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

PROJECT -

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. All the aspects of this project with loss calculation, device selection, simulation etc are industry relevant.

AIM - Design the Simulink model of Boost converter and determine the Losses of Converter and efficiency of the Converter.

THEORY -

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).

CALCULATION -

Let us consider Vin = 100V and we need the output voltage of 200V and the output power needed is Pout =2Kw.

Then the Output current would be Iout =20A

Then the load resistance is

Rload = Vout/Iout = (200/20) = 10 ohm

let us see the block diagram of the boost converter that we want to design

Let us design the inductor for the above Boost converter

For calculating the inductor we need to know what is the ripple current, a good estimation for the inductor ripple current is 20% to 40% of the output current

let us assume 20% then,

ΔIL = 0.2*Iout*(Vout/Vin)

=0.2*20*(200/100)

ΔIL = 8A

Now let us calculate the inductance of the boost converter

L = [Vin *(Vout-Vin)] / [ΔIL*fs*Vout]

Now again we need to know the minimum switching frequency, higher switching frequency lowers the ripple current but a higher frequency also gives the noise so let us assume the switching frequency

fs =10000Hz

Then the inductance would be

L = [Vin *(Vout-Vin)] / [ΔIL*fs*Vout]

= [100*(200-100)] / [8*10000*200]

L = 0.625mH

Now we have to calculate the value of capacitance we need to calculate firstly the Duty cycle .

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

Then Capacitance value is calculated by formula ,

C = (Iout*D) / (fs*ΔVout)

considering the change in ripple voltage as 20% so we are getting ΔVout=0.2

Now the Output Capacitance is given by

Cout = (20*0.5)/(10000*0.2)

Cout = 5mF

Considering rl (resistance of inductance) = 1mH

rc (resistance of capacitance)=2mF

SIMULINK MODEL -

A DC Voltage source is used to supply the DC voltage of 100V. A RL Branch is used to connect in series with DC supply.

Switching Device used is MOSFET which is connected across the DC Voltage Source. A Diode and RC circuit is connected across the MOSFET. RC branch is used to remove the voltage ripples across load and make the output voltage smooth with less distortions in waveform. A Resistive load of 10 ohms is Connected at output.

A constant block is is used for giving the Duty cycle. The output of this block is given to the PWM(DC - DC) generator whose frequency is 10Khz. This modulated signal or pulse is applied to the MOSFET as shown in the above model. The remaining components are arranged as per the circuit of Boost converter.

Here the Output voltage obtained from the converter is 195.5V which is near to 200V that we are desired.

RESULT -

The Current and Voltage across MOSFET and Diode is shown in the below graph. From the Output Voltage graph it is observed that the voltage goes near to 300V for short duration so, we want to select the MOSFET above this range .

The below MOSFET is connected in series to reach the operating region.

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. Capacitance Losses

1. 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) = Im^2 * Rdson*D

where,

Im=Current flowing through MOSFET

Rdson=Resistance of drain to source during ON state (0.023 in above MOSFET datasheet)

D=Duty cycle Vout/Vin

Im=Pi/Vi

Pi=V^2 / R =100^2/10=1000W

Then, Im=1000/100=10A

Pc =10^2*0.023*0.5

Pc = 1.15 W

2. 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(Pd) = Vf*Io*(1-D)

where ,

Vf= Forward voltage

I0= Output current

D= Duty cycle

Vf of diode =0.8V

Pd = 0.8*20*(1-0.5)

Pd = 8 W

3. Induction Losses:-

The Induction losses depend on the load current. With heavier loads, the induction 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 -

Pl= (I^2*rl)+ (rl/12)(D*Vin/f*L)

where

I= Current output

D= Duty cycle

f= Frequency

L= Inductance

Pl = (20^2*1e-3) +{(1e-3/12) (0.5*100/50*0.625e-3) }

Pl = 0.53 W

4. Capacitance 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 -

Pcap=rc*D(1-D)Io^2 + ((rc/12)*(1-D)*(D *Vin/C*f)

where,

Io= Output current

D= Duty cycle

f= Frequency

C= Capacitance

Pcap = 2e-3*0.5*(1-0.5) 20^2 + {2e-3 / 12(1-0.5)}*(0.5*100 / 5e-3*50)

Pcap = 0.2 + 0.06

Pc = 0.26 W

Total Power Losses = Pc+ Pd + Pl + Pcap

= 1.15+8+0.53+0.26

Total Power Losses = 9.94 W

Efficiency = Output Power / Input Power

= (Input Power - Power loss) / Input Power

= (2000-9.94) / 2000

Efficiency = 99.5%

CONCLUSION -

In this Project, Boost converter is designed and the Switching devices are selected according to the results. Also the power losses of different components and Efficiency of Boost converter is calculated.