electronic circuits project report audio amplifier circuit by using bjt

LIST OF FIGURES Figure 1: Circuit Figure 2: DC equivalent circuit of stage 1 and stage 2 Figure 3: AC equivalent circuit of stage 1 and stage 2 Figure 4: Complete circuit Figure 5: DC si

HANOI UNIVERSITY OF SCIENCE AND TECHNOLOGY

ELECTRONIC CIRCUITS PROJECT REPORT

Audio amplifier circuit by using BJT

HA N , 2OI /2024

Teacher sign

2.2.1 Voltage divider EC amplifier 2

III General Design 3

IV Select best alternative 7

4.1 Choose BJT 7

4.2 Choose other components: 8

V Simulation 10

VI PCB Design……….12

LIST OF FIGURES Figure 1: Circuit

Figure 2: DC equivalent circuit of stage 1 and stage 2 Figure 3: AC equivalent circuit of stage 1 and stage 2 Figure 4: Complete circuit

Figure 5: DC simulation Figure 6: Bode Diagram Figure 7: PCB Design

LIST OF TABLES Table 1: Table of specifications

Table 2: Table of BJT

Table 3: Compared between simulations and calculations

I Introduction

This report is about a major assignment in analog electronics The content of this project is about using Bipolar Junction Transistor to design an audio amplifier circuit We use LT spice software to simulate the circuit and produce predicted results to compare with previous calculations Regarding calculations, we sequentially follow the 9-step design process to choose the combinations, operations, and components that bessuit the proposed requirements Design steps include Determine need, Specification, Plan, Block Diagram, Design subblock, Select best alternative, Testing, Mass production, Delivery This project helps students better understand the design processes and thinking of a circuit engineer

II Method

2.1 Determine need and specifications

With the knowledge we have learned, we have to design an audio amplifier circuit using BJT The circuit must fully satisfy the requirements of Input Impedance (Zin), Gain (Av), Power consumption (Pcon) which will be clearly stated in Table 1

Table 1: Table of specifications

The conditions for input impedance, gain, and power loss are the initial requirements given We added the remaining 4 conditions about Power supply, Bandwidth, Output power and load resistance so that the circuit can operate.

2.2 Choose topology 2.2.1 Voltage divider EC amplifier

We will explain why choose common emitter with voltage divider bias First, we choose voltage divider bias at first stage because it has stable Q point operation, its stability is both better than Fix bias and Collector feedback bias Second, this configuration offers high voltage gain, which is essential for amplifying weak signals from microphones or other audio sources

2.2.2 Darlington pair

But if only the first stage is installed, the circuit will have problems because the common emitter depends too much on the load resistance, and the speaker's load resistance is very small, so the voltage gain is changed and not amplified Therefore, on the 2nd stage we install Darlington pair because it has 2 Common collector stages because it is less dependent on load resistance Not only that, but the Darlington Pair also offers very high current gain which means it can drive high current loads like speakers or headphones with ease This is important in audio applications that require high power output Only then will we get the output, and the circuit can be used effectively

2.3 Block diagram

We built a diagram of the operating principle of this audio amplifier to have a more general view of how the circuit works as well as see the blocks that need to be processed The input will be the Microphone and then connected to the audio amplifier, then the transmission line will go through the Darlington pair and finally the output will be the Speaker The supplied DC voltage of 12V will be connected to the audio amplifier and Darlington pair

DC voltage supply

Input (Micro) Audio amplifier Darlington pair Output (Speaker)

𝐼𝐶𝑄1=12𝐼𝐶1𝑚𝑎𝑥= 12

𝑉𝐶𝐶𝑅𝐶+ 𝑅𝐸𝑉𝐶𝐸𝑄1=1

• 𝑉𝐸1= 𝑉𝐵1− 𝑉𝐵𝐸1= 𝐼𝐶𝑄1𝑅𝐸• Pcons= VCC (IC+ IR1) = VCC2

R1+R2+ VCCIC- Stage 2

• Determine DC load line: (𝐼𝐶3≈ 𝐼𝐸3)

𝑉𝐶𝐸𝑄3= 𝑉𝐶𝐶− 𝐼𝐶3𝑅𝐸3• Q-point in the middle of DC loadline:

𝑉𝐶𝐶𝑅𝐸3

• 𝐼𝐵2=𝑉𝐶𝐶−𝑉𝐵𝐸2−𝑉𝐵𝐸3𝑅𝐵2+𝛽2𝛽3𝑅𝐸3

• 𝐼𝐶3≈ 𝐼𝐸3= 𝛽3𝐼𝐵3= 𝛽3𝐼𝐸2= 𝛽3𝛽2𝐼𝐵2➔ 𝐼𝐶𝑄3= 𝛽3𝛽2

3.2 AC equivalent circuit without load

Figure 3: AC equivalent circuit of stage 1 and stage 2

- Stage 1:

• 𝑣𝑖 = 𝑖𝑏1𝛽1𝑟𝑒1+ 𝑖𝑒1𝑅𝐸1= 𝑖𝑏1𝛽1𝑟 𝑒1+ 𝑖𝑏1(𝛽1+ 1) 𝑅𝐸1= 𝑖𝑏1𝛽1(𝑟 𝑒1+ 𝑅𝐸1) • 𝑣𝑜1= −𝑖𝑜1𝑅𝑐= −𝑖𝑏1𝛽1𝑅𝐶

➔ 𝑨𝒗𝑵𝑳𝟏 =𝒗𝒐𝟏

𝒗𝒊𝟏= − 𝑹𝑪

𝒓𝒆𝟏+𝑹𝑬𝟏• 𝑨𝒊𝑵𝑳𝟏 =𝒊𝒐𝟏

𝒗 /𝒁𝒊𝟏 𝒊𝟏= 𝑨𝑽𝟏𝒁𝒊𝟏

𝒊𝒐𝟏= 𝑹𝑪

- Stage 2

▪ 𝑣𝑖 = 𝑖𝑏2𝛽2𝑟𝑒2+ 𝑖𝑏3𝛽3𝑟𝑒3+ 𝑖𝑒3𝑅𝐸3• 𝑖𝑏3= 𝑖𝑒2= (𝛽2+ 1) 𝑖𝑏2

• 𝑖𝑒3= (𝛽3+ 1)𝑖𝑏3= (𝛽2+ 1)(𝛽3+ 1) 𝑖𝑏2

• 𝑣𝑖 = 𝑖𝑏2𝛽2𝑟𝑒2+ (𝛽2+ 1)𝛽3𝑟 𝑒3+ (𝛽2+ 1)(𝛽3+ 1) 𝑖𝑏2= 𝑖𝑏2(𝛽2(𝑟 𝑒2+ 𝛽3(𝑟𝑒3+𝑅𝐸3)) = 𝑣𝑏

▪ 𝑍𝑏=𝑣𝑏

𝑖𝑏2= 𝛽2(𝑟𝑒2+ 𝛽3(𝑟𝑒3+ 𝑅𝐸3)) • 𝑍𝑖 = 𝑅𝐵2// 𝑍𝑏

➔ 𝒁𝒊𝟐= 𝑹𝑩𝟐// 𝜷𝟐(𝒓𝒆𝟐+ 𝜷𝟑(𝒓𝒆𝟑+ 𝐑𝐄𝟑)) ▪ 𝑣𝑜2= −𝑖𝑒3𝑅𝐸2= −𝛽2𝛽3𝑖𝑏2𝑅𝐸2

➔ 𝑨𝒗𝑵𝑳𝟐 =𝒗𝒐𝟐

𝒗𝒊𝟐= − 𝜷𝟐𝜷𝟑𝑹𝑬𝟐

𝜷𝟐(𝒓 +𝜷 (𝒓 +𝑹𝒆𝟐 𝟑 𝒆𝟑 𝑬𝟐))= − 𝑹𝑬𝟐𝒓𝒆𝟐𝜷𝟑+𝒓𝒆𝟑+𝑹𝑬𝟐

(𝛽2𝑟𝑒2+(𝛽2+1)𝛽3𝑟𝑒3)

• 𝑖𝑜2= 𝐼𝑅𝐸2− 𝑖𝑒3= 𝐼𝑅𝐸2− (𝛽3+ 1)𝑖𝑏3=𝑣𝑜2

𝑅𝐸2− (𝛽3+ 1)(𝛽2+ 1)𝑖𝑏2➔ 𝑖𝑜2= 𝑣𝑜2

𝑅𝐸2+ 𝑣𝑜2(𝛽3+1)(𝛽2+1)

(𝛽2𝑟𝑒2+(𝛽2+1)𝛽3𝑟𝑒3)= 𝑣𝑜2(1

𝑅𝐸2+ 𝑟𝑒2 1

𝛽3+𝑟𝑒3) ➔ 𝒁𝒐𝟐= 𝟏 𝟏

𝑹𝑬𝟐+ 𝟏𝒓𝒆𝟐𝜷𝟑+𝒓𝒆𝟑

= 𝑹𝑬𝟐//(𝒓𝒆𝟐

𝜷𝟑+ 𝒓𝒆𝟑)

3.3 AC Equivalent with load - 𝐴𝑣𝑜𝑣𝑒𝑟𝑎𝑙𝑙= 𝑍𝑖1

𝑅 +𝑍𝑠𝑖1𝐴𝑣1𝑁𝐿𝑍𝑖2𝑍𝑜1+𝑍𝑖2𝐴𝑣2𝑁𝐿

𝑍𝑜2+𝑅𝑙𝑜𝑎𝑑- With 𝑅𝑠→ 0: 𝐴𝑣𝑜𝑣𝑒𝑟𝑎𝑙𝑙= 𝐴𝑣𝑁𝐿1

2𝜋(𝑍𝑜1+𝑍𝑖2)𝐶1• 𝑓𝐿𝑂 = 1

2N2222 TO92 Trans NPN

0.6A 40V

BC337-25 TO92 Trans NPN 0.8A

45V

BC548 TO-92 TRANS NPN 0,1A 30V

TIP41C TO220 NPN 6A 100V

V_CE 40V 45V 30V 100V V_BC 60V 50V 30V 100V V_BE 5V 1.2V 6V 5V

I_C 600mA 800mA 100mA 6A PRICE 1.500đ 1.000đ 3.000đ 3.000đ

DC CURRENT

It is easy to see that 2N2222 is the best choice for the first and second BJT because it has low cost, wide voltage range, low power consumption, stability for EC and easy to stimulate

For the final BJT, we choose TIP41C since dc current gain (Beta) of TIP41C is less dependent on Ic Moreover, Tip41C can go through high 6A-current

4.2 Choose other component : s 4.2.1 First stage

• In this circuit, we take advantage of the available 12V power source In the next section, we will choose the appropriate resistors and capacitors for the power supply and the circuit’s requirements

• Consider VCC= 12(𝑉); β =210

• Q point in the middle of DC load line → IC= V𝐶𝐶

2(R +RC E)= 6RC+RE• Pcons= VCC(IC+ I𝑅1) ≤ 0,2𝑊

→ IC+ I𝑅1≤ 16,67𝑚𝐴 → 6

RC+RE≤ 16 67, 𝑚𝐴 → RC+ RE≥ 0,36𝑘Ω • Choose 𝑉𝐸= 1

10 𝑉𝑐𝑐= 1,2𝑉, choose Ic= 1𝑚𝐴 → IC= V𝐶𝐶

2(RC+ RE)= 1𝑚𝐴 ≈ 𝐼𝐸