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Chapter 4

Power Amplifiers

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Chapter Objectives
 The differences between classes A, AB, and C
amplifiers
 What causes amplifier distortion
 Efficiency of various classes of amplifiers
 Power calculations for various class amplifiers

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Introduction
 An amplifier receives a signal from some pickup
transducer or other input source and provides a larger
version of the signal to some output device or to
another amplifier stage. An input transducer signal is
generally small (a few mv from a cassette or CD input,
or a few mV from an antenna) and needs to be
amplified sufficiently to operate an output device
(speaker or other power-handling device).
 In small-signal amplifiers, the main factors are usually
amplification linearity and magnitude of gain.

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Introduction
 Since signal voltage and current are small in a small-
signal amplifier, the amount of power-handling capacity
and power efficiency are of little concern. A voltage
amplifier provides voltage amplification primarily to
increase the voltage of the input signal.
 Large-signal or power amplifiers, provide sufficient
power to an output load to drive a speaker or other
power device, typically a few watts to tens of watts.
 During operation, a power amplifier takes power from a
dc power source and converts it into useful ac signal
power. It is used to handle large-voltage signals at
moderate to high current levels.
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Introduction
 The main features of a large-signal amplifier are the
circuit’s power efficiency, the maximum amount of
power that the circuit is capable of handling, and the
impedance matching to the output device.
 One method used to categorize amplifiers is by class.
Basically, amplifier classes represent the amount the
output signal varies over one cycle of operation for a
full cycle of input signal.
 A brief description of amplifier classes is provided
next.

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Class A Power Amplifier

 Figure 4.1: Class A Power Amplifier characteristics

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Class A Power Amplifier

 The output signal varies for a full 360° of the cycle i.e.
amplifying devices operating in class A conduct over
the whole of the input cycle.
 Figure 4.1 shows that this requires the Q-point to be
biased at a level so that at least half the signal swing
of the output may vary up and down without going to a
high-enough voltage to be limited by the supply
voltage level or too low to approach the lower supply
level, or 0 V in this description.

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Class B Power Amplifier

A class B circuit provides an
output signal varying over
one-half the input signal
cycle, or for 180° of signal,
as shown in Fig. 4.2
The dc bias point for class
B is therefore at 0 V, with
the output then varying
from this bias point for a
half cycle.
Figure 4.2: Class B Power amplifier characteristics
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Class B Power Amplifier

 Obviously, the output is not a faithful reproduction of
the input if only one half-cycle is present. Two class B
operations—one to provide output on the positive
output half-cycle and another to provide operation on
the negative-output half-cycle are necessary.
 The combined half-cycles then provide an output for a
full 360° of operation. This type of connection is
referred to as push-pull operation. Note that class B
operation by itself creates a much distorted output
signal since reproduction of the input takes place for
only 180° of the output signal swing.
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Class AB Power Amplifier

 An amplifier may be biased at a dc level above the
zero base current level of class B and above one-half
the supply voltage level of class A; this bias condition
is class AB.
 Class AB operation still requires a push-pull
connection to achieve a full output cycle.
 For class AB operation, the output signal swing occurs
between 180° and 360° and is neither class A nor
class B operation.

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Class C Power Amplifier

 The output of a class C amplifier is biased for
operation at less than 180° of the cycle and will
operate only with a tuned (resonant) circuit, which
provides a full cycle of operation for the tuned or
resonant frequency.
 This operating class is therefore used in special areas
of tuned circuits, such as radio or communications.

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Class D Power Amplifier

 This operating class is a form of amplifier operation
using pulse (digital) signals, which are on for a short
interval and off for a longer interval. Using digital
techniques makes it possible to obtain a signal that
varies over the full cycle (using sample-and-hold
circuitry) to recreate the output from many pieces of
input signal.
 The major advantage of class D operation is that the
amplifier is on (using power) only for short intervals
and the overall efficiency can practically be very high.

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Amplifier Efficiency
 The power efficiency of an amplifier, defined as the
ratio of power output to power input, improves (gets
higher) going from class A to class D.
 In general terms, a class A amplifier, with dc bias at
one-half the supply voltage level, uses a good amount
of power to maintain bias, even with no input signal
applied. This results in very poor efficiency, especially
with small input signals, when very little ac power is
delivered to the load. In fact, the maximum efficiency
of a class A circuit is only 25% with a direct or series
fed load connection and 50% with a transformer
connection to the load. © Niwareeba Roland
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Amplifier Efficiency
 Class B operation, with no dc bias power for no input
signal, can be shown to provide a maximum efficiency
that reaches 78.5%.
 Class D operation can achieve power efficiency over
90% and provides the most efficient operation of all
the operating classes.
 Since class AB falls between class A and class B in
bias, it also falls between their efficiency ratings—
between 25% (or 50%) and 78.5%.

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Amplifier Efficiency
 In class B operation, a push-pull connection is
obtained using either a transformer coupling or by
using complementary (or quasi-complementary)
operation with npn and pnp transistors to provide
operation on opposite polarity cycles.
 While transformer operation can provide opposite
cycle signals, the transformer itself is quite large in
many applications. A transformerless circuit using
complementary transistors provides the same
operation in a much smaller package.

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SERIES-FED CLASS A AMPLIFIER

Figure 4.3: Series-fed class A large-signal amplifier.

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SERIES-FED CLASS A AMPLIFIER

 The simple fixed-bias circuit connection shown in Fig.
4.3 can be used to discuss the main features of a
class A series-fed amplifier.
 The dc bias set by VCC and RB fixes the dc base-bias
current at
𝑉𝑉𝑐𝑐𝑐𝑐 −0.7𝑉𝑉
𝐼𝐼𝐵𝐵 =
𝑅𝑅𝐵𝐵
 With the collector current then being
𝐼𝐼𝐶𝐶 = 𝛽𝛽𝐼𝐼𝐵𝐵
 With the collector–emitter voltage then being
𝑉𝑉𝐶𝐶𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶 − 𝐼𝐼𝐶𝐶 𝑅𝑅𝐶𝐶
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SERIES-FED CLASS A AMPLIFIER

Figure 4.4 Transistor characteristic showing load line and Q point.

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SERIES-FED CLASS A AMPLIFIER

AC Operation
 When an input ac signal is applied to the amplifier of
Fig. 4.3, the output will vary from its dc bias operating
voltage and current. A small input signal, as shown in
Fig. 4.5, will cause 𝐼𝐼𝐵𝐵 to vary above and below the dc
bias point, which will then cause 𝐼𝐼𝐶𝐶 (output) to vary
from the dc bias point set as well as 𝑉𝑉𝐶𝐶𝐶𝐶 to vary
around its dc bias value. As the input signal is made
larger, the output will vary further around the
established dc bias point until either the current or the
voltage reaches a limiting condition.
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SERIES-FED CLASS A AMPLIFIER

AC Operation
 For the current, this limiting condition is either zero
current at the low end or VCC/RC at the high end of its
swing.
 For the collector–emitter voltage, the limit is either 0 V
or the supply voltage, VCC.

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SERIES-FED CLASS A AMPLIFIER

Figure 4.5 Amplifier input and output signal variation.

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SERIES-FED CLASS A AMPLIFIER

Power Considerations
 The power into an amplifier is provided by the supply.
 With no input signal, the dc current drawn is the
collector bias current, ICQ. The power then drawn from
the supply is
𝑃𝑃𝑖𝑖 𝑑𝑑𝑑𝑑 = 𝑉𝑉𝑐𝑐𝑐𝑐 𝐼𝐼𝐶𝐶𝐶𝐶
 Even with an ac signal applied, the average current
drawn from the supply remains the same, so this
equation represents the input power supplied to the
class A series fed amplifier.
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SERIES-FED CLASS A AMPLIFIER

Output Power
 The ac power delivered to the load (RC) can be
expressed in a number of ways. i.e using RMS, peak
and peak-to-peak signals as:
𝑉𝑉𝐶𝐶2 (𝑟𝑟𝑟𝑟𝑟𝑟)
 𝑃𝑃𝑜𝑜 𝑎𝑎𝑎𝑎 = 𝑉𝑉𝐶𝐶𝐶𝐶 𝑟𝑟𝑟𝑟𝑟𝑟 𝐼𝐼𝐶𝐶 𝑟𝑟𝑟𝑟𝑟𝑟 = 𝐼𝐼𝐶𝐶2 (𝑟𝑟𝑟𝑟𝑟𝑟)𝑅𝑅𝐶𝐶 =
𝑅𝑅𝐶𝐶
𝑉𝑉𝐶𝐶𝐶𝐶 (𝑝𝑝)𝐼𝐼𝐶𝐶 (𝑝𝑝) 𝐼𝐼𝐶𝐶2 (𝑝𝑝)𝑅𝑅𝐶𝐶 2
𝑉𝑉𝐶𝐶𝐶𝐶 (𝑝𝑝)
 𝑃𝑃𝑜𝑜 𝑎𝑎𝑎𝑎 = = =
2 2 2𝑅𝑅𝐶𝐶
𝑉𝑉𝐶𝐶𝐶𝐶 (𝑝𝑝−𝑝𝑝)𝐼𝐼𝐶𝐶 (𝑝𝑝−𝑝𝑝) 𝐼𝐼𝐶𝐶2 (𝑝𝑝−𝑝𝑝) 2
𝐼𝐼𝐶𝐶𝐶𝐶 (𝑝𝑝−𝑝𝑝)
 𝑃𝑃𝑜𝑜 𝑎𝑎𝑎𝑎 = = 𝑅𝑅𝐶𝐶 =
8 8 8𝑅𝑅𝐶𝐶

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SERIES-FED CLASS A AMPLIFIER

Efficiency
 The efficiency of an amplifier represents the amount of
ac power delivered (transferred) from the dc source.
The efficiency of the amplifier is calculated using
𝑃𝑃𝑜𝑜 (𝑎𝑎𝑎𝑎)
 %𝜂𝜂 = × 100%
𝑃𝑃𝑖𝑖 (𝑑𝑑𝑑𝑑)
Maximum Efficiency
 For the class A series-fed amplifier, the maximum
efficiency can be determined using the maximum
voltage and current swings.

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SERIES-FED CLASS A AMPLIFIER

For the voltage swing it is
 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑉𝑉𝐶𝐶𝐶𝐶 𝑝𝑝 − 𝑝𝑝 = 𝑉𝑉𝐶𝐶𝐶𝐶
For the current swing it is
𝑉𝑉𝐶𝐶𝐶𝐶
 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝐼𝐼𝐶𝐶 𝑝𝑝 − 𝑝𝑝 =
𝑅𝑅𝐶𝐶
Therefore,
2
𝑉𝑉𝐶𝐶𝐶𝐶 (𝑉𝑉𝐶𝐶𝐶𝐶 /𝑅𝑅𝐶𝐶 ) 𝑉𝑉𝐶𝐶𝐶𝐶
 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑃𝑃𝑜𝑜 𝑎𝑎𝑎𝑎 = =
8 8𝑅𝑅𝐶𝐶
 The maximum power input can be calculated using the
dc bias current set to one-half the maximum value:

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SERIES-FED CLASS A AMPLIFIER

2
𝑉𝑉𝐶𝐶𝐶𝐶 /𝑅𝑅𝐶𝐶 𝑉𝑉𝐶𝐶𝐶𝐶
 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑃𝑃𝑖𝑖 𝑑𝑑𝑑𝑑 = 𝑉𝑉𝐶𝐶𝐶𝐶 𝑚𝑚𝑚𝑚𝑚𝑚 𝐼𝐼𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶 =
2 2𝑅𝑅𝐶𝐶
2
𝑚𝑚𝑚𝑚𝑚𝑚𝑃𝑃𝑜𝑜 (𝑎𝑎𝑎𝑎) 𝑉𝑉𝐶𝐶𝐶𝐶 /8𝑅𝑅𝐶𝐶
 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 % 𝜂𝜂 = × 100% = 2 /2𝑅𝑅 ×
𝑚𝑚𝑚𝑚𝑚𝑚 𝑃𝑃𝑖𝑖 (𝑎𝑎𝑎𝑎) 𝑉𝑉𝐶𝐶𝐶𝐶 𝐶𝐶
100% = 25%
 The maximum efficiency of a class A series-fed
amplifier is thus seen to be 25%. Since this maximum
efficiency will occur only for ideal conditions of both
voltage swing and current swing, most series-fed
circuits will provide efficiencies of much less than
25%.
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SERIES-FED CLASS A AMPLIFIER

 Example
Calculate the input power, output power, and efficiency
of the amplifier circuit in Fig. 4.6 for an input voltage that
results in a base current of 10 mA peak.

Figure 4.6 Operation of a

series-fed circuit for
Example

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SERIES-FED CLASS A AMPLIFIER

Solution
The Q-point can be determined as follows
𝑉𝑉𝐶𝐶𝐶𝐶 −0.7𝑉𝑉
 𝐼𝐼𝐵𝐵𝐵𝐵 = = 19.3 𝑚𝑚𝑚𝑚
𝑅𝑅𝐵𝐵
 𝐼𝐼𝐶𝐶𝐶𝐶 = 𝛽𝛽𝐼𝐼𝐵𝐵 = 25 19.3 𝑚𝑚𝑚𝑚 = 482.5 𝑚𝑚𝑚𝑚 = 0.483 𝐴𝐴
 𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶 − 𝐼𝐼𝐶𝐶 𝑅𝑅𝐶𝐶 = 20 𝑉𝑉 − 0.483 × 20 = 10.4 𝑉𝑉
 The ac variation of the output signal can be obtained
graphically using the dc load line drawn on figure 4.6 b
by connecting 𝑉𝑉𝐶𝐶𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶 = 20 𝑉𝑉 with 𝐼𝐼𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶�𝑅𝑅𝐶𝐶 =
1000 𝑚𝑚𝑚𝑚 = 1𝐴𝐴, as shown.
© Niwareeba Roland
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SERIES-FED CLASS A AMPLIFIER

When the input ac base current increases from its dc
bias level, the collector current rises by
 𝐼𝐼𝐶𝐶 𝑝𝑝 = 𝛽𝛽𝐼𝐼𝐵𝐵 𝑝𝑝 = 25 10 𝑚𝑚𝑚𝑚 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 250 𝑚𝑚𝑚𝑚 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
2
𝐼𝐼𝐶𝐶2 (𝑝𝑝) 250×10−3 𝐴𝐴
 𝑃𝑃𝑜𝑜 𝑎𝑎𝑎𝑎 = 𝑅𝑅𝐶𝐶 = 20𝛺𝛺 = 0.625 𝑊𝑊
2 2
 𝑃𝑃𝑖𝑖 𝑑𝑑𝑑𝑑 = 𝑉𝑉𝐶𝐶𝐶𝐶 𝐼𝐼𝐶𝐶𝐶𝐶 = 20 𝑉𝑉 0.48 𝐴𝐴 = 9.6 𝑊𝑊
The amplifier’s power efficiency can be calculated as
𝑃𝑃𝑜𝑜 (𝑎𝑎𝑎𝑎)
 %𝜂𝜂 = × 100%= 6.5%
𝑃𝑃𝑖𝑖 (𝑑𝑑𝑑𝑑)

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TRANSFORMER-COUPLED CLASS
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A AMPLIFIER

Figure 4.7: Transformer-coupled audio power amplifier.

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 TRANSFORMER-COUPLED CLASS
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A AMPLIFIER
 For maximum power transfer to the load, impedance
matching is necessary. However, it is had when the
loads have very small impedances.
 This, thankfully, can be eliminated by using a
transformer to couple the output signal to the load
achieving a maximum efficiency of 50 % as shown in
Fig. 4.7.

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A AMPLIFIER
Transformer Action
 A transformer can increase or decrease voltage or
current levels according to the turns ratio.
 In addition, the impedance connected to one side of a
transformer can be made to appear either larger or
smaller (step up or step down) at the other side of the
transformer, depending on the square of the
transformer winding turns ratio.

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A AMPLIFIER
CURRENT , VOLTAGE AND IMPEDANCE
TRANSFORMATION
𝑉𝑉2 𝑁𝑁2 𝐼𝐼1 1
= = =
𝑉𝑉1 𝑁𝑁1 𝐼𝐼2 𝑎𝑎
a is the turns ratio.
𝐼𝐼2 𝑅𝑅𝐿𝐿 𝐼𝐼1
′ =
𝐼𝐼1 𝑅𝑅𝐿𝐿 𝐼𝐼2
𝑅𝑅′𝐿𝐿 = 𝑎𝑎2 𝑅𝑅𝐿𝐿
𝑅𝑅′𝐿𝐿 is the load resistance reflected to the primary side.
Note: Since the transformer is to be used for impedance
matching, it has to be a step down transformer.
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A AMPLIFIER
Examples
 Calculate the effective resistance seen looking into the
primary of a 15: 1 transformer connected to an 8 Ω
load. (1.8 kΩ)
 What transformer turns ratio is required to match a 16
Ω speaker load so that the effective load resistance
seen at the primary is 10 kΩ? (25:1)

© Niwareeba Roland
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A AMPLIFIER
DC Load Line
 The transformer (dc) winding resistance determines
the dc load line for the circuit of Fig. 4.7 . Typically,
this dc resistance is small (ideally 0 Ω) and, as shown
in Fig. 4.8 , a 0 Ω dc load line is a straight vertical line.
A practical transformer winding resistance would be a
few ohms.
 There is no dc voltage drop across the 0 Ω dc load
resistance, and the load line is drawn straight vertically
from the voltage point, 𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶

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TRANSFORMER-COUPLED CLASS
36

A AMPLIFIER

Figure 4.8: Load lines for class A transformer-coupled amplifier.

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 TRANSFORMER-COUPLED CLASS
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A AMPLIFIER
DC power input
 This is supplied by the dc voltage supply when there is
no input ac signal. The dc current drawn in this case
will be 𝐼𝐼𝐶𝐶𝐶𝐶
 𝑃𝑃𝐷𝐷𝐷𝐷 = 𝑉𝑉𝐶𝐶𝐶𝐶 𝐼𝐼𝐶𝐶𝐶𝐶
AC operation
 The load on the secondary is 𝑅𝑅𝐿𝐿 . Hence, the reflected
load on the primary will be 𝑅𝑅′𝐿𝐿 . Therefore, the ac load
line will be drawn through the operating point with a
1
slope of- as shown in fig 4.8.
𝑅𝑅′𝐿𝐿
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 TRANSFORMER-COUPLED CLASS
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A AMPLIFIER
AC Output Power
𝑉𝑉𝐶𝐶𝐶𝐶 (𝑝𝑝−𝑝𝑝)𝐼𝐼𝐶𝐶 (𝑝𝑝−𝑝𝑝)
 𝑃𝑃𝑜𝑜 𝑎𝑎𝑎𝑎 =
8
Efficiency
𝑃𝑃𝑜𝑜 (𝑎𝑎𝑎𝑎)
 %𝜂𝜂 = × 100%
𝑃𝑃𝑖𝑖 (𝑑𝑑𝑑𝑑)
Maximum Efficiency
𝑉𝑉𝐶𝐶𝐶𝐶 𝑝𝑝−𝑝𝑝 𝑚𝑚𝑚𝑚𝑚𝑚𝐼𝐼𝐶𝐶 𝑝𝑝−𝑝𝑝 𝑚𝑚𝑚𝑚𝑚𝑚 2𝑉𝑉𝐶𝐶𝐶𝐶 ×2𝐼𝐼𝐶𝐶𝐶𝐶
 𝑃𝑃𝑜𝑜 𝑎𝑎𝑎𝑎 𝑚𝑚𝑚𝑚𝑚𝑚 = = =
8 8
𝑉𝑉𝐶𝐶𝐶𝐶 𝐼𝐼𝐶𝐶𝐶𝐶
2
𝑃𝑃𝑜𝑜 𝑎𝑎𝑎𝑎 𝑚𝑚𝑚𝑚𝑚𝑚
 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = = 50%
𝑃𝑃𝑖𝑖 𝑑𝑑𝑑𝑑 © Niwareeba Roland
TRANSFORMER-COUPLED CLASS
39

A AMPLIFIER
 Example
• Calculate the ac power
delivered to the 8-Ω
speaker for the circuit of
Fig. 4.12.
• The circuit component
values result in a dc
base current of 6 mA,
and the input signal (Vi)
results in a peak base
current swing of 4 mA
Figure 4.12
© Niwareeba Roland
 TRANSFORMER-COUPLED CLASS
40

A AMPLIFIER
Solution
 The dc load line is drawn vertically (figure 4.13)
 𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑉𝑉𝐶𝐶𝐶𝐶 = 10 𝑉𝑉
 For IB=6 mA, the operating point on figure 4.13 is
 𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶 = 10𝑉𝑉, 𝐼𝐼𝐶𝐶𝐶𝐶 = 140 𝑚𝑚𝑚𝑚
The effective ac resistance seen at the primary is
 𝑅𝑅𝑅𝐿𝐿 = 𝑎𝑎2 𝑅𝑅𝐿𝐿 = 72𝛺𝛺
 The ac load line can then be drawn of slope -1/72
going through the indicated operating point. To help
draw the load line, consider the following procedure.
© Niwareeba Roland
 TRANSFORMER-COUPLED CLASS
41

A AMPLIFIER
𝑉𝑉𝐶𝐶𝐶𝐶
 For a current swing of 𝐼𝐼𝐶𝐶 = = 139 𝑚𝑚𝑚𝑚 𝐼𝐼𝐶𝐶𝐶𝐶𝐶𝐶 + 𝐼𝐼𝐶𝐶 =
𝑅𝑅′𝐿𝐿
140 𝑚𝑚𝑚𝑚 + 139 𝑚𝑚𝑚𝑚 = 279 𝑚𝑚𝑚𝑚 along the y-axis (Point A)
 Connect point A through the Q-point to obtain the ac
load line. For the given base current swing of 4 mA
peak, the maximum and minimum collector current
and 𝑉𝑉𝐶𝐶𝐶𝐶 obtained from Fig. 4.12 are
 𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 1.7 𝑉𝑉,
𝑉𝑉𝐶𝐶𝐶𝐶 𝑚𝑚𝑚𝑚𝑚𝑚 = 18.3 𝑉𝑉 𝐼𝐼𝐶𝐶 𝑚𝑚𝑚𝑚𝑚𝑚 = 25 𝑚𝑚𝑚𝑚, 𝐼𝐼𝐶𝐶 𝑚𝑚𝑚𝑚𝑚𝑚 = 255 𝑚𝑚𝑚𝑚
 Giving 𝑃𝑃𝑜𝑜 𝑎𝑎𝑎𝑎 = 0.477𝑊𝑊

© Niwareeba Roland
TRANSFORMER-COUPLED CLASS
42

A AMPLIFIER

Figure 4.13

© Niwareeba Roland
 TRANSFORMER-COUPLED CLASS
43

A AMPLIFIER
 For the circuit of Fig. 4.12 above, calculate the dc
input power, power dissipated by the transistor, and
efficiency of the circuit for the input signal in the
example above.
 𝑃𝑃𝑖𝑖 𝑑𝑑𝑑𝑑 = 𝑉𝑉𝐶𝐶𝐶𝐶 𝐼𝐼𝐶𝐶𝐶𝐶 = 10 (140𝑚𝑚) = 1.4 𝑊𝑊
 𝑃𝑃𝑄𝑄 = 𝑃𝑃𝑖𝑖 𝑑𝑑𝑑𝑑 − 𝑃𝑃𝑜𝑜 𝑎𝑎𝑎𝑎 = 1.4𝑊𝑊 − 0.477𝑊𝑊 = 0.92𝑊𝑊
 The efficiency of the amplifier is then
𝑃𝑃𝑜𝑜 𝑎𝑎𝑎𝑎
 %𝜂𝜂 = × 100% = 34.1%
𝑃𝑃𝑖𝑖 𝑑𝑑𝑑𝑑

© Niwareeba Roland
 Advantages of class-A
44

amplifiers
 Class-A designs are simpler than other classes; for
example class-AB and -B designs require two devices
(push-pull output) to handle both halves of the
waveform; class A can use a single device single
ended.
 The amplifying element is biased so the device is
always conducting to some extent, normally implying
the quiescent (small-signal) collector current (for
transistors ; drain current for FETs or anode/plate
current for vacuum tubes) is close to the most linear
portion of its transconductance curve.
© Niwareeba Roland
 Advantages of class-A
45

amplifiers
 Because the device is never shut off completely there
is no "turn on" time, little problem with charge storage,
and generally better high frequency performance and
feedback loop stability (and usually fewer high-order
harmonics).
 The point at which the device comes closest to being
cut off is not close to zero signal, so the problem of
crossover distortion associated with class-AB and -B
designs is avoided.

© Niwareeba Roland
 Disadvantage of class-A
46

amplifiers
 They are very inefficient. A theoretical maximum of
50% is obtainable with inductive output coupling and
only 25% with capacitive coupling, unless deliberate
use of nonlinearities is made (such as in square-law
output stages).
 In a power amplifier, this not only wastes power and
limits battery operation, increase costs and may
restrict the output devices that can be used.

© Niwareeba Roland
 47

CLASS B AMPLIFIER OPERATION

 In class B, the transistor is biased just off. The AC signal
turns the transistor on.
 The transistor only conducts when it is turned on by one-
half of the AC cycle.
 In order to get a full AC cycle out of a class B amplifier,
you need two transistors:
• An npn transistor that provides the negative half of the
AC cycle
• A pnp transistor that provides the positive half.

© Niwareeba Roland
 48

CLASS B AMPLIFIER OPERATION

FIG. 4.14: Block representation of push–pull operation.

© Niwareeba Roland
 49

CLASS B AMPLIFIER OPERATION

 Input (DC) Power

FIG. 4.16: Connection of push–pull amplifier to load: (a) using

two voltage supplies; (b) using one voltage supply.

© Niwareeba Roland
 50

CLASS B AMPLIFIER OPERATION

 The power supplied to the load by an amplifier is
drawn from the power supply (or power supplies; see
Fig. 4.16 ) that provides the input or dc power. The
amount of this input power can be calculated using
 𝑃𝑃𝑖𝑖 (𝑑𝑑𝑑𝑑) = 𝑉𝑉𝐶𝐶𝐶𝐶 𝐼𝐼𝑑𝑑𝑑𝑑
 𝐼𝐼𝑑𝑑𝑑𝑑 is the average current drawn from the supplies.
 In class B operation, the current drawn from a single
power supply has the form of a full-wave rectified
signal, whereas that drawn from two power supplies
has the form of a half-wave rectified signal from each
supply.
© Niwareeba Roland
 51

CLASS B AMPLIFIER OPERATION

 In either case, the value of the average current drawn
can be expressed as
2
 𝐼𝐼𝑑𝑑𝑑𝑑 = 𝐼𝐼(𝑝𝑝)
𝜋𝜋
 𝐼𝐼(𝑝𝑝) is the peak value of the output current waveform.
2
 𝑃𝑃𝑖𝑖 (𝑑𝑑𝑑𝑑) = 𝑉𝑉𝐶𝐶𝐶𝐶 𝐼𝐼(𝑝𝑝)
𝜋𝜋

© Niwareeba Roland
 52

CLASS B AMPLIFIER OPERATION

Output (AC) Power
 The power delivered to the load (usually referred to as
a resistance, RL can be calculated using any one of a
number of equations.
𝑉𝑉𝐿𝐿2 (𝑟𝑟𝑟𝑟𝑟𝑟) 𝑉𝑉𝐿𝐿2 (𝑝𝑝−𝑝𝑝) 𝑉𝑉𝐿𝐿2 (𝑝𝑝)
 𝑃𝑃𝑜𝑜 𝑎𝑎𝑎𝑎 = = =
𝑅𝑅𝐿𝐿 8𝑅𝑅𝐿𝐿 2𝑅𝑅𝐿𝐿

© Niwareeba Roland
 53

CLASS B AMPLIFIER OPERATION

Efficiency
 The efficiency of the class B amplifier can be
calculated using the basic equation
𝑃𝑃𝑜𝑜 𝑎𝑎𝑎𝑎 𝑉𝑉𝐿𝐿2 (𝑝𝑝)/2𝑅𝑅𝐿𝐿
%𝜂𝜂 = × 100% = × 100%
𝑃𝑃𝑖𝑖 (𝑑𝑑𝑑𝑑) 2
𝑉𝑉𝐶𝐶𝐶𝐶 𝐼𝐼(𝑝𝑝)
𝜋𝜋
𝜋𝜋 𝑉𝑉𝐿𝐿 (𝑝𝑝)
= × 100%
4 𝑉𝑉𝐶𝐶𝐶𝐶
 𝐼𝐼 𝑝𝑝 = 𝑉𝑉𝐿𝐿 (𝑝𝑝)/𝑅𝑅𝐿𝐿
 𝑉𝑉𝐿𝐿 (𝑝𝑝)𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑉𝑉𝐶𝐶𝐶𝐶
 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 %𝜂𝜂 = 78.5%
© Niwareeba Roland
 54

CLASS B AMPLIFIER OPERATION

Power Dissipated by Output Transistors
 The power dissipated (as heat) by the output power
transistors is the difference between the input power
delivered by the supplies and the output power
delivered to the load,
 𝑃𝑃2𝑄𝑄 = 𝑃𝑃𝑖𝑖 (𝑑𝑑𝑑𝑑) − 𝑃𝑃𝑜𝑜 (𝑎𝑎𝑎𝑎)
 where 𝑃𝑃2𝑄𝑄 is the power dissipated by the two output
power transistors. The dissipated power handled by
each transistor is then
𝑃𝑃2𝑄𝑄
 𝑃𝑃𝑄𝑄 =
2
© Niwareeba Roland
 55

CLASS B AMPLIFIER OPERATION

Example 1
 For a class B amplifier providing a 20-V peak signal to
a 16-Ω load (speaker) and a power supply of VCC = 30
V, determine the input power, output power, and
circuit efficiency.
(23.9 W, 12.5 W, 52.3%)
Example 2
 For a class B amplifier using a supply of VCC= 30 V
and driving a load of 16-Ω, determine the maximum
input power, output power, and transistor dissipation.
(35.81 W ,28.125 W, 5.7 W)
© Niwareeba Roland
 56

CLASS B AMPLIFIER CIRCUITS

 Transformer-Coupled Push–Pull Circuits

FIG. 4.17: Push–pull circuit.

© Niwareeba Roland
 57

CLASS B AMPLIFIER OPERATION

 The circuit of Fig. 4.17 uses a center-tapped input
transformer to produce opposite-polarity signals to the two
transistor inputs and an output transformer to drive the
load in a push–pull mode of operation described next.
 During the first half-cycle of operation, transistor Q1 is
driven into conduction, whereas transistor Q2 is driven off.
The current I1 through the transformer results in the first
halfcycle of signal to the load. During the second half-cycle
of the input signal, Q2 conducts, whereas Q1 stays off, the
current I2 through the transformer resulting in the second
halfcycle to the load. The overall signal developed across
the load then varies over the full cycle of signal operation.
© Niwareeba Roland
Complementary-Symmetry
58

Circuits

FIG. 4.18: Complementary-symmetry push–pull circuit

© Niwareeba Roland
 Complementary-Symmetry
59

Circuits
 Using complementary transistors ( npn and pnp ) it is
possible to obtain a full cycle output across a load
using half-cycles of operation from each transistor.
 Whereas a single input signal is applied to the base of
both transistors, the transistors, being of opposite
type, will conduct on opposite half-cycles of the input.
The npn transistor will be biased into conduction by
the positive half-cycle of signal, with a resulting
halfcycle of signal across the load as shown in Fig.
4.18 b. During the negative half-cycle of signal, the
pnp transistor is biased into conduction when the input
goes negative, as shown in Fig. 4.18 c. © Niwareeba Roland
 Complementary-Symmetry
60

Circuits
 During a complete cycle of the input, a complete cycle
of output signal is developed across the load. One
disadvantage of the circuit is the need for two
separate voltage supplies.
 Another, less obvious disadvantage with the
complementary circuit is shown in the resulting
crossover distortion in the output signal (see Fig. 4.18
d).
 Crossover distortion refers to the fact that during the
signal crossover from positive to negative (or vice
versa) there is some nonlinearity in the output signal
© Niwareeba Roland
 61

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.

Figure 4.19: Crossover distortion

© Niwareeba Roland
 62

AMPLIFIER DISTORTION
 If the output of an amplifier is not a complete AC sine
wave, then it is distorting the output. The amplifier is non-
linear.
 This distortion can be analyzed using Fourier analysis.
 In Fourier analysis, any distorted periodic waveform can be
broken down into frequency components.
 These components are harmonics of the fundamental
frequency.

© Niwareeba Roland
 63

Harmonic Distortion
 A signal is considered to have harmonic distortion
when there are harmonic frequency components (not
just the fundamental component). If the fundamental
frequency has an amplitude A1 and the nth frequency
component has an amplitude 𝐴𝐴𝑛𝑛 , a harmonic distortion
can be defined as
𝐴𝐴𝑛𝑛
 % 𝑛𝑛𝑛𝑛𝑛 ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = % 𝐷𝐷𝑛𝑛 = × 100%
𝐴𝐴1
 The fundamental component is typically larger than
any harmonic component.

© Niwareeba Roland
 64

Harmonic Distortion
Example
 Calculate the harmonic distortion components for an
output signal having fundamental amplitude of 2.5 V,
second harmonic amplitude of 0.25 V, third harmonic
amplitude of 0.1 V, and fourth harmonic amplitude of
0.05 V (10, 4, 2%)

© Niwareeba Roland
 65

Total Harmonic Distortion

 When an output signal has a number of individual
harmonic distortion components, the signal can be
seen to have a total harmonic distortion based on the
individual elements as combined by the relationship of
the following equation:

 % 𝑇𝑇𝑇𝑇𝑇𝑇 = 𝐷𝐷22 + 𝐷𝐷32 + 𝐷𝐷42 + ⋯ × 100%

 where THD is total harmonic distortion.
EXAMPLE Calculate the total harmonic distortion for the amplitude
components given in the previous example. (10.95 %)

© Niwareeba Roland
 66

Second Harmonic Distortion

FIG. 4.20: Waveform for obtaining second harmonic distortion.

© Niwareeba Roland
 67

Second Harmonic Distortion

 An equation that approximately describes the distorted
signal waveform is
 𝑖𝑖𝐶𝐶 ≈ 𝐼𝐼𝐶𝐶𝐶𝐶 + 𝐼𝐼0 + 𝐼𝐼1 cos 𝜔𝜔𝜔𝜔 + 𝐼𝐼2 cos 2𝜔𝜔𝜔𝜔
 𝐼𝐼𝐶𝐶𝐶𝐶 is the original quiescent current.
 𝐼𝐼0 is an additional dc current, due to the nonzero
average of the distorted signal.
 𝐼𝐼1 is the fundamental component of the distorted ac
signal.
 𝐼𝐼2 is the component of the second harmonic.

© Niwareeba Roland
 68

Second Harmonic Distortion

 At point 1 (𝜔𝜔𝜔𝜔 = 0)
 𝑖𝑖𝐶𝐶 = 𝐼𝐼𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 𝐼𝐼𝐶𝐶𝐶𝐶 + 𝐼𝐼0 + 𝐼𝐼1 cos 0 + 𝐼𝐼2 cos 0 = 𝐼𝐼𝐶𝐶𝐶𝐶 + 𝐼𝐼0 +
𝐼𝐼1 + 𝐼𝐼2
 At point 2 (𝜔𝜔𝜔𝜔 = 𝜋𝜋/2)
 𝑖𝑖𝐶𝐶 = 𝐼𝐼𝐶𝐶𝐶𝐶 + 𝐼𝐼0 + 𝐼𝐼1 cos 𝜋𝜋/2 + 𝐼𝐼2 cos 2𝜋𝜋/2 = 𝐼𝐼𝐶𝐶𝐶𝐶 + 𝐼𝐼0 +
𝐼𝐼1 − 𝐼𝐼2
 𝐼𝐼𝐶𝐶𝐶𝐶 = 𝐼𝐼𝐶𝐶𝐶𝐶 + 𝐼𝐼0 − 𝐼𝐼2
 At point 3 (𝜔𝜔𝜔𝜔 = 𝜋𝜋)
 𝑖𝑖𝐶𝐶 = 𝐼𝐼𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 𝐼𝐼𝐶𝐶𝐶𝐶 +𝐼𝐼0 + 𝐼𝐼1 cos 𝜋𝜋 + 𝐼𝐼2 cos 2𝜋𝜋 = 𝐼𝐼𝐶𝐶𝐶𝐶 + 𝐼𝐼0 −
𝐼𝐼1 + 𝐼𝐼2
© Niwareeba Roland
 69

Second Harmonic Distortion

 Sovling the preceding three equations simultaneously gives
𝐼𝐼𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 +𝐼𝐼𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 −2𝐼𝐼𝐶𝐶𝐶𝐶 𝐼𝐼𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 −𝐼𝐼𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
 𝐼𝐼0 = 𝐼𝐼2 = , 𝐼𝐼1 =
4 2
 We can express the definition of second harmonic distortion
as
1
𝐼𝐼2 𝐼𝐼 +𝐼𝐼 −𝐼𝐼𝐶𝐶𝐶𝐶
2 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶
 𝐷𝐷2 = × 100% = × 100%
𝐼𝐼1 𝐼𝐼𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 −𝐼𝐼𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶

 In a similar manner, 𝐷𝐷2 can be expressed in terms of

measured collector-emitter voltages
1
𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 +𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 −𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶
 𝐷𝐷2 = 2
× 100%
𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 −𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶

© Niwareeba Roland
 70

Second Harmonic Distortion

EXAMPLE
 Calculate the second harmonic distortion if an output
waveform displayed on an oscilloscope provides the
following measurements:
I. 𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 1 𝑉𝑉, 𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 22 𝑉𝑉, 𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶 = 12 𝑉𝑉
II. 𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 4 𝑉𝑉, 𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 20 𝑉𝑉, 𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶 = 12 𝑉𝑉
(2.38 %, 0%-no distortion)

© Niwareeba Roland
 Power of a Signal Having
71

Distortion
 When distortion does occur, the output power calculated
for the undistorted signal is no longer correct. When
distortion is present, the output power delivered to the
load resistor RC due to the fundamental component of
the distorted signal is
𝐼𝐼12 𝑅𝑅𝐶𝐶
 𝑃𝑃1 =
2
 The total power due to all the harmonic components of
the distorted signal can then be calculated using
𝑅𝑅𝐶𝐶
 𝑃𝑃 = 𝐼𝐼12 + 𝐼𝐼22 +⋯
2

© Niwareeba Roland
 72

Second Harmonic Distortion

 The total power can also be expressed in terms of
the total harmonic distortion,
2 𝑅𝑅𝐶𝐶
 𝑃𝑃 = 1 + 𝐷𝐷22 + 𝐷𝐷32 +⋯ 𝐼𝐼1 = 1 + 𝑇𝑇𝑇𝑇𝑇𝑇 2 𝑃𝑃1
2
Example
 For a harmonic distortion reading of D2 = 0.1, D3 =
0.02, and D4 = 0.01, with I1 = 4 A and RC = 8 Ω,
calculate the total harmonic distortion, fundamental
power component, and total power.
(0.1, 64 W, 64.64 W)

© Niwareeba Roland
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Power Transistor Derating Curve

 Power transistors
dissipate a lot of
power in heat. This
can be destructive to
the amplifier as well as
to surrounding
components.

Figure 4.21: Power transistor derating curve

© Niwareeba Roland
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Class C Amplifier

Class C (Figure 4.22) amplifiers are biased into

conduction much less than 180o. They are usually used in
RF applications, such as RF oscillators and modulators.
+VCC
The transistor is on when the
input signal exceeds |VBB| + VBE. RC

Because class C amplifiers are Vout

C
biased on for a small percentage
of time, they can be very RB
Vin
efficient. – VBB

Figure 4.22: Class C PA

© Niwareeba Roland
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Class C Amplifier
Class C operation is useful in oscillators. The collector circuit has a
parallel resonant circuit (“tank”) and oscillations are sustained by
the short pulse of collector current on each cycle.

+VCC

C2 L
Ic C3

C1 Vout

RB
Vout
Vin

– VBB

Figure 4.23: Class C PA

© Niwareeba Roland
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Class C Amplifier
The circuit can be set up with clamping bias, where the bias
resistor is connected to ground. C1 will charge (through the
base-emitter diode) to 0.7 V less than the positive peak.
+VCC

Clamping action causes

the transistor to be cut C2 L

off except at the positive ≈Vp – 0.7 V

Vp
peak of the input. The 0 + –
Q
–Vp
R1C1 time constant needs Q conducts Q conducts
C1 R1

to be long compared to 0.7 V

–Vp
the period of the signal.
Figure 4.24: Class C PA
© Niwareeba Roland
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Class D Amplifier
 A class D amplifier amplifies pulses, and requires a
pulsed input.

 There are many circuits that can convert a sinusoidal

waveform to a pulse, as well as circuits that convert a
pulse to a sine wave. This circuit has applications in
digital circuitry.

© Niwareeba Roland
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Class D Amplifier

Figure 4.25
© Niwareeba Roland