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

A boost converter simulink model is created

Dc voltage, IGBT, Diode, inductor, filter cap and load parameters are defined as below

Output voltages a and currents are

Output voltage of boost converter for any given duty cycle can be derived as

Vout / Vin =1/ (1-D)

as given in question

vout =200

vin = 100

200/100= 1/1-D

D= 1-1/2 = 1/2 = 0.5 or 50%

Losses in the overall converter can be can be breifly classified according to losses across various devices in the circuit

a)switching device losses

b)diode losses

c) capacitor losses

d) inductor losses

we are neglecting parasitic losses in the wiring and considering an ideal source

switching device losses= conduction losses + switching losses

Pcon= Rce X Iout2 X (Vout/Vin)

Pout = IoutxRout = 20 x 10 = 200

Iout = Pout/Vout = 200/200 = 1A

= 0.001 x 1 x 1 x 200/100

= 0.002 Watts

Psw= 0.5 x Vin X Iout X (trise + tfall) X fsw

= 0.5 x 100 x 1 x (0.001+0.001)x 10000

= 5 watts

Diode losses

Pdi = (1-D)(Iout2 X Rdi + Vfd x Iout)

vfd= forward voltage drop

Pdi = (1-0.5)(1 x 1 x 0.001 + 0.8 x 5)

= 0.5x(0.001+3.2)

= 1.6 watts

Inductor losses

Pind = RLXIin2 + (RL/12)(DXVin/fxL)2

= 1 x 1e-6 x10x10 + (1x1e-6/12)(0.5x100/50x200xe-3)2

= 1e-3 + (1e-5/12)(1/200xe-6)2

= 2.08Xe-3

= 0.002 Watts

Capacitor Losses

Pcap= Rc X D(1-D) X Iin2 + (Rc/12)(1-D)(DXVin/fxC)

= 2x 1e-6x 0.5 x0.5 x 10x10 + (2xe-6/12)(0.5)(0.5x100/50x100xe-3)

= 5 xe-5 + (e-6/12)(10)

= 0.00005 + 0.000008

= 0.000058

Total losses = 0.002+5+1.6+0.002+0.000058 = 6.6W approx

efficiency = output/input = (200-6.6)/200 = 0.96 = 96%