Project 1

FINAL PROJECT 14A

AIM:- To develope model of any three topologies from the below 14 topologies.

Here i will choose the following three topologies as per my choice.

1. Buck converter in CCM. (500W)
2. Boost converter in CCM. (600W)
3. Cuk converter in CCM. (5000W)

500w Buck converter in CCM

In buck converter we are getting the output voltage less than the input voltage

with same power on input and output side.

Here thge given parameters are.

Input voltage Vin = 40 - 50V

Output voltage Vout = 30V.

Power P = 500W

we will assume the switching frequency as Fs = 200Khz.

First we will calculate the range of Duty cycle D .

if Vin = 40v, D = Vout/Vin = 30/40 = 0.75

if Vin = 50v , D = Vout/Vin = 30/50 = 0.6

So, Range of D is

0.6<=D<=0.75

Now, we will Take the input voltage as 45 V

D(45) = Vout/Vin = 30/45 = 0.67

Inductor current Il = P/Vout = 500/30 = 16.67A

R = Vout^2/P = #0^2/500 = 1.8ohm

I will design the inductor for 1% ripple, so

Delta Il = 1*Il/100 = 0.1667A

=> L = Vout(1-D)/Fs*Delta Il

L = 30(1-0.67)/200Khz*0.167

L = 0.2964mH = 29.64e-4

Also

We will design the caoacitor of 20% rippel as

Delta Vout = 20*Vout/100 = 6

=> Delta Vout / Vout = 6/30 = 0.2

Cout = (1-D)/8*L*Fs*(Delta Vout/Vout)

Cout = 1.75e-5mF = 1.75e-8 F

Now we will build the model in Simulink as

The necessery waveforms of Component current and voltages are as. (Zoomed out)

Zoomed in

600w BOOST converter in CCM

In boost converter we are getting the output voltage higher than the input voltage

with same power on input and output side.

Here thge given parameters are.

Input voltage Vin = 20 - 30V

Output voltage Vout = 40V.

Power P = 600W

we will assume the switching frequency as Fs = 200Khz.

First we will calculate the range of Duty cycle D .

if Vin = 20v, D = (Vout-Vin)/Vout =(40-20)/40 = 0.5

if Vin = 30v , D = (Vout-Vin)/Vout =(40-30)/40=0.25

So, Range of D is

0.25 <= D <= 0.5

Now, we will Take the input voltage as 25 V

D(25) = (Vout-Vin)/Vout =(40-25)/40=0.375

Load R = Vout^2/P = 40^2/600 = 2.67ohm

Inductor current (Il) = Vin/R(1-D)^2 = 23.97A

Now we will design the inductor with 20% ripple

Delta (Il) = 20*Il/100 = 4.79A

L = Vin/(Fs*Delta(Il)) = 25/200Khz(4.79) = 0.02609mH = 2.069e-5 H

Also we will design a capacitpr with 1% ripple as

Delta (Vout) = 1*Vout/ !00 = 0.25

Also Delta(Vout)/Vout = 0.0025

Cout = D/(R*Fs*(DeltaVout/Vout)) = 0.2808mF = 28.089F

Now we will design the boost converter in Simulink as

The nacessery output input and component current and voltage waveforms are as

(Zoomed out)

Zoomed in

5000W CUK converter in CCM

A CUK converter transforms a DC voltage at the input to a DC voltage at the output

with reversed polarity. Compared to the Buck, Boost and Buck-Boost converters the CUK

converter uses an additional inductor and capacitor to store energy.

Here thge given parameters are.

Input voltage Vin = 300 - 600V

Output voltage Vout = -450V.

Power P = 5000W

we will assume the switching frequency as Fs = 200Khz.

First we will calculate the range of Duty cycle D .

if Vin = 300v, D = Vout/(Vout-Vin)=-450/(-450-300) = 0.6

if Vin = 30v , D = Vout/(Vout-Vin)=-450/(-450-600) = 0.4285

So, Range of D is

0.4285 <= D <= 0.6

Now, we will Take the input voltage as 500 V

D(500) = Vout/(Vout-Vin)=-450/(-450-500) = 0.473

Now the value of Currents in inductors.

Il1 = Iin = P/Vin = 5000/500 = 10A

Il2 = Iout = P/Vout = 5000/-450 = -11.11A

R = Vout^2/P = -450^2/5000 = 40.5ohm

Lets design the CUK inductors at 1% ripple

Delta(Il1) = Il1*1/100 = 0.1A

Delta(Il2) = Il2*1/100 = 0.11A

Now

L1 =D*(Vin/Fs*Delta(Il1)) = 11.825mH = 0.011825H

L2 =D*(Vin/Fs*Delta(Il2)) = 10.65mH = 0.01065H

We need to find the Cap. C1 voltage for Capacitor calculation as

Vc1 = Vin - Vout = 500-(-450) = 950V

We will design the capacitors for 20% ripple

Delta (Vc1) = 20*Vc1/100 = 190v

now C1 = (Vout*D)/(R*Fs*Delta(Vc1)) = 0.000138mF = 1.383F

Also we will design Cout for 20% ripples as

Delta (Vout) = 20*(Vout)/100 = 90

Delta (Vout)/Vout = 90/450 = 0.2v

Cout = (1-D)/(8*L2*Fs*(Delta (Vout)/Vout)) = 7.65e-7mF = 7.65e-10F

Now we will design the Cuk converter in Simulink as.

Nacessery waveforms of input, output, Component current and voltage as (Zoomed out)

Zoomed in

Now i will show you the Description of few commenly used components used in these converters.

MOSFET USED

DIODE USED

PWM GENERATOR USED

That is all for the project 14

Kindly find the attached .mdl files below.

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