterthat steps up voltage (while stepping down current) from its input (supply) to its output (load).

         It is a class of Switched mode power supply (SMPS) containing at least two semiconductors (a diode and a Transisiter) and at least one energy storage element: a capacitor, inductor, or the two in combination.

         To reduce voltage ripple ,filters made of capacitors (sometimes in combination with inductors) are normally added to such a converter's output (load-side filter) and input (supply-side filter).

                                                       Fig--> Boost converter schematic

Simulink Model :-

 for designing the Boost converter in Simulink we ares used the following blocks:

1. DC Voltage source
2. MOSFET
3. Diode
4. series RLC branch
5. Voltage measurment block
6. PWM generator
7. Display
8. Bus selector
9. Power gui
10. Scope

Design parameters :

1. Vin = 100 V
2. Vout = 200 V
3. L = 200uH 
4. C = 400uF
5. R= 10 Ohms
6. rL = 10 mOhms
7. rC = 2 mOhms
8. rd=5 mOhms
9. switching frequency=

For duty cycle:-

We know that the 

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

were,
vin= Input voltage
Vo= Output Voltage
D= Duty cycle

from the given data

Vin = 100 V

Vout = 195.5 V

Then from the above equatation 

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

200/100 = (1/ (1-D) )

2*(1-D)= 1

(1-D) = 0.5

          D=0.5

therefore 

Duty cycle = 0.5

Simulink Model :

The simulink model of Boost conveter will be as followes:

Procedure:-

For creating the simulink model of BOOSt converter The supply voltage and Series RL branch will be connected in series mannaer

And MOSFET" is actes as as power switch and it is connected to the Parallel manner

.After that a Cpacitor and load(Resistive Load) are connected parallel with using the "Series RLC branch" block.

For on and of Switch we are send the Pulses with switching frequnecy ()Using DC-DC PWM generator Block>>we are assigning the Duty cycle ration as 0.5 by using the Constant block.

And the output voltage will be displayed by using the display block.

The voltage and currents drwan by both "MOSFET " and Diode" will be plotting by using the Scope and Bus selector Blocks.

Design parameters values : 

MOSFET parameters :

 Pulse generator parameter Block:

Diode Parameters:

Step Parameters : 

Results : 

MOSFET VOLTAGE AND CURRENT:

From the above graph the top graph will be the Mosfet Current Graph

Bottom Graph will be the Voltage graph

we can observe that the 0.5 Duty cycle ration in the graph

The Minimum And maximum Currents in MOSFET will be determined by using Signal statistics Below 

DIODE VOLTAGE AND CURRENT:

From the above graph the top graph will be the DIODE Current Graph

Bottom Graph will be the Voltage graph

we can observe that the 0.5 Duty cycle ration in the graph

The Minimum And maximum Currents in Diode will be determined by using Signal statistics Below 

Maximum Current =3.994e+01 Amps

>>Minimum current = 0.00e+00 Amps

Output voltage :

rom the above graph we can see the Out put voltage is grater than the Input voltage 

Hence the Working of Boost Converter is working in proper manner

the Minmum and maximum Voltage will be displayed by using the Below Signal statistics block.

Determining the Losses in Power electronic Converters:-

In any DC-DC converter majorly consider  4 types of losses are as followes:-

1. Conduction Losses
2. Diode losses
3. Induction Losses
4. Capacitor Losses 

Conduction Losses:-

            Conduction losses are a result of device parasitic resistances impeding the DC current flow in a DC/DC converter. These losses are in direct relationship with the duty cycle.

       When the integrated high-side MOSFET turns on, the load current flows through it. The drain-to-source channel resistance (R_DSON) causes power dissipation.

Formula :-

                    Conduction Losses:- Pc= I² * R *D-----------------(1)

were,
p= Power Dissipation
I= Current through MOSFET
R= Resistance in Ohm
D=Duty Cycle= Vout / Vin

Diode losses:-

         The diode is forward-biased when the integrated MOSFET turns off. During this time, the inductor current ramps down through the output capacitors, the load and the forward-biased diode. Since the load current is now conducting through the diode, there will be power dissipation in Diode.

This will be calculated by using below Formula:-

                   Diode Losses:- P= Vf * I* (1-D) ------(2)        

were,
Vf= forward voltage
I= Current output
D=Duty Cycle= Vout / Vin

Induction Losses:-

  The conduction losses depend on the load current. With heavier loads, the conduction loss in the MOSFET increases and is the dominating factor. Conduction losses plus switching, driver and internal low-dropout regulator (LDO) losses lead to a considerable generation of heat.

 This losses can be calculated by using the below formula:

             Pd= I² * R + { (rl / 12) (D* Vin / F*L)}-------(3)

were,
I= Current output
D=Duty Cycle= Vout / Vi
F= Frequency
L= Inductnce value

Capacitence losses:-

        While getting the output throught the capacitor we can get some power dissipation in capacitor also.

this power dissipation is determined by using the following Formula:-

             Pc= {rc* D*(1-D) I² + {rc / 12(1-D)}}*{(Din / C*F)}----(4)

were,
I= Current output
D=Duty Cycle= Vout / Vi
F= Frequency
C= Capacitance value

From the given data we can find the losses for Designed DC-DC Converter:-

1. Vin = 100 V
2. Vout = 200 V
3. L = 200uH 
4. C = 400uF
5. R= 10 Ohms
6. rL = 10 mOhms
7. rC = 2 mOhms
8. rd=5 mOhms
9. Duty cyclee= 0.5

Conduction Losses:-

 from the equatation (1)

Conduction Losses:- Pc= I² * R *D

                              Pc= (1)² * 10*0.5

Conduction Losses (Pc)=5W

Diode losses:-

From the equatation (2)

 Diode Losses:- Pd= Vf * I* (1-D)

                             = 19.55 * 1 * (1-0.5)

                             = 19.55*0.5

 Diode Losses(Pd) =9.75 w

Inductior Losses:-

From the equatation (3)

       PI= I² * R + { (rl / 12) (D* Vin / F*L)}

              =1² * 10 +(0.0008*5000)

            = 10+4

          PI  =14 W

Capacitance Losses:-

 From the equatation (4)

Pc= {rc* D*(1-D) I² + {rc / 12(1-D)}}*{(Din / C*F)}

      =2*0.5*0.5+(2*0.0001*0.5*0.5*100 / 400*50*0.000001)

       Pc= 0.7085 W

Total Losses= Pcon+Pd+Pi+Pc

                   =5+9.775+1.4+0.7085

Total losses=11.74585 W

Efficiency Of Converter:-

 The efficiency Of converter is determined by using below formula:

Formula:-

                       Efficiency =  (Pin -Ploss / Pin ) 100-----------(5)

From the data 

Pin= Vin * I

I= Vin / R

   =100 / 10

  I= 10 Amps

Therefore,

Pin = vin * I

       =100 * 10 

       Pin=1000 W

From above equatation(5)

Efficiency = (Pin -Ploss / Pin )100

               = 1000-11.75 / 1000 100

Efficiency =  98.825 %

Results :-

The total Power losses Ploss= 11.75 W

The efficiency of Converter = 98.825 %