The document is a final report on a boost converter module designed using a microcontroller for PWM control. It was created by three students for their Industrial Electrical Engineering lecture. The report includes an introduction to boost converters and their operating principles. It then details the design of the specific boost converter module including calculating component values like the inductor, MOSFET, diode and capacitor based on the design parameters of input voltage range, output voltage, current, and switching frequency.

1. FINAL REPORT
BOOST CONVERTER MODULE : MICROCONTROLLER
PWM
Group Members:
1. Abdillah Aziz Munthasir (1310161018)
2. Mohammad Agung Dirmawan (1310161024)
3. Gilang Andaru Trinanda (1310161027)
Class :
3-D4 ELIN A
Lecture :
Ir. Moh. Zaenal Effendi, M.T.
INDUSTRIAL ELECTRICAL ENGINEERING
ELECTRO DEPARTEMENT
ELECTRONICS ENGINEERING POLYTECHNIC INSTITUTE OF
SURABAYA
2019

2. 1. Introduction
A boost converter (step-up converter) is a DC-to-DC power converter that 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 transistor) 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).
Figure 1.1. Boost Converter Schematic
Operation
The key principle that drives the boost converter is the tendency of an inductor to resist
changes in current by creating and destroying a magnetic field. In a boost converter, the
output voltage is always higher than the input voltage. A schematic of a boost power stage
is shown in Figure 1.
a. When the switch is closed, current flows through the inductor in clockwise direction
and the inductor stores some energy by generating a magnetic field. Polarity of the
left side of the inductor is positive.
b. When the switch is opened, current will be reduced as the impedance is higher. The
magnetic field previously created will be destroyed to maintain the current towards
the load. Thus the polarity will be reversed (meaning the left side of the inductor will
become negative). As a result, two sources will be in series causing a higher voltage
to charge the capacitor through the diode D.
If the switch is cycled fast enough, the inductor will not discharge fully in between
charging stages, and the load will always see a voltage greater than that of the input source
alone when the switch is opened. Also while the switch is opened, the capacitor in parallel
with the load is charged to this combined voltage. When the switch is then closed and the
right hand side is shorted out from the left hand side, the capacitor is therefore able to provide
the voltage and energy to the load. During this time, the blocking diode prevents the

3. capacitor from discharging through the switch. The switch must of course be opened again
fast enough to prevent the capacitor from discharging too much.
Figure 1.2 The two current paths of a boost converter, depending on the state of the switch S.
The basic principle of a Boost converter consists of 2 distinct states (see figure 2):
 in the On-state, the switch S (see figure 1) is closed, resulting in an increase in the
inductor current;
 in the Off-state, the switch is open and the only path offered to inductor current is
through the flyback diode D, the capacitor C and the load R. This results in
transferring the energy accumulated during the On-state into the capacitor.
 The input current is the same as the inductor current as can be seen in figure 2. So it
is not discontinuous as in the buck converter and the requirements on the input filter
are relaxed compared to a buck converter.
 When switch S is ON
When switch in ON the diode will be open circuited since the n side of diode is at
higher voltage compared to p side which is shorted to ground through the switch. Hence the
boost converter can be redrawn as follows.During this state the inductor charges and the
inductor current increases. The current through the inductor is given as
𝐼𝐿 = (
1
𝐿
)∫ 𝑉 𝑑𝑡
Assume that prior to the opening of switch the inductor current is I’L, off. Since the input
voltage is constant

4. 𝐼𝐿.𝑜𝑓𝑓 = (
1
𝐿
)∫( 𝑉𝑖𝑛) 𝑑𝑡 + 𝐼′
𝐿,𝑜𝑓𝑓
Assume the switch is open for ton seconds which is given by D 𝑥 𝑇𝑆 where D is duty cycle
and Ts is switching time period. The current through the inductor at the end of switch on
state is given as
𝐼𝐿.𝑜𝑛 = (
1
𝐿
) 𝑥 𝑉𝑖 𝑛 𝑥 𝐷 𝑥 𝑇𝑆 + 𝐼′
𝐿,𝑜𝑛 (equation 1); Hence Δ𝐼𝐿 = (
1
𝐿
) 𝑥 𝑉𝑖𝑛 𝑥 𝐷 𝑥 𝑇𝑆.
 When switch is off
When switch in OFF the diode will be short circuited and the boost converter circuit
can be redrawn as follows. The inductor now discharges through the diode and RC
combination. Assume that prior to the closing of switch the inductor current is I’’L, off. The
current through the inductor is given as
𝐼′′′
𝐿,𝑜𝑓𝑓 = −(
1
𝐿
) ∫( 𝑉𝑖𝑛 − 𝑉𝑜𝑢𝑡 ) 𝑑𝑡 + 𝐼′′
𝐿,𝑜𝑓𝑓
Note the negative sign signifies that the inductor is discharging. Assume the switch is open
for toff seconds which is given by (1-D) 𝑇𝑆 where D is duty cycle and Ts is switching time
period. The current through the inductor at the end of switch off state is given as
𝐼′′′
𝐿,𝑜𝑓𝑓 = − (
1
𝐿
)( 𝑉𝑖𝑛 − 𝑉𝑜𝑢𝑡 )(1 − 𝐷) 𝑇𝑆 + 𝐼′′
𝐿,𝑜𝑓𝑓 (equation 2)
In steady state condition as the current through the inductor does not change abruptly, the
current at the end of switch on state and the current at the end of switch off state should be
equal. Also the currents at the start of switch off state should be equal to current at the end
of switch on state. Hence
𝐼′′′
𝐿,𝑜𝑓𝑓 = 𝐼𝐿,𝑜𝑛 , also 𝐼′′
𝐿,𝑜𝑓𝑓 = 𝐼′′
𝐿,𝑜𝑓𝑓
Using the equations 1 and 2 we get
(
1
𝐿
) 𝑥 𝑉𝑖𝑛 𝑥 𝐷 𝑥 𝑇𝑆 = −(
1
𝐿
) 𝑥 ( 𝑉𝑖𝑛 − 𝑉𝑜𝑢𝑡 ) 𝑥 (1− 𝐷) 𝑇𝑆
𝑉𝑖𝑛 𝑥 𝐷 = −( 𝑉𝑖𝑛 − 𝑉𝑜𝑢𝑡 ) 𝑥 (1− 𝐷)
𝑉𝑖𝑛 𝑥 (𝐷 − 1 + 𝐷) = 𝑉𝑜𝑢𝑡 𝑥 (1 − 𝐷)
𝑉𝑖𝑛
𝑉𝑜𝑢𝑡
=
1
1 − 𝐷

5. Since D < 1 𝑉𝑜𝑢𝑡 > 𝑉𝑖𝑛. Assuming no losses in the circuit and applying the law of
conservation of energy : 𝑉𝑜𝑢𝑡 𝑥 𝐼𝑜𝑢𝑡 = 𝑉𝑖𝑛 𝑥 𝐼𝑖𝑛
This implies 𝐼𝑜𝑢𝑡 /𝐼𝑖𝑛 = (1-D), Thus 𝐼𝑜𝑢𝑡 < 𝐼𝑖𝑛. As the duty cycle increases the output voltage
increases and output current decreases. But due to parasitic elements in the lumped elements
resistor, inductor, capacitor the step up ratio Vout/Vin decreases at higher duty cycles and
approaches zero at unit duty cycle.
2. Inductor
Inductors are components that are simple in their construction, consisting of coils of
insulated copper wire wound around a former that will have some type of core at its centre.
This core might be a metal such as iron that can be easily magnetised; or in high frequency
inductors, it will more likely to be just air.
Inductors depend for their action on the magnetic field that is present around any
conductor when it is carrying a current. If the wire coil is wound around a core made of a
material that is easily magnetised, such as iron, then the magnetic field around the coil is
concentrated within the core; this greatly increases the efficiency of the inductor.
An inductor is characterized by its inductance, which is the ratio of the voltage to the
rate of change of current. In the International System of Units (SI), the unit of inductance is
the henry (H) named for 19th century American scientist Joseph Henry. In the measurement
of magnetic circuits, it is equivalent to weber/ampere. Inductors have values that typically
range from 1 µH (10−6 H) to 20 H. Many inductors have a magnetic core made of iron or
ferrite inside the coil, which serves to increase the magnetic field and thus the inductance.
Along with capacitors and resistors, inductors are one of the three passive linear circuit
elements that make up electronic circuits. Inductors are widely used in alternating current
(AC) electronic equipment, particularly in radio equipment. They are used to block AC while
allowing DC to pass; inductors designed for this purpose are called chokes.
A choke is an inductor designed specifically for blocking high-frequency alternating
current (AC) in an electrical circuit, while allowing DC or low-frequency signals to pass. It
usually consists of a coil of insulated wire wound on a magnetic core, although some consist
of a donut-shaped "bead" of ferrite material strung on a wire. Like other inductors, chokes
resist changes in current passing through them increasingly with frequency. The difference
between chokes and other inductors is that chokes do not require the high Q factor
construction techniques that are used to reduce the resistance in inductors used in tuned
circuits.

6. 3. Design of Boost Converter Module
Design of Boost Converter based on calculate some parameters and reference of data
sheet component like diode, MOSFET, capacitor, resistor etc.
The boost converter has following parameters :
𝑉𝑠(𝑚𝑎𝑥) = 30 Volt
𝑉𝑠(𝑚𝑖𝑛) = 24 Volt
𝑉𝑜 = 60 Volt
𝐼𝑜 = 2.5 Ampere
Switching Frequency (𝑓𝑠) = 40 kHz
Components :
Q : MOSFET IRF460
D : MUR 1560 (Ultra Fast Recovery Diode)
Inductor (L) : Ferrit core PQ 3535 with Cross sectional are (𝐴 𝑐 = 1.96 𝑐𝑚2
);
Bobbin diameter ( 𝐷 𝑏𝑜𝑏 = 17 𝑚𝑚)
𝑅 𝑠 : Snubber resistor (..... Ohm, 5 – 10 watt)
𝐶𝑠 : Snubber resistor (..... nF, 1 kVolt)
𝐷 𝑠 : Snubber diode (FR3017)
 Duty Cycle :
𝐷 = 1 − (
𝑉 𝑠(𝑚𝑖𝑛 )
𝑉𝑜
)
𝐷 = 1 − (
24
60
)
𝐷 = 1 − (
2
5
) 0.6 = 60%

7.  The Inductor Value :
𝑅 =
𝑉𝑜
𝐼𝑜
=
60
2.5
= 24 Ω
𝐼𝐿( 𝑎𝑣𝑔) =
𝑉𝑠
(1 − 𝐷) 2 𝑅
=
24
(1 − 0.6) 2 × 24
= 6.25 𝐴𝑚𝑝𝑒𝑟𝑒
∆𝐼𝐿 = 20% × 𝐼𝐿( 𝑎𝑣𝑔) = 0.2 × 𝐼𝐿( 𝑎𝑣𝑔) = 0.2 × 6.25 = 1.25 𝐴𝑚𝑝𝑒𝑟𝑒
𝐿 = (
1
𝑓
) × [( 𝑉𝑜 + 𝑉𝐹 ) − 𝑉𝑠(𝑚𝑖𝑛)]× (
𝑉𝑠(𝑚𝑖𝑛)
( 𝑉𝑜 + 𝑉𝐹 )
) × (
1
∆𝐼𝐿
)
𝐿 = (
1
40 × 103
) × [(60 + 1.2) − 24] × (
24
(60 + 1.2)
) × (
1
1.25
)
𝐿 = 2.5 × 10−5
× 37.2 × 0.3921568 × 0.8 = 2.91764× 10−4
= 291.764 𝜇𝐻
≈ 292 𝜇𝐻
 The Maximum Inductor Value :
𝑖 𝐿( 𝑚𝑎𝑥) = 𝐼𝐿( 𝑎𝑣𝑔) +
∆𝐼𝐿
2
= 6.25 +
1.25
2
= 6.25 + 0.625 = 6.875 𝐴𝑚𝑝𝑒𝑟𝑒
 Winding Number of Inductor :
𝑛 =
𝐿 × 𝑖 𝐿( 𝑚𝑎𝑥)
𝐵 𝑚𝑎𝑥 × 𝐴 𝑐
× 104
=
2.917647058× 10−4
× 6.875
0.25 × 1.96
× 104
= 40.93637
≈ 41
 Winding Number of Inductor :
𝑖 𝐿( 𝑟𝑚𝑠) 𝑡 = √(𝑖 𝐿( 𝑎𝑣𝑔))2 + (
∆𝐼 𝐿
2
√3
)
2
= √6.252 + (
1.25
2
√3
)
2
= 6.260408001Ampere
 Calculation of Wire Size :
 Cross Sectional Area of Wire (𝑞 𝑤)
𝑞 𝑤(𝑡) =
𝑖 𝐿( 𝑟𝑚𝑠) 𝑡
𝐽
=
6.260408001
4.5
= 1.391201778 𝑚𝑚2
 Diameter of Wire (𝑑 𝑤)
𝑑 𝑤(𝑡) = √
4
𝜋
× 𝑞 𝑤(𝑡) = √
4
𝜋
× 1.391201778 = 1.330914392 𝑚𝑚
 Recalculate by assuming number of split wire (∑Split) = 15
𝑖 𝐿( 𝑟𝑚𝑠) 𝑠𝑝𝑙𝑖𝑡 =
𝑖 𝐿( 𝑟𝑚𝑠) 𝑡
∑Split
=
6.260408001
15
= 0.41736 𝐴𝑚𝑝𝑒𝑟𝑒
𝑞 𝑤( 𝑡) 𝑠𝑝𝑙𝑖𝑡 =
𝑖 𝐿( 𝑟𝑚𝑠) 𝑠𝑝𝑙𝑖𝑡
𝐽
=
0.41736
4.5
= 0.0927467852 𝑚𝑚2
≈ 0.1𝑚𝑚2

8. 𝑑 𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = √
4
𝜋
× 𝑞 𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = √
4
𝜋
× 0.1 = 0.35 𝑚𝑚
 Wire Size (∑𝑺𝒑𝒍𝒊𝒕 𝑪𝒉𝒐𝒐𝒔𝒆 ) = 15+3
Diameter of bobbin PQ3535 ( 𝑑 𝑏𝑜𝑏) = 17 𝑚𝑚 = 1.7 𝑐𝑚
Circumference of Bobin ( 𝐾𝑏𝑜𝑏) = 𝜋 × 𝑑 𝑏𝑜𝑏 = 𝜋 × 1.7 = 5.340707511 cm
Total Wire Length = (𝑛( 𝑤𝑖𝑛𝑑𝑖𝑛𝑔) × 𝐾𝑏𝑜𝑏 × ∑Split) + 40% × (𝑛( 𝑤𝑖𝑛𝑑𝑖𝑛𝑔) 𝜋 ×
𝐾𝑏𝑜𝑏 × ∑Split)
Total Wire Length = (41 × 5.341 × 18) + [40% × (41 × 5.341 × 18)]
Total Wire Length = (3941.65)+ [1576.66] = 5518.31 𝑐𝑚 = 55.18𝑚 ≈ 56 𝑚
 Output Capacitance
𝐶 𝑜 =
𝑉𝑜 × 𝐷
𝑅 × ∆𝑉𝑜 × 𝑓
=
60 × 0.6
24 × 0.001 × 60 × 40 × 103
= 6.25 × 10−4
𝐹 = 625 𝜇𝐹
≈ 1000 𝜇𝐹,100 𝑉𝑜𝑙𝑡
 Snubber Circuit
𝐶𝑠𝑛𝑢𝑏𝑏𝑒𝑟 ≈
𝐼 𝑂𝑁 × 𝑡𝑓𝑎𝑙𝑙
2 × 𝑉𝑂𝐹𝐹
𝐶𝑠𝑛𝑢𝑏𝑏𝑒𝑟 =
𝐼 𝑂 ×
𝑉𝑜
𝑉𝑆
× 𝑡𝑓𝑎𝑙𝑙
2 × 𝑉𝑂
=
2.5 ×
60
24
× 58 × 10−9
2 × 60
= 3.020833333× 10−9
𝐹
= 3.02 𝑛𝐹
𝐶𝑠𝑛𝑢𝑏𝑏𝑒𝑟 𝐶ℎ𝑜𝑜𝑠𝑒 ≈ 4.7𝑛𝐹, 1 Kvolt
𝑅 𝑠𝑛𝑢𝑏𝑏𝑒𝑟 <
𝐷𝑇
2 × 𝐶𝑠𝑛𝑢𝑏𝑏𝑒𝑟
𝑅 𝑠𝑛𝑢𝑏𝑏𝑒𝑟 <
0.6 × 2.5 × 10−5
2 × 3.020833333 × 10−9
𝑅 𝑠𝑛𝑢𝑏𝑏𝑒𝑟 < 2482.758621 Ω
𝑅 𝑠𝑛𝑢𝑏𝑏𝑒𝑟 𝐶ℎ𝑜𝑜𝑠𝑒 =
1
2
× 2482.758621 = 1241.379311 Ω ≈ 1 𝐾Ω, 10 watt

9. 4. DesignProcedure
 Acrylic Template Design
 Schematic Circuit of Boost Converter
 Board Circuit of Boost Converter

10.  Schematic Circuit of PWM with ARM
 Board Circuit of PWM with ARM

11. 5. Experimental Result
 Inductor Test
Inductance of inductor ≈ 290 𝜇𝐻
Quality factor (Q) = 368,2
Figure 5.1. Inductance of Design Inductor
Figure 5.2. Quality Factor of Design Inductor
Winding Steps :
1. Firstly, decide the parameters of boost converter like Vs, Vo, Is, Io, frequency
switching etc. Then calculate value of duty cycle, inductor, capacitor, resistor,
snubber circuit etc.
2. To design the inductor, we must calculate the winding and decide split of inductor.
In laboratory, we got ferrit core PQ 35/35 with cross sectional are (𝐴 𝑐 = 1.96 𝑐𝑚2
)
and bobbin diameter ( 𝐷 𝑏𝑜𝑏 = 17 𝑚𝑚).
3. After calculate the winding and decide split of inductor, determine the lenght of wire
that we need for design boost converter.
4. Take the wire and roll on wire rolling.
5. Next step, divide of the wire based on number of split. Wire must on strained
condition when we divide it. At the same time, get a message of wire use Majun.
6. After that, use wire roll to rolling wire into bobbin.

12. 7. Give isolated (gap) for eash stack (layer). When we roll the wire must be carefull,
because email of wire (laminated) easy get peeled off.
8. For the last stack (layer) give black isolated on the surface.
9. Next, remove email of wire (laminated) for each strand of wire. Make sure that no
email left on strand wire.
10. Wire direction set to the foot of bobbin in same side entered and exit then soldered
wire.
11. Then take the core (ferrit) into bobbin and give air gap by using clear isolate (tape) .
Addition of isolated (tape) until we got the value of design inductor for boost
converter. To check the inductance value use LCR meter.
12. After we got inductance value for design inductor, checking of quality factor (Q) use
quality factor meter.
13. Record the value of inductance and quality factor of design inductor.
 PWM Test
Figure 5.3. Pulse Width Modulation (PMW) Test
PWM Test Steps
1. Make circuit of PWM used Microcontroller STM32F4 as in the schematic PWM.
2. After that, start testing circuit of PWM. First, prepare the equipment needed such as
DC power supply 2 units, one for supply microcontroller 5 VDC and 12 VDC for driver
PWM, oscilloscope, multimeter digital, power cable and probe cable.
3. Then connect input terminal PTR 1 to the DC power supply, inject DC power supply
5VDC.
4. Then connect input terminal PTR 2 to the DC power supply, inject DC power supply
12VDC.
5. Then put in the potensiometer 50K and 100K to terminal PTR 3 and PTR 4.

13. 6. Then connect output terminal PTR 5 to the probe cable of oscilloscope.
7. Next set potensiometer 50K until wave picture in oscilloscope show frequency 40
KHz with time/div 5 µs. Then set potensiometer 100K to change duty cycle.
8. List data result of PWM Microcontroller testing and take a picture VGS wave that
show in oscilloscope.
Table 5.1. Various Duty Cycle
Frequency Duty Cycle PWM
40 Khz 30%
40 Khz 50%
40Khz 70%

14.  Functional Test
Figure 5.4. Functional Boost Converter Test
Functional Test Steps
1. Connecting Boost converter, PWM microcontroller module and other component to
the acrylic.
2. Then, install connection of cable PWM microcontroller and Boost converter for good
connection, PWM microcontroller modul on side acrylic and Boost converter in side
acrylic.
3. After that, start testing, connect the boost converter module microcontroller cable to
input 220 VAC. Push on switch on the module when the PWM switch is OFF and DC
power supply is also OFF.
4. Turn OFF the PWM switch and put the oscilloscope probe to test point PWM to view
VPULSE wave. Set the frequency 40 KHz and PWM duty cycle 30%.
5. Turn ON the PWM switch and put the oscilloscope probe to view VGS. DC power
supply is also OFF.
6. Then connect series ampermeter DC analog with load (lamp).
7. Turn ON DC power supply, set value of DC power supply into VIN = 5 VDC. Put the
oscilloscope probe to view VDS wave.
8. Measure the VIN, IIN, VOUT, IOUT and view wave using oscilloscope.
9. Repeat step 2 until until 8 for another duty cycle and input voltage.

15. Table 5.2. Experimental Data of Boost Converter
Duty
Cycle
(%)
Vs
(Volt)
Is
(A)
Vo
Prak
(Volt)
Vo
Teori
(Volt)
Io
(A)
Ps
(Watt)
Po
Prak
(Watt)
Po
Teori
(Watt)
η
(%)
Error
(%)
30
5 0,29 6,17 7,14 0,19 1,45 1,17 1,35 80,84 13,16
10 0,39 13,25 14,28 0,24 3,9 3,18 3,42 81,53 7,01
24 0,61 32,94 34,28 0,42 14,64 13,83 14,39 94,46 3,89
50
5 0,45 8,45 10 0,22 2,25 1,859 2,2 82,62 15,5
10 0,64 18,36 20 0,3 6,4 5,5 6 85,93 8,3
24 1,03 45,5 48 0,5 24,72 22,75 24 92,03 5,2
60
5 0,62 10,67 12,5 0,24 3,1 2,56 3 82,58 14,6
10 0,87 22,43 25 0,34 8,7 7,62 8,5 87,58 10,35
24 1,43 55 60 0,5 34,32 27,5 30 80,12 8,3
Where :
𝑉𝑜𝑢𝑡 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 ∶ 𝑉𝑜 = −𝑉𝑠 × (
𝐷
1 − 𝐷
)
𝐸𝑟𝑟𝑜𝑟 ∶ %𝑒𝑟𝑟𝑜𝑟 = |
𝑉𝑜 𝑇𝑒𝑜𝑟𝑖 − 𝑉𝑜 𝑃𝑟𝑎𝑘
𝑉𝑜 𝑇𝑒𝑜𝑟𝑖
| × 100%
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑃 𝑜𝑢𝑡
𝑃 𝑖𝑛
× 100%
Table 5.3. Duty cycle 30%
Vin Vduty cycle VGS VDS Vout

16. Table 5.4. Duty cycle 50%
Vin Vduty cycle VGS VDS Vout
Table 5.5. Duty cycle 60%
Vin Vduty cycle VGS VDS Vout
 Capability Test
Figure 5.5. Capability Boost Converter Test

17. Capability Test Steps
1. First, start testing. Connect the boost converter microcontroller module cable to input
220 V, push on switch in the module when the PWM switch is OFF and DC power
supply is also OFF.
2. Turn OFF the PWM switch and put the oscilloscope probe to test point PWM to view
VPULSE. Set the frequency into 40 KHz and PWM duty cycle into 50%.
3. Turn ON the PWM switch and put the oscilloscope probe to Gate and Source point
test to view VGS. DC power supply is also OFF. Then connect series ampermeter DC
analog with variable resistor load.
4. Turn ON DC power supply. Set value of DC power supply VIN into 24 VDC to the
module. Put the oscilloscope probe to Drain and Source test point to check the VDS.
5. Then change the variable resistor load until ampermeter DC analog show the output
current is 0,5 A.
6. Measure the VIN, IIN, VOUT, IOUT , then put the oscilloscope probe to the load to check
the Vout.
7. Repeate step 3-6 with same input voltage, until output current maximal (2,5 A).
8. If the output current cannot reach output current maximum, don’t continued.Because,
if shows that the reability of your modul only reaches that value.
Table 5.3. Experimental Data of Boost Converter
Duty cycle
(%)
Vs
(Volt)
Is
(Ampere)
Vout
(Volt)
Vout cal
(Volt)
Io
(Ampere)
Efficiency (%)
50 24,03 1,03 46,4 48 0,5 93,73
50 24,09 2,17 44,8 48 1 85,7
50 24 3,05 36,45 48 1,5 74,69
𝑊ℎ𝑒𝑟𝑒 ∶ 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
𝑃 𝑜𝑢𝑡
𝑃 𝑖𝑛
× 100%

18. 6. Analysis
In this project, we make boost converter with PWM microcotroller STM32F4. The
boost converter is DC-DC converter use to increase output voltage. It is called a boost
converter because the output voltage is lager than the input voltage. The boost converter that
operates by periodically switching components that is MOSFET. In this project we use
MOSFET IRFP460 to operate switching on high frequency. Design of boost converter based
on calculate some parameters and reference of data sheet component like diode, MOSFET,
capacitor, resistor etc. The boost converter has following parameters :
𝑉𝑠(𝑚𝑎𝑥) = 30 Volt
𝑉𝑠(𝑚𝑖𝑛) = 24 Volt
𝑉𝑜 = 60 Volt
𝐼𝑜 = 2.5 Ampere
Switching Frequency (𝑓𝑠) = 40 kHz
After get the parameters of boost converter, calculate the duty cycle, inductor,
capacitor, resistor, snubber circuit, winding etc. To design the boost converter, choose
Vs(min) to get lager rasio output voltage. After the calculate value some components choose
value larger than value of design or suitable in the value generally components, for example
C calculate = 6.25× 10−4
𝐹 = 625 𝜇𝐹 𝑎𝑛𝑑 𝑐ℎ𝑜𝑜𝑠𝑒 𝐶 ≈ 1000 𝜇𝐹, 100 𝑉𝑜𝑙𝑡 .
To design the inductor, we must calculate the winding and decide split of inductor.
In laboratory, we got ferrit core PQ 35/35 with cross sectional are (𝐴 𝑐 = 1.96 𝑐𝑚2
) and
bobbin diameter ( 𝐷 𝑏𝑜𝑏 = 17 𝑚𝑚). To check the inductance value use LCR meter and to
check of quality factor (Q) use quality factor meter. After that, we got value of design
inductor that is Inductance of inductor ≈ 290 𝜇𝐻 and Quality factor (Q) = 368,2.
Pulse Width modulation or also called PWM is a method of reducing the average
power delivered by an electrical signal, by effectively chopping it up into discrete parts. In
this project, microcontroller STM32-F4 using to created the PWM. For test this design of
PWM, setting frequency on 40 Khz and duty 30%,50% and 70%. Then check manual by
look the duty on oscilloscope.
To complate this project, boost converter must be checking by functional test and
capability test. On functionl test , source voltage setting by 5V, 10V, and 24V with duty
cycle 30%, 50% and 60%. Based on functional test, error output volatge reach 15,5% on
duty cycle 50% with Vs 5 volt and efficiency reach 80,12% – 94,4%. The output voltage,
error, and efficiency can be calculate from this formula :
𝑉𝑜 = (
𝑉𝑠(𝑚𝑖𝑛)
1 − 𝐷
) ; %𝑒𝑟𝑟𝑜𝑟 = |
𝑉𝑜 𝑇𝑒𝑜𝑟𝑖 − 𝑉𝑜 𝑃𝑟𝑎𝑘
𝑉𝑜 𝑇𝑒𝑜𝑟𝑖
|× 100% ; 𝐸𝑓𝑓 =
𝑃 𝑜𝑢𝑡
𝑃 𝑖𝑛
× 100%

19. Based on capability test, this converter can endure until 1,5 A. If this boost convereter give
current more than 1,5 A MOSFET can’t be able to switching, this case can demaged
MOSFET. Based on this test, efficiency of this converter reach 93,73% when 0,5A and
74,69% on 1,5A.
7. Conclusion
From the experiment we can get a conclusion :
1. To use PWM with microcontroller STM32F4, this micro must be update to able
create PWM and using optocopler FOD 3182.
2. The quality factor (Q) of design inductor is 368,2 because in design use more split,
so resistance bigger than inductance.
3. Based on functional test, error output volatge reach 15,5% on duty cycle 50% with
Vs 5 volt and efficiency reach 80,12% – 94,4%.
4. Based on capability test, efficiency of this converter reach 93,73% when 0,5A and
74,69% on 1,5A.
5. This module (Boost Converter) is capable to operate until 1,5A.
8. Suggestion
In this section, we will give some suggestion to design boost converter using ARM
STM32 F4 Microcontroller:
1. Using a good isolation of winding the inductor is like your design to get a same value
and best quality factor of inductor.
2. Design the boost converter with mathematic calculation and you should know your
component design and cost based o general market.
3. Read a datasheet of every component correctly to operate it, to take a data of
experiment, please do as SOP.
4. To have very good perform of module, make sure that all components are on circuit
is good and ready to use. Because not good condition of components can cause a big
error of result.

20. 9. Reference
1. http://www.learnabout-electronics.org/ac_theory/inductors.php
2. https://en.wikipedia.org/wiki/Inductor
3. http://www.electronoobs.com/eng_circuitos_tut10_1.php
4. https://en.wikipedia.org/wiki/Boost_converter
5. http://ecetutorials.com/electrical/boost-converter-principle-of-operation-
applications/
6. Hart, Daniel W, Power Electronics, New York : The McGraw-Hill Comparison,
inc, 2011.
7. Rashid, Muhammad H, Power Electronics Handbook :Devices, Circuits, and
Apllications Handbook, Burlington : Elsevier inc, 2011.

21. 10. Appendix
Figure 10.1. Boost Converter Module

23. Figure 10.2.Template and Design Schematic - Board Boost Converter

24. Figure 10.3. Design Schematic - Board PWM