Chapter 5portal.unimap.edu.my/portal/page/portal30/Lecturer Notes... · Collector-Emitter Loop From...
Transcript of Chapter 5portal.unimap.edu.my/portal/page/portal30/Lecturer Notes... · Collector-Emitter Loop From...
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Chapter 5
Transistor bias circuits
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Objectives
Discuss the concept of dc biasing of a transistor for
linear operation
Analyze voltage-divider bias, base bias, emitter bias
and collector-feedback bias circuits.and collector-feedback bias circuits.
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Biasing
Biasing: The DC voltages applied to a transistor in
order to turn it on so that it can amplify the AC
signal.
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Operating Point
The DC input establishes
an operating or
quiescent point called the
Q- point.
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The Three States of Operation
• Active or Linear Region Operation
• Base–Emitter junction is forward biased Base–
Collector junction is reverse biased
• Cutoff Region Operation
• Base–Emitter junction is reverse biased• Base–Emitter junction is reverse biased
• Saturation Region Operation
• Base–Emitter junction is forward biased Base–
Collector junction is forward biased
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DC Biasing Circuits
• Fixed-bias circuit
• Emitter-stabilized bias circuit
• Collector-emitter loop
• Voltage divider bias circuit• Voltage divider bias circuit
• DC bias with voltage feedback
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Fixed Bias
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The Base - Emitter Loop
From Kirchhoff’s voltage law:
+VCC – IBRB – VBE = 0
Solving for base current:Solving for base current:
B
− VBEVCC
RI B ====
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Collector-Emitter Loop
Collector current:
I ==== ββββIBC
From Kirchhoff’s voltage law:From Kirchhoff’s voltage law:
VCE ==== VCC −−−− ICRC
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Saturation
When the transistor is operating in saturation, current
through the transistor is at its maximum possible value.
VCCI Csat ====
R CCsat
R C
VCE ≅≅≅≅ 0 V
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Load Line Analysis
The end points of the load line
are: ICsat
IC = VCC / RC
VCE = 0 V
VCEcutoffVCEcutoff
VCE = VCC
IC = 0 mA
The Q-point is the operating point:
• where the value of RB sets the value
of IB
• that sets the values of VCE and IC
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Circuit Values Affect the Q-Point
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Circuit Values Affect the Q-Point
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Circuit Values Affect the Q-Point
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Emitter-Stabilized Bias Circuit
Adding a resistor (RE)
to the emitter circuit
stabilizes the biasstabilizes the bias
circuit.
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Base-Emitter Loop
From Kirchhoff’s voltage law:
+ VCC - IERE - VBE - IERE ====0
Since IE = (ββββ + 1)IB:
VCC - IBRB - (ββββ ++++1)IBRE ==== 0VCC - IBRB - (ββββ ++++1)IBRE ==== 0
Solving for IB:
EB
VCC - VBE
++++ (ββββ ++++1)RIB ==== R
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Collector-Emitter Loop
From Kirchhoff’s voltage law:
CE C C CC+ I R − V = 0I
ER
E + V
Since IE ≅≅≅≅ IC:
VCE ==== VCC – IC(RC ++++ RE )VCE ==== VCC – IC(RC ++++ RE )
Also:
VE ==== IERE
VC ==== VCE ++++ VE ==== VCC - ICRC
VB ==== VCC – IRRB ==== VBE ++++ VE
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Improved Biased Stability
Stability refers to a circuit condition in which the currents
and voltages will remain fairly constant over a wide range of
temperatures and transistor Beta (ββββ) values.
Adding RE to the emitter improves the stability of aAdding RE to the emitter improves the stability of a
transistor.
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Saturation Level
The endpoints can be determined from the load line.
VCEcutoff: ICsat:
VCE ==== VCC
IC ==== 0mA VCC
RC ++++ RIC ====
VCE ==== 0 V
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Voltage Divider Bias
This is a very stable
bias circuit.
The currents and The currents and
voltages are nearly
independent of any
variations in ββββ.
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Approximate Analysis
Where IB << I1 and I1 ≅≅≅≅ I2:
21B
R ++++ R
R2VCCV ====
Where ββββRE > 10R2:
I ====VE
EE
R
VE ==== VB −−−− VBE
From Kirchhoff’s voltage law:
VCE = VCC − ICRC −IERE
IE ≅ IC
VCE =VCC−IC(RC + RE )
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Voltage Divider Bias Analysis
Transistor Saturation Level
VCCIEC
==== ICmax ====CsatR ++++ R
Load LineAnalysis Load LineAnalysis
Cutoff: Saturation:
VCE ==== VCC
IC ==== 0mA
CE
VCC
RC ++++ RE
V ==== 0V
IC ====
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DC Bias with Voltage Feedback
Another way to improve
the stability of a bias
circuit is to add a
feedback path from
collector to base.collector to base.
In this bias circuit the Q-
point is only slightly
dependent on the
transistor beta, ββββ.
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Base-Emitter Loop
From Kirchhoff’s voltage law:
VCC – I′′′′CRC – IBRB – VBE – IERE ==== 0
Where IB << IC:
CI' = I
C + I
B ≅ I
C
Knowing I = ββββI and I ≅≅≅≅ I , the loop Knowing IC = ββββIB and IE ≅≅≅≅ IC, the loop
equation becomes:
VCC – ββββIBRC −−−− IBRB −−−− VBE −−−− ββββIBRE ==== 0
Solving for IB:
RB ++++ ββββ(RC ++++ RE )
VCC −−−− VBEIB ====
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Collector-Emitter Loop
Applying Kirchoff’s voltage law:
IE + VCE + I’CRC – VCC = 0
Since I′′′′′′′′C ≅≅≅≅ IC and IC = ββββIB:
IC(RC + RE) + VCE – VCC=0
Solving for VCE:
VCE = VCC – IC(RC + RE)
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Base-Emitter Bias Analysis
Transistor Saturation Level
ECCsat
VCCIR ++++ R
==== ICmax ====
Load LineAnalysis
Cutoff: Saturation:
V= VCCVCE
IC = 0mA ECC
I
VCE = 0 V
R + R= CC
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PNP Transistors
The analysis for pnp transistor biasing circuits is the same as that for
npn transistor circuits. The only difference isthat the currents are
flowing in the opposite direction.
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Base voltage
Emitter voltage
By Ohm’s Law,
Analysis of Voltage Bias for PNP Transistor
EE
EDC
B VRRR
RV
+=
ββββ21
1
BEBE VVV +=
VRIRIVV +++= By Ohm’s Law,
And,
DC
B
E
BEBEE
E RR
VVVI
β+
−−=
EECCCCEC
CCC
RIRIVV
RIV
−−=
=
BEEEBBBEEVRIRIVV +++=
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Evaluate IC and VEC for pnp transistor circuit in Figure below.
Given VEE = +15V, R1 = 63kΩ, R2 = 27kΩ, RC = 1.8kΩ, RE =
2.6kΩ, βDC =120.
Example 1
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Figure below shown the schematic with a negative supply
voltage, determine IC and VCE for a pnp transistor circuit with
given values: R1 = 25kΩ, R2 = 60kΩ, RC = 6kΩ, RE = 9kΩ,
VCC = -12V, and βDC = 90
Example 2
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Construct a complete circuit required to replace the transistor in
Figure below with a pnp transistor. Given VCC = 10V, R1 =
78kΩ, R2 = 100kΩ, RC = 18kΩ, RE = 8kΩ.
Example 3
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