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

AIM:

• Extending the learned lessons in previous module, you will be diving into another power electronics concept on your own.
• 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.

Solution:

Boost Converter (step-up converter):

• The 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).

Fig.Boost converter circuit diagram.

For designing the Boost converter model in Simulink environment we have taken following blocks:

• DC Voltage source
• MOSFET
• Diode
• series RLC branch
• Voltage measurment block
• PWM generator
• Display
• Bus selector
• Power gui
• Scope

We have given design parameters from the standard datasheet for calculations and modelling puprpose are as follows,

• Vin = 100 V
• Vout = 200 V
• L = 200uH
• C = 400uF
• R= 10 Ohms
• rL = 10 mOhms
• rC = 2 mOhms
• rd=5 mOhms

To calculate duty cycle:-

Vo / Vin = (1/ (1-D))

Where, Vo= Output voltage

Vin= Input voltage

D= Duty cycle

from the given data ,Vin = 100 V & Vout = 200 V

Put this values in the above equation, we get

Vo / Vin =(1/ (1-D) )

200/100 = (1/ (1-D) )

2*(1-D)= 1

(1-D) = 0.5

D=0.5

Duty cycle = 0.5

Simulink Model :-

The simulink model of Boost conveter is in shown below,

Procedure:

• For designing the simulink model of DC-DC Boost converter the supply voltage and Series RL branch are connected in series and MOSFET is act as power switch and it is connected across the supply voltage.
• After that a Capacitor Cand load resistance Rc are connected in series and the whole combination is connected parallel in the circuit.
• For ON and OFF Switch operation, we have connected a pulse generator as input of pulses having duty cycle of 50%.
• The output voltage will be displayed by using the display block.
• The voltage and currents drawn by both "MOSFET " and Diode" is plotted with the help of scope and signals from this blocks are connected to the scope via bus selector.

Design Block Parameters Values:

Dc Voltage Source

MOSFET Parameters

Diode Parameters

Pulse Generator Parameter Block

Series RL branch:

Series RC branch:

Resistance:

Output of Boost Converter :

Output voltage :-

• From the above graph we can see the Output voltage is grater than the Input voltage
• Hence the Working of Boost Converter is working in proper manner
• The Minmum and maximum Voltage will be displayed by using the Below Signal statistics block.

• The Minimum Voltage is 1.146 V
• The Maximum voltage is 198.1 V

MOSFET Voltage and Current:

• In the above graph the top graph shows the Mosfet Current Graph and Bottom Graph shows MOsfet Voltage graph.
• From the graph we analyzed that the Duty cycle of the signal is 0.5 or 50%.
• Maximum Current =39.71 Amps
• Minimum current = 0.0Amps
• The minimum Voltage in MOSFET = 1.946 V
• the maximm voltage in MOSFET= 198.9 V

DIODE VOLTAGE AND CURRENT:-

• Above graph shows current passes through the diode and voltage appears across the diode resp.
• The Duty cycle of the signal is 50% shown the graph.
• Maximum Current =39.71 Amps
• Minimum current = 11.46 Amps

Calculation for Losses in Power electronic Converters:

From the given data we can find the losses for designed DC-DC boost Converter:

• Vin = 100 V
• Vout = 200 V
• L = 200uH
• C = 400uF
• R= 10 Ohms
• rL = 10 mOhms
• rC = 2 mOhms
• rd=5 mOhms
• Duty cycle= 0.5

In any DC-DC converter majorly there are four types of losses are as follows,

• Conduction Losses
• Diode losses
• Induction Losses
• 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.

Where,

Conduction Losses Pc = (1)² * 10*0.5

Conduction Losses (Pc)=5W

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:-

Where,

Diode Losses Pd = 19.55 * 1 * (1-0.5) = 19.55*0.5

Diode Losses (Pd) = 9.75 W

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.

Where,

Induction Losses Pd=1² * 10 +(0.0008*5000) = 10+4

Induction Losses Pd= 14 W

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,

Where,

Capacitance Losses Pc=2*0.5*0.5+(2*0.0001*0.5*0.5*100 / 400*50*0.000001)

Capacitance Losses Pc= 0.7085 W

Total Losses= Pcon+Pd+Pi+Pc

=5+9.775+1.4+0.7085

The total Power losses Ploss= 11.75 W

Efficiency Of Boost Converter:

The efficiency of boost converter is determined by using below formula:

Efficiency = (Pin -Ploss / Pin )100

So, from the given data

Pin= Vin * I

I= Vin / R =100 / 10

I= 10 Amps

Therefore, Pin = Vin * I=100 * 10

Pin=1000 W

From above equations, we have putting values of Pin, Ploss and we get,

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 %