LAB PART3: POWER ELECTRONICS
Lab 2: DC-DC Converter
  1. OBJECTIVES
     1. To apply concepts of DC/DC Converter
     2. To learn how to implement basic power electronic circuit
     3. To get familiar with a power electronic simulation platform
Required Component
  a. MATLAB simulation software.
                 READ THE ENTIRE DUIDE BEFORE PROCEEDING
          Follow the Question Provided in Lab to Avoid Unnecessary Loss of Marks
  2. BUCK CONVERTER
     A buck converter (step-down converter) is a DC-to-DC power converter which steps
     down voltage (while stepping up current) from its input (supply) to its output (load). The
     Buck Converter circuit is shown in Fig.1.
     Simulation Procedure
     a.   Build the model to simulate this circuit using the following values of circuit
          components:
              Vin = 36 V
              Inductance L = 40 uH
              Capacitance C = 5.21 uF
                Resistance R = 1.2 Ω
                MOSFET gate signal 100 kHz
a.   For duty cycle equal to 25%, observe output voltage and current then record the
     waveform. Repeat when duty cycle equal to 50% and 75%.
                                                                     For D = 25%
   For D = 50%
   For D = 75%
      b. For what value of duty cycle that make Vout = 0.5 Vin?
                      Vout
         We have D=
                       Vin
                      1
         Where V out = V ¿
                      2
                   1
         Then D = =0.5=50 %
                   2
      c. Compare the result from simulation with the result from the formula.
         We have D = 50%
                 V out =D× V ¿
                         50 ×32
                 V out =        =16 V
                          100
          And from signal statistic we get
                  V out =V mean=15.64 V
Thus Vout from formula = 16V
     Vout from simulation = 15.64 V
       d. Calculate output voltage ripple.
                                                    ∆ V o 1−D
           We have Output ripple Voltage formula:        =
                                                     V o 8 LC f 2
           and L = 40 uH
               C = 5.21 uF
                f = 100kHz
                  For D = 25%
                   ∆Vo          1−0.25
                       =                            =0.044
                    V o 8∗40e-6∗5.21e-6∗( 100e+3 )2
                  For D = 50%
                   ∆Vo           1−0.5
                       =                            =0.029
                    V o 8∗40e-6∗5.21e-6∗( 100e+3 )2
                  For D = 75%
                   ∆Vo          1−0.75
                       =                            =0.015
                    V o 8∗40e-6∗5.21e-6∗( 100e+3 )2
   3. BOOST CONVERTER
      A boost converter (step-up converter) is a DC-to-DC power converter which steps up
      voltage (while stepping down current) from its input (supply) to its output (load). The
      Boost Converter circuit is shown in Fig.2.
      Simulation Procedure
a. Build the model to simulate this circuit using the following values of circuit components:
               Vin = 9 V
               Inductance L = 38 uH
   Capacitance C = 50 uF
   Resistance R = 12 Ω
   MOSFET gate signal 100 kHz
b. For duty cycle equal to 25%, observe output voltage and current then record the waveform.
   Repeat when duty cycle equal to 50% and 75%.
               For D = 25%
                 For D = 50%
                 For D = 75%
c. For what value of duty cycle that make Vout = 2 Vin?
                             V¿
         We have D=1−
                           V out
         Where V out =2V ¿
         Then D=50 %
d. Compare the result from simulation with the result from the formula.
                For D = 25%
                                 V
                  Then V out = ¿
                               1− D
                  Where ¿ V
                         V   =9
                  Thus V out =12V
                   For D = 50%
                                 V
                    Then V out = ¿
                                1− D
                    Where V ¿ =9 V
                    Thus V out =18V
                   For D = 75%
                                 V
                    Then V out = ¿
                                1− D
                    Where V ¿ =9 V
                    Thus V out =36 V
e. Calculate output voltage ripple.
                                           ∆Vo    D
   Voltage ripple for Boost converter is       =
                                            Vo   RCf
   Where R = 12 Ohm
         C = 50 uF
         f = 100k Hz
     D = 25%
            ∆Vo        0.25
                =                =0.0041
             V o 12∗50e-6∗100e+3
      D = 50%
            ∆Vo        0.5
                =                =0.0083
             V o 12∗50e-6∗100e+3
      D = 75%
            ∆Vo        0.75
                =                =0.0125
             V o 12∗50e-6∗100e+3
4. BUCK-BOOST CONVERTER
The buck–boost converter is a type of DC-to-DC converter that has an output voltage
magnitude that is either greater than or less than the input voltage magnitude. The Buck/Boost
converter circuit is shown in Fig.3.
   Simulation Procedure
a. Build the model to simulate this circuit using the following values of circuit components:
            Vin = 9 V
            Inductance L = 38 uH
            Capacitance C = 50 uF
            Resistance R = 12 Ω
            MOSFET gate signal 100 kHz
b. For duty cycle equal to 25%, observe output voltage and current then record the
   waveform. Repeat when duty cycle equal to 50% and 75%.
       For D = 25%
   For D = 50%
          For D = 75%
c. For what value of duty cycle that make Vout = 0.5Vin and Vout = 2 Vin?
                          −D
      We have V o =V s (        )
                         1−D
                        1
       Where V out = V ¿
                        2
                        V out
                 D=
                     V out −V ¿
                    1
         Then D= =33.33 %
                    3
       Where V out =2V ¿
                    2
         Then D= =66.66 %
                    3
d. Compare the result from simulation with the result from the formula.
       For D = 25%
                           D∗V ¿
         We have V out =
                           1−D
    Where V ¿ =9 V
    Thus V out =3V
    And from simulation we goV out =2.43V
   For D = 50%
    We get V out =9V
    And from Simulink we go V out =8.089 V
   For D = 75%
    We get V out =27 V
    And from Simulink we got V out =23.78V
e. Calculate output voltage ripple
                                                ∆Vo    D
   Voltage ripple for Buck-Boost converter is       =
                                                 Vo   RCf
   Where R = 12 Ohm
           C = 50 uF
           f = 100k Hz
      D = 25%
            ∆Vo        0.25
                =                =0.0041
             V o 12∗50e-6∗100e+3
      D = 50%
            ∆Vo        0.5
                =                =0.0083
             V o 12∗50e-6∗100e+3
      D = 75%
            ∆Vo        0.75
                =                =0.0125
             V o 12∗50e-6∗100e+3
5. CONVERER WITH CONTROLLER
   To get better output voltage waveform, controller is used to help the system. PI controller
   is the most suitable one for this of application. The Buck/Boost converter circuit is shown
   in Fig.4.
a. Build the model to simulate this circuit using the following values of circuit components:
          Vin = 36 V
          Inductance L = 40 uH
          Capacitance C = 5.21 uF
          Resistance R = 1.2 Ω
          MOSFET gate signal 100 kHz
          Repeating sequence: time values: [0 1/100000], output values: [0 1]
.
    b. For PI controller, K p and K i value can take as trial and error.
    c. For what value of duty cycle that make V out =0.5 V ¿ ? observe output voltage and current
       and record the waveform.
                     T on
       We have D=          ×100 %
                     T off
       From scope of PWM we get
T on=2.241 μs
T off =1.943 μs
          2.242
→ D=               ×100 %=53.56 %
       2.242+1.943
       From scope of Vout we got V out =17.99V
           For D=53.56 V out =17.99
           For D=x V out =0.5V ¿=18
                 18 ×53.56
              x=           =53.58 %
                   17.99
Thus when we want to make V out =0.5 V ¿ , we take duty cycle D=53.58 %
   d. Calculate output voltage ripple and compare with result from question ‘d’ point 2.
                                             ∆ V o 1−D
      We have Output ripple Voltage formula:      =
                                              V o 8 LC f 2
          From question ‘d’ point 2
              L = 40 uH
              C = 5.21 uF
               f = 100kHz
          For D = 50%
                  ∆Vo           1−0.5
                      =                            =0.029
                   V o 8∗40e-6∗5.21e-6∗( 100e+3 )2
             ‘d’ point 5
                L = 40 uH
                C = 5.21 uF
                MOSFET gate signal 100 kHz
                  ∆Vo           1−0.54
                       =                           =0.027
                   V o 8∗40e-6∗5.21e-6∗( 100e+3 )2
           When we value of D increased, thus the value of output voltage ripple is
           discrease.
e. For what value of K p and K i that provide better result? Provide that output waveform
   To Provide better result we need to take Value of K p ≫ k i
   Then we get:
The signal of Voutput have value of V(peak to peak) = 0.0978V, thus we have get a
smooth DC voltage.
Conclusion: the average output Voltage from Simulink have a small value difference
from Calculation by formular, because in experiment in Simulink or reality there is some
voltage drop on cable and diode.