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 converter that 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 Transistor) 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).

 

Simulink Model: -

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

1. DC Voltage source
2. MOSFET
3. Diode
4. series RLC branch
5. Voltage measurement 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 equitation 

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

Procedure: -

For creating the Simulink model of Boost converter, the supply voltage and Series RL branch will be connected in series manner

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

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

For on and off Switch, we are sending the Pulses with switching frequency () 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 drawn by both "MOSFET " and Diode" will be plotting by using the Scope and Bus selector Blocks.

Mosfet voltage and current graph: -

The above graph we can see the Output voltage is greater than the Input voltage 

Hence the Working of Boost Converter is working in proper manner

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

 

Diode voltage and current graph: -

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.

 These 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= Inductance value

 

Capacitance losses: -

        While getting the output through 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 cycle= 0.5

 

Conduction Losses: -

 from the equitation (1)

Conduction Losses: - Pc= I² * R *D

                              Pc= (1) ² * 10*0.5

Conduction Losses (Pc)=5W

Diode losses: -

From the equitation (2)

 Diode Losses: - Pd= Vf * I* (1-D)

                             = 19.55 * 1 * (1-0.5)

                             = 19.55*0.5

 Diode Losses (Pd) =9.75 w

 

Inductor Losses: -

From the equitation (3)

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

              =1² * 10 +(0.0008*5000)

            = 10+4

          PI =14 W

Capacitance Losses: -

 From the equitation (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 equitation (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 %