COMSATS Institute of Information Technology
Virtual campus
Islamabad
Dr. Nasim Zafar
Electronics 1 EEE 231
Fall Semester – 2012
PotentialDividerBiasing Circuits:
Lecture No:
19
Examples and Exercises.
.
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References:
Ø Microelectronic Circuits:
Adel S. Sedra and Kenneth C. Smith.
Ø
Integrated Electronics :
Jacob Millman and Christos Halkias (McGrawHill).
Ø Introductory Electronic Devices and Circuits
Robert T. Paynter
Ø
Electronic Devices :
Nasim Zafar
Thomas L. Floyd ( Prentice Hall ).
3
Basic Circuits of BJT:
NPN Transistor
IE = IC + IB
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Transistor Output Characteristics:
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Transistor Output Characteristics:
Load Line – Biasing and Stability:
Ø
Active region:
–
–
Ø
Ø
Ø
BJT acts as a signal amplifier.
BE junction is forward biased and CB junction is reverse biased.
Graphical construction for determining the dc collector current IC and the
collectortoemitter voltage VCE .
The requirement is to set the Qpoint 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
v
Active region Amplifier: BJT acts as a Signal Amplifier.
1. BE Junction Forward Biased
C
C
IC
VBE ≈ 0.7 V for Si
2. BC Junction Reverse Biased
IC
B
B
IB
IB
E
IE
E
3. KCL: IE = IC + IB
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IE
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The DC Operating Point:
Biasing and Stability
Slope of the Load Line:
VCC = VCE + VRC
VCE = VCC VRC
VCE = VCC IC
RC
Ic
1
( )VCE
Rc
VCC
RC
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Current Equations in a BJT:
NPN Transistor
Ø
Ø
Ø
Collector Current
iC
iB
Base Current
Emitter Current
iE
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In
I se
iC
Is
iC
Is
v BE
e
v BE
e
9
VT
VT
v BE
VT
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1. FixedBiased Transistor Circuits.
BaseBiased (Fixed Bias) Transistor Circuit:
Single Power Supply
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BaseBiased (Fixed Bias) Transistor Circuit:
Circuit Characteristics 1:
VCC
IC
RC
RB
Circuit Recognition: A single resistor RB
between the base terminal and VCC. No emitter
resistor.
Output
IB
Input
Q1
+0.7 V
VBE
Advantage: Circuit simplicity.
Disadvantage: Qpoint shifts with temp.
IE
Applications: Switching circuits only.
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VCC
Load line equations:
VCC
I C (sat ) ≅
RC
BaseBiased (Fixed Bias) Transistor Circuit:
IC
RC
RB
Circuit Characteristics 2:
Output
Qpoint equations:
IB
Input
Q1
+0.7 V
VBE
VCE (off ) = VCC
IE
VCC − VBE
IB =
RB
I C = hFE I B
VCE = VCC − I C RC
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BaseBiased (Fixed Bias) Transistor Circuit:
Qpoint equations:
VCC
1. Base–Emitter Loop:
VCC = VBE + IB RB
IC
RC
RB
Output
IB
Input
Q1
+0.7 V
VBE
VCC − VBE
IB =
RB
I C = βI B
IE
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BaseBiased (Fixed Bias) Transistor Circuit:
2. Collector–Emitter Loop:
VCC
VCC = VCE +
IC
VRC
VCE = VCC IC R
RC
RB
Output
IB
Input
Q1
+0.7 V
VBE
IE
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I C = βI B
β = dc current gain = hFE
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Circuit 19.1; Example 19.1
+8 V
IC
RB
360 k
VCC − 0.7V 8V − 0.7V
IB =
=
RB
360kΩ
RC
2k
hFE = 100
VBE
I C = hFE I B = ( 100 ) ( 20.28μA )
= 2.028mA
IB
+0.7 V
= 20.28μA
IE
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VCE = VCC − I C RC
= 8V − ( 2.028mA ) ( 2kΩ)
= 3.94V
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Example 19.2
Construct the DC Load line for circuit 19.1; shown in slide 12, and
plot the Qpoint from the values obtained in Example 19.1.
Determine whether the circuit is midpoint biased.
I C (mA)
I C (sat )
4
VCC
8V
=
=
= 4mA
RC 2kΩ
VCE ( off ) = VCC = 8V
3
Q
2
The circuit is midpoint biased.
1
2
4
6
8
10
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VCE (V)
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Example 19.3 (Qpoint 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 Qpoint values of IC and VCE
at both of these temperatures.
+8 V
I
RB C
360 k
IB
+0.7 V
VBE
Temp(°C)
IB (mA)
IC (mA)
VCE (V)
25
20.28
2.028
3.94
100
20.28
3.04
1.92
RC
2k
hFE = 100 (T = 25 C)
hFE = 150 (T = 100 C)
IE
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3.
2. VoltageDividerBias Circuits.
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VoltageDivider Bias Circuits:
NPN Transistor.
Ø
Ø
Voltagedivider 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 emitterbiased
circuit.
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VoltageDivider Bias Characteristics(1)
+VCC
I1
R1
IC
RC
IB
R2
Advantages: The circuit Q
point values are stable against
changes in hFE.
Output Disadvantages: Requires more
components than most other
biasing circuits.
Input
I2
Circuit Recognition: The
voltage divider in the base
circuit.
IE
RE
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Applications: Used primarily to
bias linear amplifier.
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VoltageDivider Bias Characteristics(2)
+VCC
I1
R1
IC
The Thevenin voltage:
RC
IB
Output
Input
I2
R2
IE
RE
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VoltageDivider Bias Characteristics(3)
+VCC
Load line equations:
VCC
I C (sat) =
RC + RE
VCE (off ) = VCC
I1
R1
IC
RC
IB
Qpoint equations (assume
that hFERE > 10R2):
Output
VB = VCC
R2
R1 + R2
VE = VB − 0.7V
Input
I2
R2
IE
RE
I CQ ≅ I E =
VE
RE
VCEQ = VCC − I CQ ( RC + RE )
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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:
+10 V
VB = VCC
R2
R1 + R2
4.7kΩ
= 2.07V
22.7kΩ
VE = VB − 0.7V
= 2.07V − 0.7V = 1.37V
= ( 10V )
I1
R1
18 k
IC
RC
3k
IB
Because ICQ @ IE (or hFE >> 1),
hFE = 50
I2
R2
4.7 k
IE
RE
1.1 k
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I CQ ≅
VE 1.37V
=
= 1.25mA
RE 1.1kΩ
VCEQ = VCC − I CQ ( RC + RE )
= 10V − ( 1.25mA ) ( 4.1kΩ) = 4.87V
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Circuit 19.2; Example 19.4 (b).
Verify that I2 > 10 IB.
+10 V
I1
R1
18 k
IC
RC
3k
IB
hFE = 50
I2
R2
4.7 k
IE
VB 2.07V
I2 =
=
= 440.4μA
R2 4.7kΩ
IE
1.25mA
IB =
=
hFE + 1
50+1
= 24.51μA
∴ I 2 > 10 I B
RE
1.1 k
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Example 19.5
A voltagedivider 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.
VB = VCC
R2
680Ω
= ( 10V )
= 3.12V
R1 + R2
2180Ω
VE = VB − 0.7V = 3.12V − 0.7V = 2.42V
ICQ ≅ I E =
VE 2.42V
=
= 10mA
RE 240Ω
hFE ( ave ) = hFE (min) hFE (max) = 100 300 = 173
IB =
IE
10mA
= 57.5μA
hFE (ave) + 1 174
=
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