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

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: To design of boost converter, and to calculate the loss of different components including power switching device and deriving efficiency of boost converter.

BOOST CONVERTER:

A boost converter (also known as step-up converter) is one of the simplest types of switch-mode converters. As the name suggests, the converter takes an input voltage and boosts it. In other words, its like a step up transformer i.e it step up the level of DC voltage (while transformer step up / down the level of AC voltage) from low to high while decreases the current from high to low while the supplied power is same.

All it has is an inductor, a semiconductor switch, a diode and a capacitor. The boost converter is very simple and requires very few components.The biggest advantage of a boost converter is it offers very high efficiency. Some of the boost converters can go up to 99% efficiency. That means of the input voltage only 1% of the power is wasted.

The input voltage source is connected to an inductor. The solid-state device which operates as a switch is connected across the source. The second switch used is a diode. The diode is connected to a capacitor, and the load and the two are connected in parallel as shown in the figure above.

The inductor connected to input source leads to a constant input current, and thus the Boost converter is seen as the constant current input source. And the load can be seen as a constant voltage source. The controlled switch is turned on and off by using Pulse Width Modulation(PWM). PWM can be time-based or frequency based. Frequency-based modulation has disadvantages like a wide range of frequencies to achieve the desired control of the switch which in turn will give the desired output voltage. Time-based Modulation is mostly used for DC-DC converters. It is simple to construct and use. The frequency remains constant in this type of PWM modulation. The Boost converter has two modes of operation. The first mode is when the switch is on and conducting.

Mode I : Switch is ON, Diode is OFF

The Switch is ON and therefore represents a short circuit ideally offering zero resistance to the flow of current so when the switch is ON all the current will flow through the switch and back to the DC input source. Let us say the switch is on for a time T_ON and is off for a time T_OFF. We define the time period, T, asand the switching frequency,

Let us now define another term, the duty cycle,

Let us analyze the Boost converter in steady state operation for this mode using KVL.

Since the switch is closed for a time T_ON = D_T we can say that Δt = DT.

While performing the analysis of the Boost converter, we have to keep in mind that

1. The inductor current is continuous and this is made possible by selecting an appropriate value of L.
2. The inductor current in steady state rises from a value with a positive slope to a maximum value during the ON state and then drops back down to the initial value with a negative slope. Therefore the net change of the inductor current over anyone complete cycle is zero.

Mode II : Switch is OFF, Diode is ON

In this mode, the polarity of the inductor is reversed. The energy stored in the inductor is released and is ultimately dissipated in the load resistance, and this helps to maintain the flow of current in the same direction through the load and also step-up the output voltage as the inductor is now also acting as a source in conjunction with the input source. But for analysis, we keep the original conventions to analyze the circuit using KVL.

Let us now analyse the Boost converter in steady state operation for Mode II using KVL.

Since the switch is open for a timewe can say that.

It is already established that the net change of the inductor current over any one complete cycle is zero.

Boost Converter Parameters:

Designing Boost Converter for converting 100V to 200V. The Output power will be 2KW.

So by using P = V*I

`I = P/V`

`I = 2000/100`

`I_L = 20A`

So Load Resistor will be

`R_(Load) = V/I`

`R_(Load) = 200/20`

`R_(Load) = 10 Ohm`

1. Duty Cycle:

Considering efficiency of 85%

Then D = 1 - 100*0.85/200

        D = 0.575

2. Inductor Value:

Considering Switching frequency of 20KHz

To calculate Inductor Ripple Current good estimation for the ripple current is 20% to 40% of the output current.

Considering 20% of Ripple Current

`ΔI_L = 0.2 * 20 * 200/100`

`ΔI_L = 8A`

So L = `(100 * (200-100))/(8*20*1000*200)`

L = 3.125mH

3. Capacitor:

`ΔV_(out) = (20/(1-0.575) + 8/2)`

`ΔV_(out) = 51.05V`

`C_(out) = (20*0.575)/(20*1000*51.05)`

C = 11.26microH

rL = 10 mOhms

rC = 2 mOhms

BOOST COVERTER SIMULINK MODEL:

Blocks Used:

• DC Voltage source
• MOSFET
• Diode
• Voltage Measurement block
• Current Measurement block
• PWM Generator
• Series RLC branch
• Display
• Bus selector
• Power gui
• Scope

• 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 10Ohms is Connected at output. 
• A voltage measurement and current measurement block is used to measure the Output voltage and output current.
• A closed loop feedback is used to generate the gate pulses for MOSFET. The output voltage subtracted from Reference voltage genertates the error and it is fed to PI controller which rectifies the error and generates the Gate pulses. These signal is given to DC-DC PWM Generator block which generates the High Switching Frequency gate pulses.

• The 1st image shows the Input voltage, Output current and Output Voltage of Boost Convereter. The 100V input is converted to 200V and the output gets stable in about 0.05seconds. The output current flowing is 20A which matches with the calculated value.
• The 2nd image shows the MOSFET Current and Voltage. When the MOSFET turns ON then the peak current rise is upto 160A and then the current flowing is about 50A. The maximum voltage flowing through Voltage is 200V.
• The last image shows the Diode Voltage and Diode Current Characteristics.

Losses in Converter:

In any DC-DC converter there are 4 types of losses:-

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.

where, I = current through MOSFET

           R = Resistance in ohm

           D = duty cycle

`I_m = P_i/V_i`

Pi = `V^2/R`

Pi = `100^2/10`

Pi = 1000

`I_m = 1000/100`

Im = 10A

Pc = `10^2 * 0.023*0.575`

Pc = 1.3225W

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.

Vf = Diode Forward voltage 

I = Output Current

D = Duty cycle 

Vf of Diode = 0.8 

Iout = 19.94A

Pd = 0.8* 20* (1 - 0.575)

Pd = 6.8W

Inductor 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 I= Current

         D= duty cycle 

         f =  Frequency

         L= inductance 

 Pl = `20^2 * 10*10^-6 + (10*10^-6)/12 *(0.575*100)/(50 * 3.125*10^-3)`

Pl = 0.0043W

Capacitance losses:-

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

This power dissipation is determined by using the  Formula:-

Pc = `2*10^-6*0.575(1-0.575)*20^2+(2*10^-6)/12*(1-0.575)*(0.575*100)/(11.12*10^-6*50)`

Pc = 0.00752W

Total Power Losses = Pcond + Pdiode + Pind +Pcap

                                 = 1.3225 + 6.8 + 0.0043 + 0.00752

                                 = 8.13W

Total Power Losses = 8.13W

Efficiency = (2000 - 8.13)/2000

Efficiency = 99.5%