Challenge

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

Design of boost converter, and calculating the loss of different components including power switching device and deriving efficiency of boost converter. This is applicable in power supply industry, electric vehicles etc where DC-DC converter is required to boost the voltage from DC source.

Assumed Design Parameters :

Vin = 100V

Vout = 200V

Fsw = 20kHZ

L = 200uF

C = 400uF

R = 10 ohms

rL = 10 m ohms

rC = 2 m ohms

rd= 5 m ohm

Duty Cycle:

Vout/Vin = 1/(1-D)

200/100 = 1 / (1-D)

2 = 1 / (1-D)

2*(1-D) = 1

(1-D) = ½

(1-D) = 0.5

D = 1-0.5

D = 0.5

DC DC boost converter model:-

result:-

diode voltage & current:-

zoom in:-

mosfet voltage & current:-

zoom in:-

The Loss Calculation of boost converter:

The Boost converter has four types of losses. They are:

Conduction losses
Diode losses
Induction losses
Capacitor losses

Conduction Losses:

Conduction Losses: Pc= I² * Rds *D

p= Power Dissipation

I= Current through MOSFET

R= Resistance in Ohm

D=Duty Cycle

Im = Pi/Vi = (V^2/R)/Vi

= ((100)^2/10) / 100

Im =10

Pc= 10^2 * 0.023 * 0.5

Conduction loss Pc = 1.15w

Diode losses:

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

Vf= forward voltage

I= Current output

D=Duty Cycle= Vout / Vin

P = 19.55 * 1 * (1-0.5)

= 9.775

Diode losses P = 9.7 W

Induction Losses:

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

I= Current output

D=Duty Cycle= Vout / Vi

F= Frequency

L= Inductance value

Pd = 1^2 * 10 +{(0.0008 * 5000}

Inductor loss Pl = 10W

Capacitance losses:

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

I= Current output

D=Duty Cycle= Vout / Vi

F= Frequency

C= Capacitance value

Pc = 2*0.5*0.5 + {(2*0.0001*0.5*0.5*100) / (400*50*1*10^-6)}

Pc = 0.7085W

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:

Efficiency = (Pin -Ploss / Pin ) 100

From the data

Pin= Vin * I

I= Vin / R

=100 / 10

I= 10 Amps

Pin = vin * I

=100 * 10

Pin=1000 W

Efficiency = (Pin -Ploss / Pin )100

= 1000-11.74 / 1000 100

Efficiency = 98 %

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