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Includes theory of buck boost converter, parameter calculation of designing buck boost converter, Main circuit PCB both PWM Generator and Buck Boost Converter, Software Calculator, Visual basic Programm
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Name : Vania Kurnia Alvi
Class : 3 D4 LA
NRP : 1310171021
Date : Saturday, May 9th, 2020
Lecturer : Ir. Moh Zaenal Efendi, M.T.
It is called a buck-boost converter because the output voltage can be either higher or lower than the input voltage.
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. It is equivalent to a flyback converter using a single inductor instead of a transformer. Two different topologies are called buck–boost converter. Both of them can produce a range of output voltages, ranging from much larger (in absolute magnitude) than the input voltage, down to almost zero.
2. THEORY (CONCEPT OF THEORY)
Circuit of Buck-Boost converter Buck-Boost analysis: switch closed
When switch is closed,diode is reverse biased.The output is isolated. The input supplies energy to inductor. This energy is stored in the inductor The inductor voltage is:
dt
or di dt
V V LdiL L s L s
The change in inductor current is:
Steady–state operation
i i
o s
s o
L closed L opened
Note : Output of buck boost converter either be higher or lower than the source voltage
NOTE:
Assuming no power loss in converterInput power = Output power
V I Vo s s
2
Average source current is related to inductor current as:
I (^) s IL D
Vs-Vo
Vs
^2
2 ( )
2
s s
o s
Lavg o
s L o
Maximum inductor current
I I iL VsD s max L (^) 2 ( 1 ) 2 2
Minimum inductor current
I I iL VsD s min L (^) 2 ( 1 ) 2 2
For continous operation,
f
s s
2 min
2 min
2
min
L 10 L min
Capacitor filter value
The change in capacitor charge
Ro^ DT C Vo
Ripple factor:
RCf
r V
RCf
o
o
o o o
The Buck Boost Converter has following parameters : Vs(max) = 20 Volt Vs(min) = 9 Volt Vo = -14 Volt Io = 1.5 A R = Vo/Io= 14/1.5=9.33 Ω Switching Frequency (fs) = 40 kHz
Components:
Q : MOSFET IRFP
D : MUR 1560 (Ultra Fast Recovery Diode)
Inductor (L) : Ferrit Core PQ 3535 with Cross sectional are (Ac=1.6 1 cm^2 );
Bobbin diameter (Dbob = 16.5 mm )
Rs : Snubber resistor ( 1 𝐾 Ohm ,5- 10 Watt)
Cs : Snubber Capacitor ( 5 nF , 1 KVolt )
Ds : Snubber diode ( FR3017)
Duty Cycle :
𝑉𝑜 = −𝑉𝑠(min) × [
The Inductor Value : R = |V Ioo | = |−141.5 | = 9.33 Ω
𝐼𝐿(𝑎𝑣𝑔) = (^) 𝑅×(1−𝐷)𝑉𝑠×𝐷 2
V f = 1.5 V
𝐿 = (^1 𝑓) × [(𝑉𝑜 + 𝑉𝑓)] × ( (^) (𝑉𝑜+ 𝑉𝑉𝑠(𝑚𝑖𝑛)𝑓)+𝑉𝑠(𝑚𝑖𝑛) ) × ( (^) ∆𝐼^1 𝐿 ); assume Vo is positive
Diameter of Wire ( dw )
𝑑𝑤(𝑡) = √
= 1.04755 mm
Recalculate by assuming Number of Split Wire (∑ split = 9) o 𝐼𝐿(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 = 𝐼 ∑ 𝑠𝑝𝑙𝑖𝑡𝐿(𝑟𝑚𝑠)𝑡
o 𝑞𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = 𝐼𝐿(𝑟𝑚𝑠)𝑠𝑝𝑙𝑖𝑡 𝑗
=
= 0.0955 mm^2
o 𝑑𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡 = √^4 𝜋 × 𝑞𝑤(𝑡)𝑠𝑝𝑙𝑖𝑡
= 0.348 mm ≈ 0.35 mm
Wire Size Diameter of bobbin PQ3535 (Dbob) = 16.5 mm = 1. 65 cm Circumference of Bobin (𝐾𝑏𝑜𝑏) = π × 𝐷𝑏𝑜𝑏 (𝐾𝑏𝑜𝑏) = π × 1. (𝐾𝑏𝑜𝑏) = 5.181 cm Total Wire Length = (𝑛(𝑤𝑖𝑛𝑑𝑖𝑛𝑔) × 𝐾𝑏𝑜𝑏 × ∑ split) + 40% × (𝑛(𝑤𝑖𝑛𝑑𝑖𝑛𝑔) × 𝐾𝑏𝑜𝑏 × ∑ split) Total Wire Length = (19 × 5.181 × 9) + 40% × (19 × 5.181 × 9)
Total Wire Length = 885.951 cm + 354.3804 cm Total Wire Length = 1.240.3314 cm atau 12,4 m
Output Capacitance
𝐶𝑜 = (^) 𝑅×∆𝑉𝑉𝑜×𝐷𝑜×𝑓
Where ∆𝑉𝑜 = ±0,1% × 𝑉𝑜 = 0,001 × 𝑉𝑜
Snubber Circuit
𝑰𝑶𝑵 = 𝑰𝑳 = (^) 𝑽𝒔(𝒎𝒊𝒏)𝑷𝒐×𝑫
Vo × Io Vs(min) × D
𝑰𝑳 = (^149) ×× 01. 61.^5 𝑰𝑳 = 3. 825 A
𝑉𝑜𝑓𝑓 = 23 Volt 𝒕𝒇𝒂𝒍𝒍 = 58 ns (mosfet IRFP460)
𝐶𝑠𝑛𝑢𝑏𝑏𝑒𝑟 𝑐ℎ𝑜𝑜𝑠𝑒 ≈ 5 𝑛𝐹, 1𝐾𝑉olt
a. SCHEMATIC
b. BOARD
c. TOP
d. BOTTOM
F = 50k Hz
Duty Cycle = 60% Duty Cycle = 40%
Duty Cycle = 25%
Frequency
(KHz)
Duty
Cycle(%)
V Input
(volt)
V Output
(Volt)
V Output teori (volt)
1) Frequency = 40K Hz
Duty Cycle 60 % 𝑉𝑜𝑢𝑡 ( 𝑡ℎ𝑒𝑜𝑟𝑦 ) = 𝑉𝑖𝑛 × 𝐷𝑢𝑡𝑦 𝐶𝑦𝑐𝑙𝑒
= 12 × 0,6 = 7,2 𝑣𝑜𝑙𝑡
Duty Cycle 40 % 𝑉𝑜𝑢𝑡 ( 𝑡ℎ𝑒𝑜𝑟𝑦 ) = 𝑉𝑖𝑛 × 𝐷𝑢𝑡𝑦 𝐶𝑦𝑐𝑙𝑒
= 12 × 0,4 = 4,8 𝑣𝑜𝑙𝑡
Duty Cycle 25 % 𝑉𝑜𝑢𝑡 ( 𝑡ℎ𝑒𝑜𝑟𝑦 ) = 𝑉𝑖𝑛 × 𝐷𝑢𝑡𝑦 𝐶𝑦𝑐𝑙𝑒
= 12 × 0,25 = 3 𝑣𝑜𝑙𝑡
2) Frequency = 50K Hz
Duty Cycle 60 % 𝑉𝑜𝑢𝑡 ( 𝑡ℎ𝑒𝑜𝑟𝑦 ) = 𝑉𝑖𝑛 × 𝐷𝑢𝑡𝑦 𝐶𝑦𝑐𝑙𝑒
= 12 × 0,6 = 7,2 𝑣𝑜𝑙𝑡
Main Circuit PCB Design (Group)
1. SKEMATIK
