Basic electronics - Bi Junction Terminals 02.pptx

1. Chapter 2: BJT and its
applications
Module 2 : Transistor Biasing
Reference:
Robert L. Boylestad, Louis Nashelsky, Electronic Devices & Circuits
Theory, 11th Edition, PHI, 2012
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2. Objectives
• Discuss the concept of biasing of the transistor.
• Analyse the fixed and self bias circuits.
• Identify the circuit parameters that effect Q point.
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3. Operating Point / Quiescent Point
For the BJT to operate effectively , it has to be properly biased.
Set a certain value of I and V in the BJT→ Biasing
These values correspond to a point on the BJT output characteristics
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4. Transistor circuit in Common Emitter Configuration
Applying external dc voltages to ensure
that transistor operates in the desired region.
Applying KVL to the output loop,
𝑉𝐶𝐶 - 𝐼𝐶𝑅𝐶 -𝑉𝐶𝐸=0
𝑉𝐶𝐸 = 𝑉𝐶𝐶 - 𝐼𝐶𝑅𝐶 ------(1)
𝐼𝐶 = -
𝑉𝐶𝐸
𝑅𝐶
+
𝑉𝐶𝐶
𝑅𝐶
---------(2)
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5. From equation (1), when 𝐼𝐶=0,
𝑉𝐶𝐸 = 𝑉𝐶𝐶 -------------(3)
From equation (2) when 𝑉𝐶𝐸=0,
𝐼𝐶 =
𝑉𝐶𝐶
𝑅𝐶
-------------(4)
Equation (3) and (4) form the
extremities of 𝐼𝐶 and 𝑉𝐶𝐸
Line joining these two points → DC Load Line
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6. The point where DC load line cuts the output characteristics is called
Quiescent Point (Q Point) / Operating point
Depending on the requirement, Q point can be moved to Saturation, Active
or Cut-off region
To set the Q Point, biasing is necessary
Note: When no biasing is applied, Q point is at the origin
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8. Variation in load line with circuit parameters
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9. Fixed Bias / Base Current Bias
Applying KVL to input loop,
𝑉𝐶𝐶-𝐼𝐵𝑅𝐵-𝑉𝐵𝐸=0
𝐼𝐵 =
𝑉𝐶𝐶−𝑉𝐵𝐸
𝑅𝐵
Applying KVL to output loop,
𝑉𝐶𝐶-𝐼𝐶𝑅𝐶-𝑉𝐶𝐸=0
𝑉𝐶𝐸 = 𝑉𝐶𝐶-𝐼𝐶𝑅𝐶
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10. 𝐼𝐶 = β*𝐼𝐵
𝑉𝐶𝐶 is constant
𝑉𝐵𝐸 = 0.7 (for Si)
By selecting proper 𝑅𝐵 , 𝐼𝐵 can be
fixed.
𝑉𝐶𝐸 can also be fixed by selecting
appropriate 𝑅𝐶
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11. Q1. For the circuit shown below, draw the DC load line and mark the Q
point/ operating point.(Si transistor used)
Apply KVL to the input loop
𝑉𝐶𝐶-𝐼𝐵𝑅𝐵-𝑉𝐵𝐸=0
𝐼𝐵 =
𝑉𝐶𝐶−𝑉𝐵𝐸
𝑅𝐵
= 19.78µA
𝐼𝐶𝑄 = β*𝐼𝐵 = 1.978mA
Apply KVL to the output loop
𝑉𝐶𝐶-𝐼𝐶𝑅𝐶-𝑉𝐶𝐸= 0
𝑉𝐶𝐸𝑄 = 𝑉𝐶𝐶-𝐼𝐶𝑅𝐶 = 6.04V
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10V
100
470kΩ
2kΩ

12. To draw DC Load Line
𝑉𝐶𝐸𝑚𝑎𝑥 = 𝑉𝐶𝐶 =10V
𝐼𝐶𝑚𝑎𝑥 =
𝑉𝐶𝐶
𝑅𝐶
= 5mA
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𝑰𝑪𝒎𝒂𝒙
𝑉𝐶𝐸𝑚𝑎𝑥
5mA
10V
6.04V
1.978mA

13. Q2. In a fixed bias circuit using Germanium BJT, 𝑉𝐶𝐶=12V, 𝑅𝐶= 4kΩ, 𝑉𝐶𝐸=0.2V
β=50. Find 𝐼𝐵, 𝑅𝐵neglecting 𝐼𝐶𝐸𝑂
Apply KVL to the output loop
𝑉𝐶𝐶-𝐼𝐶𝑅𝐶-𝑉𝐶𝐸= 0
𝐼𝐶= 2.95mA
𝐼𝐵=
𝐼𝐶
β
= 59µA
Apply KVL to the input loop
𝑉𝐶𝐶-𝐼𝐵𝑅𝐵-𝑉𝐵𝐸=0
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14. Q3. For the circuit shown below, find the base current required to
establish 𝑉𝐶𝐸𝑄=6V (Si transistor used).Also find 𝑅𝐵
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12V
120
2.2kΩ

15. Self Bias / Voltage Divider Bias
Uses two resistors 𝑅1 and 𝑅2 instead
of 𝑅𝐵
𝑅𝐸 is the emitter feedback resistance
connected between emitter and
ground.
The circuit can be simplified using
Thevenin’s Theorem
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A
B

18. Thevenin’s Resistance
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𝑅𝑇𝐻 is the resistance seen between AB with
𝑉𝐶𝐶 replaced by a short circuit
𝑅𝑇𝐻 =
𝑅1𝑅2
𝑅1+𝑅2

19. Thevenin’s Voltage
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𝑉𝑇𝐻 is the open circuit voltage
between AB
𝑉𝑇𝐻 =
𝑉𝐶𝐶∗𝑅2
𝑅1+𝑅2
A
B

20. Self bias circuit with input loop replaced by Thevenin’s equivalent
circuit
Apply KVL to the input loop,
𝑉𝑇𝐻 - 𝐼𝐵𝑅𝑇𝐻 - 𝑉𝐵𝐸 - 𝐼𝐸𝑅𝐸 = 0
Substitute 𝐼𝐸 = β + 1 𝐼𝐵
𝐼𝐵=
𝑉𝑇𝐻−𝑉𝐵𝐸
𝑅𝑇𝐻 + (β+1)𝑅𝐸
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21. Since β+1 ≃ β and β𝑅𝐸>> 𝑅𝑇𝐻
𝐼𝐵=
𝑉𝑇𝐻−𝑉𝐵𝐸
β𝑅𝐸
Collector current 𝐼𝐶= β*𝐼𝐵
𝐼𝐶=
𝑉𝑇𝐻−𝑉𝐵𝐸
𝑅𝐸
𝐼𝐶 is now independent of β , where β is a temperature dependent factor
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22. Applying KVL to the output loop,
𝑉𝐶𝐶-𝐼𝐶𝑅𝐶-𝑉𝐶𝐸-𝐼𝐸𝑅𝐸=0
𝑉𝐶𝐸 = 𝑉𝐶𝐶-𝐼𝐶𝑅𝐶-𝐼𝐸𝑅𝐸
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25. Q4. For the self bias circuit shown below, draw the DC load line and
mark the Q point/ operating point.(Si transistor used)
𝑅𝐸= 200Ω , 𝑅1=10𝑅2= 10kΩ , 𝑅𝑐=2kΩ, β=200 , 𝑉𝐶𝐶=15V
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26. 𝑅𝑇𝐻 =
𝑅1𝑅2
𝑅1+𝑅2
= 0.909kΩ
𝑉𝑇𝐻 =
𝑉𝐶𝐶∗𝑅2
𝑅1+𝑅2
= 1.364V
Apply KVL to the input loop,
𝑉𝑇𝐻 - 𝐼𝐵𝑅𝑇𝐻 - 𝑉𝐵𝐸 - 𝐼𝐸𝑅𝐸 = 0
Substitute 𝐼𝐸 = β + 1 𝐼𝐵
𝐼𝐵=
𝑉𝑇𝐻−𝑉𝐵𝐸
𝑅𝑇𝐻 + (β+1)𝑅𝐸
=
1.364−0.7
0.909𝑘+ 201 200
= 16.15µA
𝐼𝐶𝑄= β𝐼𝐵= 3.23mA
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27. Apply KVL to the output loop,
𝑉𝐶𝐸 = 𝑉𝐶𝐶-𝐼𝐶𝑅𝐶-𝐼𝐸𝑅𝐸
𝑉𝐶𝐸𝑄 = 𝑉𝐶𝐶-𝐼𝐶𝑅𝐶- β + 1 𝐼𝐵𝑅𝐸 = 7.9V
To draw DC Load Line
𝑉𝐶𝐸𝑚𝑎𝑥 = 𝑉𝐶𝐶 =15V
𝐼𝐶𝑚𝑎𝑥 =
𝑉𝐶𝐶
𝑅𝐶+𝑅𝐸
= 6.81mA
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28. Q5. For the self bias circuit shown below, draw the DC load line and
mark the Q point/ operating point.(Si transistor used)
𝑅𝐸= 100Ω , 𝑅1=100k Ω, 𝑅2= 5kΩ , 𝑅𝑐=2kΩ, β=50 ,
𝑉𝐶𝐶=20V
Answer:
𝑅𝑇𝐻 =
𝑉𝑇𝐻 =
𝐼𝐵=
𝑉𝐶𝐸𝑄 =
𝑉𝐶𝐸𝑚𝑎𝑥 = 𝑉𝐶𝐶 =
𝐼𝐶𝑚𝑎𝑥 =
𝑉𝐶𝐶
𝑅𝐶+𝑅𝐸
=
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29. Q6. For a self bias circuit using Si BJT, 𝑅𝐶= 500Ω , 𝑅𝐸= 300Ω 𝑉𝐶𝐶=15V,
β=100, 10𝑅2= β𝑅𝐸. Determine the value of 𝑅1 to get 𝑉𝐶𝐸𝑄=
𝑉𝐶𝐶
2
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