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PUP Voltage Divider Biasing Using 2 N3904 Transistor Exercise

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Subject
Engineering
School
Polytechnic University of the Philippines
Type
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POLYTECHNIC UNIVERSITY OF THE PHILIPPINES
College of Engineering
ELECTRONICS ENGINEERING DEPARTMENT
Name: Aubrey B. Illurimo January 7, 2020
Section: BSECE 2-5 Electronic Circuit Analysis and Design
VOLTAGE DIVIDER BIASING using 2N3904 TRANSISTOR
I. MATERIALS
1. Resistor
R
1
= 33kΩ
R
2
= 6.8kΩ
R
C
= 2.2kΩ
R
E
= 1kΩ
2. 2N3904 NPN Transistor
3. Breadboard
4. Connecting Wires
5. 10V Battery
6. Digital Multimeter
II. SCHEMATIC DIAGRAM
10V
= 2.2KΩ
= 1KΩ
=33KΩ
=6.8KΩ
Figure 1. Schematic Design

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III. CIRCUIT LAYOUT ON BREADBOARD
Figure 1. Circuit Layout on Breadboard
IV. COMPUTATION
Calculating R
TH
:






R
TH
= 5638.190955Ω
Calculating V
TH
:





I = 0.2512562814 mA
V
TH
= I R
2
= (0.2512562814 mA) (6.8kΩ)
V
TH
= 1.708542714 V
V
TH
– I
E
R
TH
– V
BE
– I
E
R
E
= 0 since I
E
= (β + 1) I
B
V
TH
– I
E
R
TH
– V
BE
– (β + 1) I
B
R
E
= 0



 

  

  


I
C
= β I
B
= (241) ( μA) = 0.9815077111 mA
R
EQ
= R
TH

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POLYTECHNIC UNIVERSITY OF THE PHILIPPINES College of Engineering ELECTRONICS ENGINEERING DEPARTMENT Name: Section: Aubrey B. Illurimo BSECE 2-5 January 7, 2020 Electronic Circuit Analysis and Design VOLTAGE DIVIDER BIASING using 2N3904 TRANSISTOR I. MATERIALS 1. Resistor R1 = 33kΩ R2 = 6.8kΩ RC = 2.2kΩ RE = 1kΩ 2. 2N3904 NPN Transistor 3. Breadboard 4. Connecting Wires 5. 10V Battery 6. Digital Multimeter II. SCHEMATIC DIAGRAM 10V =33KΩ = 2.2KΩ =6.8KΩ = 1KΩ Figure 1. Schematic Design III. CIRCUIT LAYOUT ON BREADBOARD Figure 1. Circuit Layout on Breadboard IV. COMPUTATION Calculating RTH: REQ = RTH Calculating VTH: 𝑉𝐶𝐶 10 𝑉 𝐼= = (33𝑘𝛺) + (6.8𝑘𝛺) 𝑅1 𝑅2 I = 0.2512562814 mA 𝑅𝑇𝐻 = (33𝑘𝛺) (6.8𝑘𝛺) 𝑅1 𝑅2 = (33𝑘𝛺) + (6.8𝑘𝛺) 𝑅1 + 𝑅2 VTH = I R2 = (0.2512562814 mA) (6.8kΩ) VTH = 1.708542714 V RTH = 5638.190955Ω VTH – IE RTH – VBE – IE RE = 0 since IE = (β + 1) IB VTH – IE RTH – VBE – (β + 1) IB RE = 0 𝐼𝐵 = 𝑉𝑇𝐻 − 𝑉𝐵𝐸 1.708542714𝑉 − 0.7𝑉 = = 4.072646104𝜇𝐴 𝑅𝑇𝐻 + (𝛽 + 1) 𝑅𝐸 5638.190955𝛺 + (241 + 1)(1000𝛺) IC ...
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