Power Amplifiers
Definitions
In small-signal amplifiers the main factors are
• amplification
• linearity
• gain
Large-signal or power amplifiers function primarily to provide sufficient power to drive
the output device. These amplifier circuits will handle large voltage signals and high
current levels. The main factors are
• efficiency
• maximum power capability
• impedance matching to the output device
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Amplifier Types
•
•
•
•
•
Class A
Class B
Class AB
Class C
Class D
Class A Amplifier
The output of a Class A amplifier conducts for the full 360° of the cycle
The Q-point (bias level) must be biased towards the middle of the load line so that the AC
signal can swing a full cycle. Remember that the DC load line indicates the maximum and
minimum limits set by the DC power supply.
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Class B Amplifier
A Class B amplifier output only conducts for 180° or ½ of the input signal.
The Q-point (bias level) is at 0V on the load line, so that the AC signal can only swing for
½ of a cycle.
Class AB Amplifier
This amplifier is in between the Class A and Class B. The Q-point (bias level) is above the
Class B but below the Class A.
The output conducts between 180° and 360° of the AC input signal.
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Class C
The output of the Class C conducts for less than 180° of the AC cycle. The Q-point (bias
level) is at cutoff, the output signal is very small.
Class D
The Class D output is more like a pulse. It does not resemble the AC sine wave input.
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Amplifier Efficiency
Efficiency refers to the ratio of output to input power. The lower the amount of conduction
of the amplifier the higher the efficiency.
Series-Fed Class A Amplifier
This is similar to the small-signal amplifier except that it will handle higher voltages. The
transistor used is a high power transistor.
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AC Operation: Series-Fed Class A Amplifier
A small input signal will cause the output voltage to swing to a maximum of Vcc and a
minimum of 0V. The current can also swing from 0 mA to IC (SAT) (VCC / RC)
Input Power: Series-Fed Class A Amplifier
The power into the amplifier is from the DC supply. With no input signal, the DC
current drawn is the collector bias current, ICQ.
PCC = VCC I CQ
Power input:
Output Power: Series-Fed Class A Amplifier
PL = Po (AC) =
2
VC(rms)
RC
or
PL =
2
VCE(p
- p)
8R C
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Efficiency
η% =
PL
× 100
PCC
The maximum efficiency is at maximum output and current swings.
Transistor Power Dissipation
PQ = PCC − PL
Power dissipated as heat across the transistor:
Note: The larger the input and output signal, the lower the heat dissipation.
Figure of Merit
Figure of Merit =
PQ (max)
PL (max)
Class A Amplifier (no transfomer) : Maximum Efficiency
η % = PL / PCC x 100
PL max =
V L2max
2 RC
=
2
VCC
8 RC
where VL (max) = VCC / 2
PCC V
L =V Lmax
= VCC I CQ
max(η %) =
=
V L =V Lmax
PLmax
PCC
⎛ V ⎞ V2
= VCC ⎜⎜ CC ⎟⎟ = CC
⎝ 2R C ⎠ 2R C
× 100
PLmax
2
VCC
/ 8 RC
2
VCC
/ 2 RC
× 100
1
= × 100
4
= 25 %
The maximum efficiency of a Class A amplifier without transformer is 25%.
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Class A (no transformer): Figure of Merit
Power input from the DC source:
PCC = VCC I CQ
Power dissipated as heat across the transistor:
PQ = PCC − PL
Note: The larger the input and output signal, the lower the heat dissipation.
PQmax = VCC ICQ (max) =
PLmax =
2
VCC
2RC
where PL = 0, i.e. no AC input (& output) signal
2
VCC
8RC
Figure of Merit =
PQ (max)
PL (max)
=4
Transformer-Coupled Class A Amplifier
This circuit uses a transformer to couple to the load. This improves
the efficiency of the Class A to 50%.
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Transformer Action
The transformer improves the efficiency because of the transformation
of voltage and current through the transformer.
Voltage Ratio: V2 = N 2
V1
N1
Current Ratio: I 2 = N1
I1
N2
The transformer aids in impedance matching to the load. Remember
that transformers transform voltage, current, and impedance.
2
Impedance Ratio: R′L = R 1 = ⎛⎜⎜ N1 ⎞⎟⎟ = a 2
R
R
N
L
2
⎝
2
⎠
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Transformer-Coupled Class A Amplifier
AC Operation
DC Load Line
As in all Class A amplifiers the Q-point is established close to the
midpoint of the DC load line.
AC Load Line
The saturation point (ICmax) is now at VCC/RL and the cutoff point is
at V2 (the secondary voltage of the transformer).
This increases the maximum output swing because the minimum and
maximum values of IC and VCE are spread further apart.
Transformer-Coupled Class A Amplifier
Signal Swing and Output AC Power
The voltage swing:
VCE(p-p) = VCEmax - VCEmin
The current swing:
IC(p-p) = IC(max) – IC(min)
The AC power:
PL =VCE(rms) IC(rms)=
PLmax=
VCE(p) IC(p) VCE(p−p) IC (p−p)
=
2
8
2
VCC
2R′L
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Class A Amplifier (with transfomer) : Maximum Efficiency
η % = PL / PCC x 100
PL max =
V L2max
2 R′L
=
2
VCC
2 R′L
where VL (max) = VCC
PCC V
L =V Lmax
= VCC I CQ
PLmax
max(η %) =
=
V L =V Lmax
PCC
⎛ V ⎞ V2
= VCC ⎜⎜ CC ⎟⎟ = CC
⎝ R ′L ⎠ R ′L
× 100
PLmax
2
VCC
/ 2 R′L
× 100
2
VCC /R′L
1
= × 100
2
= 50 %
The maximum efficiency of a Class A amplifier with transformer is 50%.
Transformer-Coupled Class A Figure of Merit
Power input from the DC source:
PCC = VCC I CQ
Power dissipated as heat across the transistor:
PQ = PCC − PL
Note: The larger the input and output signal, the lower the heat dissipation.
PQ
max
= VCC ICQ (max) =
PLmax=
2
VCC
R′L
where PL = 0, i.e. no AC input (& output) signal
2
VCC
2R′L
Figure of Merit =
PQ (max)
PL (max)
=2
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Class B Amplifier
In Class B the DC bias leaves the transistor biased just off. The AC signal turns the
transistor on. This is essentially no bias.
The transistor only conducts when it is turned on by ½ of the AC cycle.
In order to get a full AC cycle out of a Class B amplifier, you need two transistors.
One is an npn transistor that provides the negative half of the AC cycle and the other is a
pnp transistor that provides the positive half.
Phase splitter circuits
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Class B Amplifier Circuits
Transformer-Coupled Push-Pull
The center-tapped transformer on the input produces opposite polarity signals to the
two transistor inputs.
And the center-tapped transformer on the output combines the two halves of the AC
waveform together.
Class B Amplifier Circuits
Push-Pull Operation
During the positive half of the AC input cycle:
Transistor Q1 is conducting and Q2 is off.
During the negative half of the AC input cycle:
Transistor Q2 is conducting and Q1 is off.
Each transistor produces ½ of an AC cycle. The transformer combines the two outputs to
form a full AC cycle.
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Complementary-symmetry push-pull circuit
Class B Amplifier: Maximum Efficiency
η % = PL / PCC x 100
PL max =
V L2max
2 RL
=
2
VCC
2 RL
for maximum power VL = VCC
PCC V
L =V Lmax
= VCC I CQ
max(η %) =
=
V L =V Lmax
PLmax
PCC
⎛ 2V ⎞ 2V 2
= VCC ⎜⎜ CC ⎟⎟ = CC
⎝ πR L ⎠ πR L
× 100
PLmax
2
VCC
/ 2 RL
2
2VCC
/π R L
× 100
π
× 100
4
≅ 78 .5 %
=
The maximum efficiency of a Class B is 78.5%.
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Class B Amplifier: Figure of Merit
2PQ = PCC − PL
2PQ =
2
π
I mVCC −
1 2
I m RL
2
PQ =
1
π
1
I mVCC − I m2 R L
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dPQ
Let us find the value of Im which gives maximum PQ by solving
dPQ
dI m
=0
PQ max =
⇒
VCC
π
−
1
I m RL = 0
2
⇒
Im
PQ (max)
dI m
=
=0
2 VCC
π RL
2
1 VCC
π 2 RL
Figure of Merit =
PQ (max)
PL (max)
=
2
1 VCC
2 R
π
L
2
VCC
2 RL
=
2
π2
≅5
Class B Amplifier Circuits
Crossover Distortion
If the transistors Q1 and Q2 do not turn on and off at exactly the same time, then there is a
gap in the output voltage.
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Class AB Amplifier Circuits
Quasi-Complementary Push-Pull Amplifier
A Darlington pair and a Feedback pair combination perform the push-pull
operation.
This increases the output power capability.
Additional Class AB amplifier circuits
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Power Transistor Heat Sinking
High power transistors dissipated a
lot of power in heat. This can be
destructive to the amplifier as well
as to surrounding components.
These transistors will require heat-sinking.
Thermal-to-electrical analogy
θJA – total thermal resistance (junction to ambient)
θJC – transistor thermal resistance (junction to case)
θCS – insulator thermal resistance (case to heat sink)
θSA – heat-sink thermal resistance (heat sink to ambient)
TJ = θ JA PD + TA
PD =
TJ − TA
θ JA
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Class C Amplifiers
A Class C amplifier conducts for less than 180°. In order to produce a full sine wave
output, the Class C uses a tuned circuit (L and C tank) to provide the full AC sine wave.
The Class C is used extensively in radio communications circuits.
Class D Amplifier
A Class D amplifier amplifies pulses. It requires a pulsed input
There are many circuits that can convert a sinewave to a pulse, as well as circuits
that convert a pulse to a sinewave. This circuit has applications in digital
circuitry.
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Block diagram of class D amplifier
2005-2006 Midterm II Question
Ex. Consider the amplifier circuits given in the figures below
For the circuit shown in Figure (1), when vi = 1V sin(ωt)
a) Draw vo,
b) Calculate the power PL dissipated over RL,
c) Calculate the efficiency of the power amplifier consisting of transistors Q1 and Q2.
For the circuit shown in Figure (2), when vi = 1V sin(ωt)
d) Draw vo.
e) Compare the two amplifier circuit designs given in Figures (1) and (2), and express
which design is more preferable. Briefly explain your answer.
HINT: Consider your answer to questions 3(a) and 3(d).
Figure (1)
Figure (2)
Q1 ≡ Q2 , hFE = 100, VBE = 0.7V, VCEsat = 0.2V
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