final report electronics circuit design 1

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<b>TRƯỜNG ĐẠI HỌC TÔN ĐỨC THẮNG KHOA ĐIỆN – ĐIỆN TỬ </b>

FINAL REPORT

ELECTRONICS CIRCUIT DESIGN 1

Lecture: ThS.TR N THÀNH NAM Ầ Students:

<i>Cao Đặng Tiến Phú - 420H0170 </i>

Nguy n Ng c Huân - 420h0190 <i>ễọ</i>

Phan Thanh Liêm - 420H0132

HO CHI MINH CITY, YEAR 202

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Table ole ole of Cf Contf Contontentsentsents

ADAPTER 2 D<small>ESIGN STEP DIAGRAM</small> ... 2

1.1.1 Transformer... 2

1.1.2 Diode Rectifier & Filter ... 3

1.1.3 Voltage regulation & Load ... 3

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𝑁₂⁼^𝑉𝑉¹₂⁼^𝑉𝐴𝐶𝑣_𝑆 ⁼²²⁰22= 10

220Vrms 50Hz

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1.1.2 Diode Rectifier & Filter

We use the Full-Bridge Rectifier to minimize the loss. Base on the figure above, we have the formulation:

𝑣<small>𝑆</small>= 2 × 𝑉<small>𝐷𝑖𝑜𝑑𝑒</small>+ 𝑣<small>𝑜𝑢𝑡−𝑟𝑒𝑐𝑡𝑖𝑓𝑖𝑒𝑟</small>

⟹ 𝑣<small>𝑜𝑢𝑡−𝑟𝑒𝑐𝑡𝑖𝑓𝑖𝑒𝑟</small>= 𝑣<small>𝑆</small>− 2 × 𝑉<small>𝐷𝑖𝑜𝑑𝑒</small>= 22√2 − 2 × 0.7 = . (𝑉) 29 71

Choose 𝐶 =100𝑚𝐹⟹ 𝑍<small>𝐶</small>=_𝜔𝐶¹= _{2𝜋𝑓.𝐶}¹ = _{2𝜋. .}_{50 100 10}¹ _× ₋₃= 0.03 Ω (𝑣𝑒𝑟𝑦 𝑠𝑚𝑎𝑙𝑙) So 𝑣_{𝑜𝑢𝑡−𝑓𝑖𝑙𝑡𝑒𝑟}≈ 29 71 𝑉 .

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Base on the figure above, we also have: 𝐼<small>𝑙𝑖𝑚𝑖𝑡</small>= 𝐼<small>𝑧</small>𝑚𝑎𝑥 + 𝐼<small>𝑜</small>

⟺^𝑣^{𝑜𝑢𝑡−𝑓𝑖𝑙𝑡𝑒𝑟}_𝑅 ^{− 𝑉}^𝑧

<small>𝑙𝑖𝑚𝑖𝑡</small> = 𝐼<small>𝑧</small>𝑚𝑎𝑥 + 𝐼<small>𝑜</small>

⟺^{29.71 − 10}_𝑅

<small>𝑙𝑖𝑚𝑖𝑡</small> = 40 10× <small>−3</small>+ 1.2 ⟹ 𝑅<small>𝑙𝑖𝑚𝑖𝑡</small>= 15 89. Ω

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Simulation 1.2.1 Transformer

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Results:

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1.2.2 Diode Rectifier

1.2.3 Filter

Add a Capacitor to reduce wave amplitude

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Results:

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1.2.4 Voltage Regulater & Load

To set the output steady, we must use the diode Zener with 𝑉<small>𝑧</small>= 10𝑉 to meet the requirement of the threads.

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Results:

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OpAmp Circuit

Thiết k m ch khuế ạ ếch đạ ử ụi s d ng OPAMP sao cho 𝒗<small>𝒐</small> = 𝟓𝒗 + 𝟏𝒗 + 𝟑𝒗 −<small>𝟏𝟐𝟑</small>

𝟐𝒗<small>𝟒</small>. V v n u bi ẽ <small>o</small> ế ết:𝑣₁ = 1 𝑠𝑖𝑛 (1000𝜋𝑡) 𝑣₂ = 2 𝑠𝑖𝑛 (2000𝜋𝑡) 𝑣<small>3</small> = 3𝑠𝑖𝑛 (3000𝜋𝑡) 𝑣<small>4</small> = 4 𝑠𝑖𝑛 (4000𝜋𝑡)

The Complete Circuit

Calculation

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𝑣₄ = 4 𝑠𝑖𝑛 (4000𝜋𝑡) ⟹ 𝑣₄= 2(𝑉); 𝑓<small>4</small>=^4000𝜋_{2𝜋 = 2000 𝐻𝑧}

<b>Parameters: </b>

R1=16kΩ R2=80kΩ R3=80/3kΩ R4=30kΩ R5=60KΩ Rf1=80kΩ Rf2=60KΩ

𝑣𝑜 = 5𝑣 + 1𝑣 + 3𝑣<small>123</small>− 2𝑣<small>4</small>Select gain: 𝑅𝑓2

𝑅5

= 1

𝑅5∗^𝑅𝑓1𝑅1= 5𝑅𝑓2

𝑅5∗^𝑅𝑓1𝑅2= 1𝑅𝑓2

𝑅5∗^𝑅𝑓1𝑅3= 3𝑅𝑓2𝑅4= 2

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Choose Rf2 = R5 = 60KΩ𝑅𝑓2

𝑅4= 2 → 𝑅4 = 30𝑘^ΩChoose Rf1 = 80kΩ 𝑅𝑓1

𝑅1= 5 → 𝑅1 = 16𝑘^Ω𝑅𝑓1

𝑅2= 1 → 𝑅2 = 80𝑘^Ω𝑅𝑓1

𝑅3= 3 → 𝑅3 = /3𝑘⁸⁰ ^Ω

Simulation 1.6.1 Input data

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1.6.2 Elements data

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1.6.3 The simulation result

The simulation results of the following 𝑣₁; 𝑣 ; 𝑣 ; 𝑣_{2 3} ₄

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The simulation result of 𝑣<small>𝑜</small>

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