3/29/2011
Summary
Electronic Devices
Ninth Edition
AC Quantities
Floyd
V
AC quantities are indicated
with a italic subscript; rms
values are assumed unless
otherwise stated.
Chapter 6: BJT Amplifiers
rms
avg
Vce
Vce Vce
VCE
The figure shows an example of
a specific waveform for the
collector-emitter voltage. Notice
the DC component is VCE and the
ac component is Vce.
Vce
vce
0
t
0
Resistance is also identified with a lower case subscript when analyzed
from an ac standpoint.
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Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Summary
AC Load Line
Ic
Vb
VBQ
I CQ
RC
R1
Vce
Rs
Ib
C1
I BQ
R2
Vs
C2
VCEQ
RE
RL
For the amplifier shown, notice that the voltage waveform is
inverted between the input and output but has the same shape.
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Electronic Devices, 9th edition
Thomas L. Floyd
Operation of the linear
amplifier can be illustrated
using an ac load line.
The ac load line is different
than the dc load line because
a capacitor looks open to dc
but effectively acts as a
short to ac. Thus the
collector resistor appears to
be in parallel with the load
resistor.
IC
Q
CC
IB
A linear amplifier produces an replica of the input signal at
the output.
+V
Ib
Linear Amplifier
Ic
ICQ
Q
0
V CE
Vce
VCEQ
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Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Summary
Transistor AC Model
The Common-Emitter Amplifier
The five resistance parameters (r-parameters) can be used
for detailed analysis of a BJT circuit. For most analysis
work, the simplified r-parameters give good results.
In the common-emitter (CE) amplifier, the input signal is
applied to the base and the inverted output is taken from the
collector. The emitter is common to ac signals.
VCC
The simplified r-parameters are
shown in relation to the transistor
model.
An important r-parameter is re'. It
appears as a small ac resistance
between the base and emitter.
C
C
βac Ib
RC
βac Ib
R1
C3
Vout
B
re′
B
re′
C1
Vin
Ib
RL
R2
25 mV
r =
IE
'
e
Electronic Devices, 9th edition
Thomas L. Floyd
E
RE
E
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Electronic Devices, 9th edition
Thomas L. Floyd
C2
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1
3/29/2011
Summary
Summary
The Common-Emitter Amplifier
The Common-Emitter Amplifier
What is re' for the CE amplifier? Assume stiff voltage-divider bias.
VCC
27 kΩ


VB = 
15 V = 4.26 V
 68 kΩ + 27 kΩ 
VE = 4.26 V – 0.7 V = 3.56 V
IE =
re' =
VE 3.56 V
=
= 1.62 mA
RE 2.2 kΩ
Notice that the ac resistance of the collector circuit is RC||RL.
What is the gain of the amplifier?
VCC
+15 V
RC
3.9 kΩ
R1
+15 V
V
R
R R
Av = out = c' = C ' L
Vin
re
re
C3
68 kΩ
C1
10 µF
Av =
1.0 µF
RL
3.9 kΩ 3.9 kΩ
= 127
15.4 Ω
C2
RE
2.2 kΩ
25 mV 25 mV
=
= 15.4 Ω
1.62 mA
IE
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C3
1.0 µF
RL
3.9 kΩ
R2
27 kΩ
The gain will be a little lower if the
input loading effect is accounted for.
100 µF
3.9 kΩ
10 µF
3.9 kΩ
R2
27 kΩ
Electronic Devices, 9th edition
Thomas L. Floyd
RC
R1
68 kΩ
C1
RE
2.2 kΩ
C2
100 µF
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Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Summary
The Common-Emitter Amplifier
The Common-Emitter Amplifier
Greater gain stability can be achieved by adding a swamping resistor
to the emitter circuit of the CE amplifier. The gain will be lower as a
result.
V
Multisim is a good way to check your calculation. For an input of
10 mVpp, the output is 378 mVpp as shown on the oscilloscope
display for the swamped CE amplifier.
CC
+15 V
What is the gain with the addition
of the swamping resistor? (Ignore
the small effect on re'.)
Av =
Vout
R
R RL
= ' c = 'C
Vin
re + RE1 re + RE1
RC
R1
3.9 kΩ
C3
68 kΩ
C1
10 µF
input
1.0 µF
RE1
RL
33 Ω
3.9 kΩ
R2
27 kΩ
3.9 kΩ 3.9 kΩ
Av =
= 38.2
15.4 Ω + 33 Ω
100 µF
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Electronic Devices, 9th edition
Thomas L. Floyd
output
C2
RE2
2.2 kΩ
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Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Summary
The Common-Emitter Amplifier
The Common-Collector Amplifier
In addition to gain stability, swamping has the advantage of increasing
the ac input resistance of the amplifier. For this amplifier, Rin(tot) is given
by Rin(tot) = R1||R2||βac(re' + RE1)
The common-collector amplifier (emitter-follower) has a voltage gain of
approximately 1, but can have high input resistance and current gain.
The input is applied to the base and taken from the emitter.
VCC
+15 V
What is Rin(tot) for the amplifier if
βac = 200?
Rin(tot) = R1||R2||βac(re' + RE1)
= 68 kΩ||27 kΩ||200(15.4 Ω + 33 Ω)
RC
R1
Electronic Devices, 9th edition
Thomas L. Floyd
+VCC
C3
68 kΩ
C1
C1
10 µF
R1
Vin
1.0 µF
RE1
3.9 kΩ
2.2 kΩ
Vout
R2
R2
RE2
C2
Iin
RL
33 Ω
27 kΩ
= 6.45 kΩ
3.9 kΩ
RE
C2
RL
100 µF
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Electronic Devices, 9th edition
Thomas L. Floyd
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2
3/29/2011
Summary
Summary
The Common-Collector Amplifier
The Common-Collector Amplifier
The power gain is the ratio of the power delivered to the input resistance
divided by the power dissipated in the load. This is approximately equal to
the current gain. That is, Ap ≈Ai.
You can also write power gain
as a ratio of resistances:
VL2
Rin (tot )
P
R
Ap = L = 2 L = Av2
Pin Vin
RL
Rin (tot )
Calculate the power gain to the load for the CC amplifier using a ratio of
resistances. Assume Av = 1 and βac = 200. Use re' = 2 Ω.
VCC
VCC
+15 V
Rin(tot) = R1||R2||βac(re' + RE||RL)
R1
Vin
C1
= 39 kΩ||220 kΩ||200(2 Ω + 500 Ω)
C2
Vout
R2
 Rin (tot )  Rin (tot )
≅ 1
=
RL
 RL 
The next slide is an example…
RL
RE
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Electronic Devices, 9th edition
Thomas L. Floyd
Vin
= 24.9 kΩ
R1
39 kΩ
C1
C2
0.22 µF
RL = 1.0 kΩ
R
24.9 kΩ
Ap = in (tot ) =
= 24.9
RL
1.0 kΩ
R2
220 kΩ
RE 3.3 µF
1.0 kΩ
Vout
RL
1.0 kΩ
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Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Summary
The Common-Collector Amplifier
The Darlington Pair
The input voltage-divider in the previous example is not “rock-solid” but
the overall power gain is good. A “rock solid” stiff voltage-divider is not
always the best design. Can you spot the problem illustrated here?
Rin(tot) = R1||R2||βac(re' + RE||RL)
VCC
+10 V
A Darlington pair is two transistors connected as shown. The two
transistors act as one “super β” transistor. Darlington transistors are
available in a single package. Notice there are two diode drops from base
to emitter.
V
CC
VCC
= 10 kΩ||10 kΩ||200(25 Ω + 3.0 kΩ)
= 4.96 kΩ
Vin
RL = 10 kΩ
Rin (tot )
C1
R1
10 kΩ
=
R2
10 kΩ
Q1
Q2
Vout
CE Amplifier
Darlington CC amplifier
Load
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Electronic Devices, 9th edition
Thomas L. Floyd
The CB Amplifier
Another high β pair is the Sziklai pair (sometimes called a complementary
Darlington), in which a pnp and npn transistor are connected as shown. This
configuration has the advantage of only one diode drop between base and
emitter.
The common-base (CB) amplifier is used in applications where a low input
impedance is acceptable. It does not invert the signal, an advantage for
higher frequencies as you will see later when you study the Miller effect.
What is the purpose of C2?
+VCC
+VCC
What is the relation between IE2 and IB1?
Vin
βDC1
IB1
βDC2
IC1
C2 forces the base
to be at ac ground.
C2
R1
RC C
3
Vout
IE2
RL
C1
RE
Vin
IE2 is approximately equal to βDC2 x
ITherefore,
C1
I ≈β β I
Electronic Devices, 9th edition
Thomas L. Floyd
RL
Summary
The Sziklai Pair
E2
Vo ut
RE
Summary
The DC currents are:
IC1 is βDC1 x IB1 and is equal to IB2
C2
R2
RL
10 kΩ
RE
4.3 kΩ
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Electronic Devices, 9th edition
Thomas L. Floyd
R1
C1
Vin
C2
4.96 kΩ
= 0.496!
RL
10 kΩ
The problem is the power gain is
less than 1!
Ap =
RC
β = 200
R2
RE
DC1 DC2 B1
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Electronic Devices, 9th edition
Thomas L. Floyd
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3
3/29/2011
Summary
Summary
Multistage Amplifiers
Differential Amplifiers
To improve amplifier performance, stages are often cascaded where the
output of one drives another. This an example of a two-stage direct-coupled
amplifier in which the input and
V
+12 V
output signals are capacitively
coupled.
A differential amplifier (diff-amp) has two inputs. It amplifies the difference
in the two input voltages. This circuit is widely used as the input stage to
operational amplifiers. Differential-mode inputs are illustrated.
CC
R1
10 kΩ
C1
VS
100 mV pp
1.0 kHz
RE3
330 Ω
RC
1.0 kΩ
C3
10 µF
Q2
2N3906
V in
+VCC
Vout
Vout 2
Vout1
RC1
RL
330 Ω
RC2
1
2
Q1
2N3904
1.0 µF
R2
4.7 kΩ
1
RE1
100 Ω
Q1
Q2
2
Vin1
RE2
330 Ω
Vin2
C2
47 µF
RE
–VEE
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Electronic Devices, 9th edition
Thomas L. Floyd
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Electronic Devices, 9th edition
Thomas L. Floyd
Summary
Selected Key Terms
Differential Amplifiers
The same amplifier as in the last slide now is shown with common-mode
inputs. Diff-amps tend to reject common-mode signals, which are usually
due to noise. Ideally, the outputs are zero with common-mode inputs.
+VCC
Vout 2
Vout1
RC1
1
2
Q1
Q2
2
Vin1
Common- A BJT configuration in which the emitter is
emitter the common terminal to an ac signal.
ac ground A point in a circuit that appears as a
ground to ac signals only.
RC2
1
r-parameter One of a set of BJT characteristic parameters
that include αac, βac, re', rb', and rc'.
Vin2
Input resistance The resistance seen by an ac source
connected to the amplifier input.
RE
–VEE
Electronic Devices, 9th edition
Thomas L. Floyd
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Electronic Devices, 9th edition
Thomas L. Floyd
Selected Key Terms
Output The ac resistance looking in at the amplifier
resistance output.
Quiz
1. The equation for finding the ac emitter resistance of a
BJT is
Common- A BJT configuration in which the emitter is
collector the common terminal to an ac signal.
Differential An amplifier in which the output is a function
amplifier of the difference between two input voltages.
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25 mV
IB
'
b. re =
25 mV
IE
re' =
0.7 V
IB
d. re' =
0.7 V
IE
c.
Common-mode A condition where two signals applied to
differential inputs are of the same phase,
frequency and amplitude.
Electronic Devices, 9th edition
Thomas L. Floyd
re' =
a.
Electronic Devices, 9th edition
Thomas L. Floyd
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3/29/2011
Quiz
Quiz
2. For a CE amplifier, a swamping resistor will
3. A well-designed CC amplifier has
a. increase the input resistance
a. voltage gain > 1
b. increase the gain
b. current gain > 1
c. both of the above
c. both of the above
d. none of the above
d. none of the above
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Electronic Devices, 9th edition
Thomas L. Floyd
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Electronic Devices, 9th edition
Thomas L. Floyd
Quiz
Quiz
4. In a CC amplifier, the power gain is approximately
5. The amplifier shown is a
a. one
a. differential amplifier
b. equal to the voltage gain
b. CE amplifier
c. equal to the current gain
c. CC amplifier
d. none of the above
+VCC
d. CB amplifier
RC C
3
R1
C2
Vout
RL
C1
Vin
RE
R2
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Electronic Devices, 9th edition
Thomas L. Floyd
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Electronic Devices, 9th edition
Thomas L. Floyd
Quiz
Quiz
6. An advantage to this amplifier is that it
7. Together, Q1 and Q2 form a
a. has high current gain
a. Swamped amplifier
b. has high input resistance
C2
d. all of the above
b. Differential pair
+VCC
c. is non-inverting
R1
Vout
d. none of the above
C1
R1
Q1
RL
C1
Q2
Vin
R2
R2
Electronic Devices, 9th edition
Thomas L. Floyd
VCC
c. Sziklai pair
RC C
3
RE
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RE
Electronic Devices, 9th edition
Thomas L. Floyd
C2
Vout
RL
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3/29/2011
Quiz
Quiz
8. A CC amplifier with a power gain less than 1 is
a. a buffer
9. An npn and a pnp transistor acting together as a single
high β transistor is a
a. Darlington pair
b. an inverting amplifier
b. Sziklai pair
c. unstable
c. Differential pair
d. an example of poor design
d. cascaded amplifier
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Electronic Devices, 9th edition
Thomas L. Floyd
Quiz
Quiz
10. If identical signals are applied to both inputs of a
differential amplifier, ideally the output will be
Answers:
a. zero
b. equal to one of the signals
c. equal to the sum of the two signals
d. very large
Electronic Devices, 9th edition
Thomas L. Floyd
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Electronic Devices, 9th edition
Thomas L. Floyd
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Electronic Devices, 9th edition
Thomas L. Floyd
1. b
6. c
2. a
7. d
3. b
8. d
4. c
9. b
5. d
10. a
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6