Download Session 11 Introduction to bipolar junction transistor - OCW

Session 11
Introduction to bipolar
junction transistor (BJT)
Electronic Components and Circuits
José A. Garcia Souto
www.uc3m.es/portal/page/portal/dpto_tecnologia_electronica/Personal/JoseAntonioGarcia
Bipolar Junction Transistor
BJT
OBJECTIVES
•
•
•
•
•
Knowing the structure of the device and the transistor
effect
Knowing and distinguishing basic parameters related to
BJT transistors: α, β, hFB, hFE, ICBO, VBE(on), VCE(sat)
Reading the current-voltage characteristic of the device
Identifying the operating regions
Active, Cut-off, Saturation, Inverse Active
Analyze basic cases in DC of BJT circuits
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CCE - Session 11
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Bipolar Transistor Structure
Emitter
n+
E
Base
p
Collector
n
C
p
C
B
p+
E
n
B
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Operation: Transistor Effect
W
iE
E
p ++
n
iC
p
C
h+
VEB
I E = IC + I B
iB
VBC
B
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I C = α ⋅ I E + I CBO
α ≈1
I B = (1 − α ) ⋅ I E − I CBO
IC ≈ I E
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Characteristic parameters
• Physical Parameters
–α
–β
– ICBO
• Daasheet:
– hFB
– hFE
– ICBO
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IC ≈ α ⋅ I E
IC ≈ β ⋅ I B
α
β=
1−α
α ≈1
β >> 1
I C = α ⋅ I E + I CBO
I C = β ⋅ I B + I CBO
Search BC547 in http://www.fairchildsemi.com/
I_Collector
I_Emitter
I_Collector
Common-Base
CCE - Session 11
I_Base
Common-Emitter
5
Dual case: Transistor NPN
iE
E
n ++
p
iC
n
C
Ie+
VBE
iB
VCB
B
E
p
B
C
n ++
n
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Interpretation: current-voltage
characteristic
IC (mA)
6
6 mA
5
5 mA
4
IE =4 mA
In SATURATION the
BC junction is ON
The output voltage VCB
is constant even the IC
current changes
3
2
1
In ACTIVE the BC junction is OFF
3 mA
The output current IC of the device is
contant even the VCB voltage changes
2 mA
The IC current is determined by IE
1 mA
0 mA
0 2
4
6 8 10 12 14
There is a curve for each input value IE
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Operating Regions
Region
Base-Emitter Junction
Base-Collector Junction
Cut-off
Reverse(OFF)
Reverse
Active
Forward (ON)
Reverse (Transistor Effect)
Saturation
Forward (ON)
Forward (Saturated)
Active Reverse
Reverse
Forward
Region
Conditions NPN
Operation NPN
Cut-off
VBE < VBE-ON o IB = 0
IB=0, IC=IE=0
Active
VBE-ON y VCE > VCE-SAT
IC ≈ hFE·IB
[VBE=VBE-ON]
Saturation
VBE-ON y VCE-SAT
VCE =VCE-SAT
[VBE=VBE-SAT]
• Datasheet VBE(on) VCE(sat) hFE
VBE(sat)
Search BC547 in http://www.fairchildsemi.com/
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Output Characteristics
(Common - Emitter)
IC (mA)
6
60 µA
5
50 µA
4
IB =40 µA
3
30 µA
2
20 µA
1
10 µA
0
2
4
6
8
10 12 14
0 µA
16
VCE (V)
What input characteristic represent?
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Transfer Charateristic
IC (mA)
60
50
40
30
20
10
0
0,2
0,4
0,6
0,8 VBE (V)
What operating regions can be identified?
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Example: Operating Regions
EXAMPLE
IC
RC = 100 Ω
RB = …
IB
VCC = 10 V
VCE
VBB = …
VBE(on) = 0,7 V
VBE
VCE(sat) = 0,2 V
IE
hFE = 100
Case 1
Case 2
Case 3
Case 4
VBB = 0 V
IB = 0
VBB = 10 V
VBB = 10 V
RB = 10 kΩ
RB = 10 kΩ
RB = 10 kΩ
RB = 1 kΩ
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Currents and Voltages of
NPN and PNP
IC
IE
IB
VEB
VCE
VEC
VBE
IB
IE
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IC
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Operating Regions
Case of PNP transistor
• Datasheet VEB(on) VEC(sat) hFE
VEB(sat)
Search BC557 in http://www.fairchildsemi.com/
Region
Emitter-Base Junction
Collector-Base Junction
Cut-off
Reverse(OFF)
Reverse
Active
Forward (ON)
Reverse (Transistor Effect)
Saturation
Forward (ON)
Forward (Saturated)
Region
Conditions PNP
Operation PNP
Cut-off
VEB < VEB-ON o IB = 0
IB=0, IC=IE=0
Active
VEB-ON y VEC > VEC-SAT
IC ≈ hFE·IB
[VEB=VEB-ON]
Saturation
VEB-ON y VEC-SAT
VEC =VEC-SAT
[VEB=VEB-SAT]
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Exercise: Operating Regions
IE
DATA
RC = 100 Ω
IB
RB = …
VEB
VEC
VCC = 10 V
VBB = …
VEB(on) = 0,7 V
VEC(sat) = 0,2 V
IC
hFE = 100
Case 1
Case 2
Case 3
Case 4
VBB = 0 V
IB = 0
VBB = 10 V
VBB = 10 V
RB = 10 kΩ
RB = 10 kΩ
RB = 10 kΩ
RB = 1 kΩ
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Proposed exercise
Voltages and currents
IE
VEB
VEC
IB
IC
Indicate what type of transistor is
Reasoning the operating region
Relate VB and VE, VB and IB, VE and IE, IE and IB.
Relate VC and IC
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