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

Title = Loss calculation for a DC/DC  Boost converter-MATLAB

Q)   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.

= BOOST CONVERTER

The boost converter is used to "step-up" an input voltage to some higher level, required by a load. This unique capability is achieved by storing energy in an inductor and releasing it to the load at a higher voltage.

The figure given below is the circuit representation of the boost converter:

In the Boost converter, the supplied fixed DC input is boosted (or increased) to adjustable DC output voltage i.e., output voltage of the boost converter is always greater than the input voltage. So, a Boost converter is also called a step-up converter or step-up chopper.

Mode I: Switch S is ON and Diode D is OFF

In this mode of operation, switch S is in closed condition i.e., ON state, and diode D is in open condition i.e., OFF state. Thus, switch S allows the flow of current through it. All the current will flow through the closed path including inductor L, switch S, and back to the dc input source. The circuit diagram for this mode is shown in the figure below.

Mode II: Switch S is OFF and Diode D is ON

In this mode of operation, switch S is in open condition i.e., OFF state and diode D is in closed condition i.e., ON state. Thus, switch diode D allows the flow of current through it, whereas switching S blocks the current flow through it. The circuit diagram for this mode is shown in the figure below.

Design Values:

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

 

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

the Boost Converter simulation model is shown below:

Model explanation:

To model the boost converter, a dc voltage source is taken as input supply with 100v. and an Inductor(L) and rl is connected in series with input supply.

For switching operations, a MOSFET is connected parallel to the inductor and diode connected in series. The capacitor and Resistive loads are connected in parallel to the diode.

MOSFET and DIODE blocks measurement ports are connected to a demux block to separate the signals and connected to scope individually.

A voltage measurement block is connected across the Load resistor to observe the output voltage and it was given to display block.

A current measurement block is connected in series with the load to observe the output current and it was given to display block.

Each block’s properties are shown below:

 Result =

output voltage

from the above display ,we can clearly observed that ,the output voltage of the boost converter is reached to 200 V.

the input 100 V is step up to 200 V by using the boost converter and the desired output current is 20 A.

 MOSFET.Diode Volatge and current scope result 

from the above scope ,we can observe that during switch off the diode exhibits the reverse voltage characteristics .

 we can observe that the peak voltage is 300 V.therefore we have to select the MOSFET that can withstand a voltage of 300 V for safer side.

The spike at the starting is eliminating snubber and a DC link capacitance.

let us go to INFINEON to select the MOSFET for this application.

datasheet link

 https://drive.google.com/file/d/1FxJ2EQzE_eVKlFkEQ9oWMEf3ZvkbvP_v/view?usp=share_link

The above 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 follows: -

1. Conduction Losses
2. Diode losses
3. Induction Losses
4. 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.

Formula: -

                     Conduction Losses: - Pc= I² * R *D-----------------(1)

 were,

p= Power Dissipation

I= Current through MOSFET

R= Resistance in Ohm

D=Duty Cycle

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: - P= Vf * I* (1-D) ------(2)       

 were,

Vf= forward voltage

I= Current output

D=Duty Cycle

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.

 These losses can be calculated by using the below formula:

             PI= I² * rl + {(rl / 12) (D* Vin / F*L)^2} -------(3)

 where,

I= Current output

D=Duty Cycle

F= Frequency

L= Inductance value

Capacitance losses: -

        While getting the output through the capacitor we can get some power dissipation in capacitor also.

this power dissipation is determined by using the following Formula: -

             Pc= {rc* D*(1-D) I² + {rc / 12*(1-D)}} *{(D*Vin / L*F)} ----(4)

 were,

I= Current output

D=Duty Cycle

F= Frequency

C= Capacitance value.

Calculation of total power converter losses and calculation of efficiency.

Conclusion:

The Boost Convertor is designed and modelled in Simulink and the switching device is selected from the retailer as well as the power losses of different components and also the switching device is determined as well as the efficiency of the designed converter is calculated.

Total power convertor losses= 34 W

Efficiency of convertor= 96.6 %