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COMSATS Institute of Information Technology Virtual campus Islamabad. Dr. Nasim Zafar Electronics 1 - EEE 231 Fall Semester – 2012. Potential-Divider-Biasing Circuits: Examples and Exercises. Lecture No: 19 Contents: Base-Biased (Fixed Bias) Transistor Circuits. - PowerPoint PPT PresentationTRANSCRIPT
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Dr. Nasim Zafar
Electronics 1 - EEE 231
Fall Semester – 2012
COMSATS Institute of Information TechnologyVirtual campus
Islamabad
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Potential-Divider-Biasing Circuits:Examples and Exercises.
.
Lecture No: 19
Contents:
Base-Biased (Fixed Bias) Transistor Circuits.
Voltage-Divider-Bias transistor Circuits.
Examples and Exercises.
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References:
Microelectronic Circuits:
Adel S. Sedra and Kenneth C. Smith.
Integrated Electronics :
Jacob Millman and Christos Halkias (McGraw-Hill).
Introductory Electronic Devices and Circuits
Robert T. Paynter
Electronic Devices :
Thomas L. Floyd ( Prentice Hall ).
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Basic Circuits of BJT: NPN Transistor
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IE = IC + IB
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Transistor Output Characteristics:
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Transistor Output Characteristics: Load Line – Biasing and Stability:
Active region:– BJT acts as a signal amplifier.– B-E junction is forward biased and C-B junction is reverse biased.
Graphical construction for determining the dc collector current IC and the
collector-to-emitter voltage VCE .
The requirement is to set the Q-point such that that it does not go into the saturation or cutoff regions when an a ac signal is applied.
Maximum signal swing depends on the bias voltage.
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The DC Operating Point:Biasing and Stability
Active region - Amplifier: BJT acts as a Signal Amplifier.
1. B-E Junction Forward Biased
VBE ≈ 0.7 V for Si
2. B-C Junction Reverse Biased
3. KCL: IE = IC + IB
C
B
E
IB
IE
IC
C
B
E
IB IE
IC
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The DC Operating Point:Biasing and Stability
Slope of the Load Line:
C
CCCE
cc R
VV
R
)
1(I
VCC = VCE + VRC
VCE = VCC - VRC
VCE = VCC - IC RC
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Current Equations in a BJT:NPN Transistor
Collector Current
Base Current
Emitter Current T
BEV
vsC
E eIi
i
T
BEV
v
snC eIIi
T
BEV
vsC
B eIi
i
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1. Fixed-Biased Transistor Circuits.
Base-Biased (Fixed Bias) Transistor Circuit:
Single Power Supply
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Base-Biased (Fixed Bias) Transistor Circuit:
RC
RB
+0.7 V
IC
IB
IE
Input
Output
VBE
VCC
Q1 Advantage: Circuit simplicity.
Disadvantage: Q-point shifts with temp.
Applications: Switching circuits only.
Circuit Recognition: A single resistor RB between the base terminal and VCC. No emitter resistor.
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Circuit Characteristics - 1:
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Base-Biased (Fixed Bias) Transistor Circuit:
Circuit Characteristics - 2:
RC
RB
+0.7 V
IC
IB
IE
Input
Output
VBE
VCC
Q1
(sat )
(off )
CCC
C
CE CC
VI
R
V V
Load line equations:
Q-point equations:
CC BEB
B
C FE B
CE CC C C
V VI
R
I h I
V V I R
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Base-Biased (Fixed Bias) Transistor Circuit:Q-point equations:
CC BEB
B
V VI
R
βC BI I
RC
RB
+0.7 V
IC
IB
IE
Input
Output
VBE
VCC
Q1
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1. Base–Emitter Loop:
VCC = VBE + IB RB
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Base-Biased (Fixed Bias) Transistor Circuit:
βC BI I
RC
RB
+0.7 V
IC
IB
IE
Input
Output
VBE
VCC
Q1
= dc current gain = hFE
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2. Collector–Emitter Loop:
VCC = VCE + VRC
VCE = VCC - IC R
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Circuit 19.1; Example 19.1
RC2 k
RB360 k
+0.7 V
IC
IB
IEVBE
+8 V
hFE = 100
0.7V 8V 0.7V
360kΩ
20.28μA
CCB
B
VI
R
100 20.28μA
2.028mAC FE BI h I
8V 2.028mA 2kΩ
3.94V
CE CC C CV V I R
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Example 19.2
Construct the DC Load line for circuit 19.1; shown in slide 12, and plot the Q-point from the values obtained in Example 19.1. Determine whether the circuit is midpoint biased.
VCE (V)2 4 6 8 10
1
2
3
4
IC (mA)
Q
(sat )
8V4mA
2kΩCC
CC
VI
R
off 8VCCCEV V
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The circuit is midpoint biased.
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Example 19.3 (Q-point Shift.)
The transistor of Circuit 19.1, has values of hFE = 100 when T = 25 °C and
hFE = 150 when T = 100 °C. Determine the Q-point values of IC and VCE at both of these temperatures.
RC2 k
RB360 k
+0.7 V
IC
IB
IEVBE
+8 V
hFE = 100 (T = 25C)hFE = 150 (T = 100C)
Temp(°C) IB (mA) IC (mA) VCE (V)
25 20.28 2.028 3.94
100 20.28 3.04 1.92
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3.
2. Voltage-Divider-Bias Circuits.
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Voltage-Divider Bias Circuits:NPN Transistor.
Voltage-divider biasing is the most common form of transistor biasing used. A thorough understanding of the dc analysis of this circuit is essential for an electronic technician.
In the Circuit, R1 and R2 set up a voltage divider on the base. Notice the similarity to the emitter-biased circuit.
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Voltage-Divider Bias Characteristics-(1)
R1
R2 RE
RC
+VCC
Input
Output
I1
I2 IE
IB
IC
Circuit Recognition: The voltage divider in the base circuit.
Advantages: The circuit Q-point values are stable against changes in hFE.
Disadvantages: Requires more components than most other biasing circuits.
Applications: Used primarily to bias linear amplifier.
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Voltage-Divider Bias Characteristics-(2)
R1
R2 RE
RC
+VCC
Input
Output
I1
I2 IE
IB
IC
The Thevenin voltage:
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Voltage-Divider Bias Characteristics-(3)
R1
R2 RE
RC
+VCC
Input
Output
I1
I2 IE
IB
IC
Load line equations:
(sat )
(off )
CCC
C E
CE CC
VI
R R
V V
Q-point equations (assume that hFERE > 10R2):
2
1 2
0.7V
B CC
E B
ECQ E
E
CEQ CC CQ C E
RV V
R R
V V
VI I
R
V V I R R
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Circuit 19.2; Example 19.4 (a).
Determine the values of ICQ and VCEQ for the circuit 19.2 shown in Fig below:
R118 k
R24.7 k
RE1.1 k
RC3 k
+10 V
I1
I2IE
IB
IC
hFE = 50
2
1 2
4.7kΩ10V 2.07V
22.7kΩ
B CC
RV V
R R
0.7V
2.07V 0.7V 1.37VE BV V
Because ICQ @ IE (or hFE >> 1),
1.37V1.25mA
1.1kΩE
CQE
VI
R
10V 1.25mA 4.1kΩ 4.87V
CEQ CC CQ C EV V I R R
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Circuit 19.2; Example 19.4 (b).
Verify that I2 > 10 IB.
R118 k
R24.7 k
RE1.1 k
RC3 k
+10 V
I1
I2IE
IB
IC
hFE = 50
22
2.07V440.4μA
4.7kΩBV
IR
1.25mA
1 50+1
24.51μA
EB
FE
II
h
2 10 BI I
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Example 19.5
A voltage-divider bias circuit has the following values: R1 = 1.5 kW, R2 = 680 W, RC = 260 W, RE = 240 W and VCC = 10 V. Assuming the transistor is a 2N3904, determine the value of IB for the circuit.
2
1 2
680Ω10V 3.12V
2180ΩB CC
RV V
R R
0.7V 3.12V 0.7V 2.42VE BV V
2.42V10mA
240ΩE
CQ EE
VI I
R
( ) (min) (max) 100 300 173FE ave FE FEh h h
(ave)
10mA57.5μA
1 174E
BFE
II
h
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Load Line for Voltage Divider Bias Circuit.Example 19.5
2 4 6 8 10 12
5
10
15
20
25
IC (mA)
VCE (V)
(sat )
10V20mA
260Ω+240ΩCC
CC E
VI
R R
(off ) 10VCE CCV V
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Which value of hFE do we use?
Transistor specification sheet may list any combination of the following hFE: max. hFE, min. hFE, or typ. hFE. Use typical value if there is one. Otherwise, use
(ave) (min) (max)FE FE FEh h h
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Stability of Voltage Divider Bias Circuit:
The Q-point of voltage divider bias circuit is less dependent on hFE than that of the base bias (fixed bias).
For example, if IE is exactly 10 mA, the range of hFE is 100 to 300. Then
10mAAt 100, 100μA and 9.90mA
1 101E
FE B CQ E BFE
Ih I I I I
h
10mAAt 300, 33μA and 9.97mA
1 301E
FE B CQ E BFE
Ih I I I I
h
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Voltage-Divider Bias Circuit: Circuit-19.3; Problem 19.6 (a).
Find the operating point Q for this circuit.
The use of Thevenin equivalent circuit for the base makes the circuit simpler.
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Determination of VBB – The Thevenin Voltage
VCC = I.(R1 + R2)
-- Eq. (1)
VThev = I.R2 Eq. (2)
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Circuit-19.3; Problem 19.6 (a)Determination of VBB
From Eq (3)
VThev = 2 Volts
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Circuit-19.3; Problem 19.6 (b).Determination of VRE
Input Loop with RE
VBB = VBE + VRE
VRE = VBB – VBE
VRE = 2V - 0.7V
VRE = 1.3V
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Circuit-19.3; Problem 19.6 (c).Determination of IE
VRE = IERE
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Circuit-19.3; Problem 19.6 (d).Determination of VRC
Since
IE ≈ IC
IE = 1.3mATherefore,
IC = 1.3mA
VRC = ICRC
= (1.3mA)(4x103Ω)
VRC = 5.2V
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Circuit-19.3; Problem 19.6 (e).Determination of VCE
Output Loop
VCC=VRC+VCE+VRE
VCE = VCC-VRC-VRE
VCE = 12V - 5.2V - 1.3V
VCE = 5.5V
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Results of Problem 19.6
IE = IC = 1.3mA
VRC = 5.2V
VCE = 5.5V
VRE = 1.3V
VBB = 2Vβdc was never used in a calculation. Hence, voltage-divider biased circuits are immune to changes in βdc.
A single voltage source supplies both voltages, VCC and VBB
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Review of equations:
In Review
VRE = VBB – VBE
IE ≈ IC
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Summary:
Voltage-divider biased circuits are immune to changes in βdc.
A single voltage source supplies both voltages, VCC and VBB
The circuit Q-point values are stable against changes in hFE.
Use of the Thevenin equivalent circuit for the base makes the
circuit simpler.
Make the current in the voltage divider about 10 times IB,
to simplify the analysis.
For design, solve for the resistor values (IC and VC specified).
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Circuit 19.4; Problem 19.7 (a).
Given: VB = 3V and I = 0.2mA.
IB
I
(a) RB1 and RB2 form a voltage divider.
Assume I >> IB
I = VCC/(RB1 + RB2)
0.2mA = 9 /(RB1 + RB2)
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Circuit 19.4; Problem 19.7 (b).
Given: VB = 3V and I = 0.2mA.
RB1 = 30KW, and RB2 = 15KW.
IB
I
VB = VCC[RB2/(RB1 + RB2)]
3 = 9 [RB2/(RB1 + RB2)],
Solve for RB1 and RB2.
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(b) Determination of the Thevenin voltage:
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Prob. 19.7 (c).
Find the operating point
The use of Thevenin equivalent circuit for the base makes the circuit simpler.
VBB = VB = 3V
• RBB = RB1|| RB2 = 30K || W 15K = W 10KW
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Problem 19.7 (d).
Write B-E loop and C-E loop
B-Eloop
C-E loop
B-E Voltage Loop:
VBB = VRBB + VBE + VRE
VBB = IBRBB + VBE + IERE
IE =2.09 mA
C-E Voltage Loop:
VCC = ICRC + VCE + IERE
VCE =4.8 V
This is how all DC circuits are analyzed
and designed!
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Example 19.7
Stage 2
• C-E loop
IE2
IC2
VCC = IE2RE2 + VEC2 +IC2RC2
15 = 2.8(2) + VEC2 + 2.8 (2.7)solve for VEC2
VCE2 = 1.84V
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Example 19.7
C-E loop
neglect IB2 because it is IB2 << IC1
IE1
IC1
VCC = IC1RC1 + VCE1 +IE1RE1
15 = 1.3(5) + VCE1 +1.3(3)
VCE1= 4.87V
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Example 19.7
Stage 2
• B-E loop
IB2
IE2VCC = IE2RE2 + VEB +IB2RBB2 + VBB2
15 = IE2(2K) + .7 +IB2 (5K) + 4.87 + 1.3(3)Use IB2 IE2/ , b solve for IE2
IE2 = 2.8mA
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Example 19.7
2-stage amplifier, 1st stage has an npn transistor; 2nd stage has an pnp transistor.
IC = bIB
IC IE
VBE = 0.7(npn) = -0.7(pnp)
b = 100
Find IC1, IC2, VCE1, VCE2
• Use Thevenin circuits.
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Example 19.7
• RBB1 = RB1||RB2 = 33K
• VBB1 = VCC[RB2/(RB1+RB2)]
VBB1 = 15[50K/150K] = 5V
Stage 1
• B-E loopVBB1 = IB1RBB1 + VBE +IE1RE1
Use IB1 IE1/ b
5 = IE133K / 100 + .7 + IE13K
IE1 = 1.3mA
IB1
IE1
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