Project 1

The project design report should include the following:

• A Design report to show the reasoning behind your choices of inductance and capacitance levels in your design.
• The full schematics of your converter including gate drivers, PWM chips, RC circuits, current sensors, voltage sensors, etc…
• A complete simulation file including all the necessary waveforms and justifying all the chosen components sizing (you have to show the system ability to track the output voltage at different dynamic conditions).

The full specs for all needed components in your circuits.

Note: Feel free to deviate from the provided specs as long as you are using the assigned converter. However, you cannot copy the specs of another assigned converter

Solution:

Three converters i.e., Buck converter, Boost converter and SEPIC converter are chosen for this project and the details of each converter along with its simulation results are discussed below.

MOSFET:

To make this simulation for industrial or practical application the MOSFET switch considered is NTB5860NL n channel power MOSFET. The specification of this MOSFET switch is given below.

1. SEPIC CONVERTER

Input Voltage= 30-60V

Average Input Voltage= 45V

Output Voltage= 45V

Power = 1kW

Switching Frequency = 50 kHz

As per given data a SEPIC Converter can be used for the application.

Therefore, Duty cycle for SEPIC converter can be calculated as,

`V_o/V_s = D/(1-D)`

So,

On solving the above equation when V_s=45V, D= 0.5

Load resistance can be calculated as

`P_o= 〖V_o〗^2/R`

P_o= 1kW

V_o= 45V

So, on solving for R,

R=2.025 ohm

Inductor Current can be calculated using

`I_L= (V_o I_0)/V_s`

`P_o= V_o I_o`

`I_o= P_o/V_o`

I_o= 22.2A

So, on solving the above equation, I_L= 22.2A

The variation in current is assumed as 10%, so = 2.22A

If the switching frequency is 50 kHz, then the inductance L₁ for continuous current is determined from Equation,

`L_1=(V_s D)/(〖∆i〗_L1 f)`

`L_1=(45*0.5)/(2.22*50000)`

On solving the above equation,

L₁= 2.02*10^-4 H

For L2, the average current is determined from Kirchhoff’s current law at the node where C1, L2, and the diode are connected. The average current in each capacitor is zero, so the average current in L2 is

I_L2=I_o

I_L2=2.22A

The variation in I_L2 is determined from the circuit when the switch is closed. Using Kirchhoff’s voltage law around the path of the closed switch, C1, and L2 with the voltage across C1 assumed to be constant Vs. The minimum variation in current of 10 percent is determined from Equation,

`〖∆i〗_L1=22.2*0.1= 2.22 A`

`L_2=(V_s D)/(〖∆i〗_L2 f)`

`L_1=(45*0.5)/(2.22*50000)`

L₁= 2.02*10^-4 H

The output stage consisting of the diode, C2, and the load resistor is the same as in the boost converter, so the output ripple voltage and variation in capacitor C₁ is assumed to be 10%.

`C_2≥D/(〖∆V〗_o/V_o R*f)`

On solving above equation, C₂

_{`C_2≥49.4*10^(-6) F`}

The voltage variation in C1 is determined from the circuit with the switch closed. Capacitor current i­_C1 is the opposite of i_L2,

_So,

_{`C_1≥D/(〖∆V〗_C1/V_o R*f)`}

On solving the above equation C₁= 49.4*10^-6F

Schematic Diagram of SEPIC converter:

Simulation:

Zoom IN Simulation Result:

Output Voltage and Output Current:

Zoom OUT Simulation Result:

Output Voltage and Output Current:

Result:

S.N	Parameter	Theoretical Result	Simulation Result
1.	Source Voltage	30-60V	45-60V
2.	Average Voltage	45V	44.9V
3.	Switching Frequency	50kW	50kW
4.	Inductor, L1	2.02e-4H	2.02e-4H
5.	Power	999.67W	1000W
6.	Inductor, L2	2.02e-4H	2.02e-4H
7.	Capacitor, C1 and C2	49.4e-6F	49.4e-6F
8.	Load Resistance	2.025 ohm	2.025 ohm
9.	Duty Cycle	0.6>D>0.428	0.5>D>0.428
10.	Output Voltage	45V	44.9V
11.	Output Current	22.2A	22.2A

Conclusion: SEPIC converter is designed as per given data and result obtained through simulation proves theoretical result.

2. BOOST CONVERTER

Input Voltage= 20-30V

Average Input Voltage= 25V

Output Voltage= 40V

Power = 600W

Switching Frequency = 50 kHz

As per the given data, a Boost Converter can be used for the application.

Therefore, Duty cycle for Boost converter can be calculated as,

`V_o/V_s = 1/(1-D)`

So,

`1-V_s/V_o = D`

On solving the above equation when V_s=25V, D= 0.375

Load resistance can be calculated as

`P_o= 〖V_o〗^2/R`

P_o= 600W

V_o= 40V

So, on solving for R,

R=2.67 ohm

Inductor Current can be calculated using

`I_L= 〖V_o〗^2/(V_s*R)`

So, on solving the above equation, I_L= 23.97A

If the switching frequency is 50 kHz, then the minimum inductance for continuous current is determined from Equation,

`L_min=(〖D(1-D)〗^2*R)/2f`

On solving the above equation,

L_min= 3.91*10^-6 H

A boost converter designed for continuous-current operation will have an inductor value greater than L_min,

To provide a margin to ensure continuous current, let L= 5*10^-6H

The minimum capacitance required to limit the output ripple voltage to 10 percent is determined from Equation,

`C≥D/(〖∆V〗_o/V_o R*f)`

`C≥56.18*10^(-6) F`

Schematic Diagram of Boost Converter:

Simulation

Zoom IN Simulation Result:

Output Voltage and Output Current:

Zoom OUT Simulation Result:

Output Voltage and Output Current:

S.N	Parameter	Theoretical Result	Simulation Result
1.	Source Voltage	20-30	25-30
2.	Duty Cycle	0.5>D>0.25	0.375>D>0.25
3.	Switching Frequency	50kW	50kW
4.	Average Voltage	25V	25V
5.	Inductor, L	3.91e-6H	5e-6H
6.	Power	600W	594.51W
7.	Capacitor, C	56.18e-6F	56.18e-6F
8.	Load Resistance	2.67 ohm	2.69 ohm
9.	Output Voltage	40V	39.9V
10.	Output Current	15A	14.9A

Conclusion: Boost converter is designed as per given data and result obtained through simulation proves theoretical result.

3. BUCK CONVERTER

Given,

Input voltage= 40-50V

Output Voltage= 30V

Power = 500 W

Assuming Switching Frequency to be 50 kHz and input voltage to be 45V

We know that,

Duty cycle D, = V_out/ V_in

So D= 30/45

D= 0.66 or Duty cycle is 66%

Resistance R= (Vout)²/P_o

R= (30)²/500 =1.8 ohms

Let us design for 10% ripple in the inductor

The switching frequency and inductor size must be selected for continuous-current operation. Let the switching frequency arbitrarily be 50 kHz, which is well above the audio range and is low enough to keep switching losses small. The minimum inductor size is determined from Equation:

So

`L_min= ((1-D)R)/〖2f〗_s`

`L= (1-0.66)1.8/(50000*2)`

`L= 6.12μH`

Let the inductor be 25 percent larger than the minimum to ensure that inductor current is continuous.

`L_min= 1.25*6.12μH`

`L= 7.65μH`

Capacitor Designing

The output voltage ripple is 5% assumed,

`C_out= ((1-D))/(8L*〖f_s〗^2* 〖δV〗_o/V_o )`

`C_out= ((1-0.66))/(8*7.65*10^(-6)*〖(50k)〗^2* 0.05)`

On solving, we get,

`C_out= 45 μF`

Schematic Diagram of Buck Converter:

Simulation:

Zoom In Simulation Result:

Output Voltage and Input Voltage:

Zoom OUT Simulation Result:

Output Voltage and Output Current:

S.N	Parameter	Theoretical Result	Simulation Result
1.	Source Voltage	40-50	45-50
2.	Duty Cycle	0.5>D>0.25	0.375>D>0.25
3.	Switching Frequency	50kW	50kW
4.	Average Voltage	45V	45V
5.	Inductor, L	6.1e-6H	7.65e-6H
6.	Power	500W	495.15W
7.	Capacitor, C	45e-6F	45e-6F
8.	Load Resistance	1.8 ohm	1.83 ohm
9.	Output Voltage	30V	30.1V
10.	Output Current	16.67A	16.45A

Conclusion: Buck converter is designed as per given data and result obtained through simulation proves theoretical result.