Power Amplifier

Power Amplifiers
Introduction

 Amplifier receives a signal from some pickup transducer or other input source
and provides larger version of the signal.

 In small signal amplifiers the main factors are usually amplification, linearity
and magnitude of gain.

Classes of Power Amplifiers

 Amplifier classes represent the amount the output signal varies over one cycle
of operation for a full cycle of input signal.

 So the following classes of PA are defined:

 Class A
 Class B
 Class AB
 Class C
 Class D

Class A amplifier

 Class A amplifying devices operate over the whole of the input cycle such that
the output signal is an exact scaled-up replica of the input with no clipping.

 Class A amplifiers are the usual means of implementing small signal

amplifiers.

 They are not very efficient; a theoretical maximum of 50% is obtainable with
inductive output coupling and only 25% with capacitive coupling.

 In a Class A circuit, the amplifying element is biased so the device is always

conducting to some extent, and is operated over the most linear portion of its
characteristic curve.
  Because the device is always conducting, even if there is no input at all, power
is drawn from the power supply. This is the chief reason for its inefficiency.

Class B amplifier

 Class B amplifiers only amplify half of the input wave cycle.

 As such they create a large amount of distortion, but their efficiency is greatly
improved and is much better than Class A.

 Class B has a maximum theoretical efficiency of 78.5% (i.e., π/4). This is

because the amplifying element is switched off altogether half of the time, and
so cannot dissipate power.

 A single Class B element is rarely found in practice, though it can be used in

RF power amplifier where the distortion levels are less important. However
Class C is more commonly used for this.

Class AB amplifier

 A practical circuit using Class B elements is the complementary pair or "push-

pull" arrangement.
  Here, complementary or quasi-complementary devices are used to each
amplify the opposite halves of the input signal, which is then recombined at
the output.

 This arrangement gives excellent efficiency, but can suffer from the drawback
that there is a small mismatch at the "joins" between the two halves of the
signal.

 Class AB sacrifices some efficiency over class B in favour of linearity, so will

always be less efficient (below 78.5%). It is typically much more efficient than
class A.

Class C amplifier

 Class C amplifiers conduct less than 50% of the input signal and the distortion
at the output is high, but high efficiencies (up to 90%) are possible.

 A much more common application for Class C amplifiers is in RF

transmitters, where the distortion can be vastly reduced by using tuned loads
on the amplifier stage.

 The input signal is used to roughly switch the amplifying device on and off,
which causes pulses of current to flow through a tuned circuit.
Class D amplifier

 Class D amplifiers are much more efficient than Class AB power amplifiers.

 As such, Class D amplifiers do not need large transformers and heavy heat
sinks, which means that they are smaller and lighter in weight than an
equivalent Class AB amplifier.

 All power devices in a Class D amplifier are operated in on/off mode.

 These amplifiers use pulse width modulation.

Comparison of Amplifier classes

Parameters A AB B C D
Operating 3600 1800-3600 1800 Less than Pulse
Cycle 1800 operation
Power 25-50% Between 78.5% Typically
efficiency 25%(50%) over 90%
and 78.5%
Series fed class A amplifiers

 It is a fixed bias circuit.

 The transistor used is a power transistor that is capable of operating in the

range of a few to tens of watts.
 This circuit is not the best to use as a large-signal amplifier because of its poor
power efficiency.

 The beta of a power transistor is generally less than 100, the overall amplifier
circuit using power transistors that are capable of handling large power or
current while not providing much voltage gain.

DC bias operation

 The dc bias set by 𝑉𝐶𝐶 and 𝑅𝐵 fixes the dc base-bias current at

𝑉𝐶𝐶 − 0.7V
𝐼𝐵 =
𝑅𝐵

with the collector current then being

𝐼𝐶 = 𝛽𝐼𝐵
 with the collector–emitter voltage then

𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶

 To appreciate the importance of the dc bias on the operation of the power

amplifier, consider the collector characteristic shown in Fig .

 The intersection of the dc bias value of 𝐼𝐵 with the dc load line then
determines the operating point (Q-point) for the circuit.

 If the dc bias collector current is set at one-half the possible signal swing
𝑉
(between 0 and 𝐶𝐶⁄𝑅 ), the largest collector current swing will be possible.
𝐶

 Additionally, if the quiescent collector-emitter voltage is set at one-half the

supply voltage, the largest voltage swing will be possible.

AC Operation

 When an input ac signal is applied to the amplifier, the output will vary from
its dc bias operating voltage and current.

 A small input signal, as shown in Fig., will cause the base current to vary
above and below the dc bias point, which will then cause the collector current
(output) to vary from the dc bias point set as well as the collector–emitter
voltage 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.

 For the current this limiting condition is either zero current at the low end or
𝑉𝐶𝐶
⁄𝑅 at the high end of its swing.
𝐶
 For the collector–emitter voltage, the limit is either 0 V or the supply
voltage𝑉𝐶𝐶 .

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

𝑃𝑖 (𝑑𝑐 ) = 𝑉𝐶𝐶 𝐼𝐶𝑄

Output power

 The output voltage and current varying around the bias point provide ac power
to the load.

 The ac signal,𝑉𝑖 , causes the base current to vary around the dc bias current and
the collector current around its quiescent level, 𝐼𝐶𝑄 .

 The larger the input signal, the larger the output swing, up to the maximum set
by the circuit.

Using rms signals

P0 (ac) = VCE (rms)IC (rms)

= IC 2 (rms)R C

VC 2 (rms)
=
RC

Using peak signals

The ac power delivered to the load may be expressed using

VCE (p)IC (p)

P0 (ac) =
2

IC 2 (p)
= RC
2
 VCE 2 (p)
=
2R C

Using peak-to-peak signals

The ac power delivered to the load may be expressed using

VCE (pp)IC (pp)

P0 (ac) =
8

IC 2 (pp)
= RC
8

VCE 2 (pp)
=
8R C

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

P0 (ac)
%η = × 100%
Pi (dc)

Maximum Efficiency

 For the class A series-fed amplifier, the maximum efficiency can be

determined using the maximum voltage and current swings.

 For the voltage swing it is

 VCE (pp) = VCC
 For the current swing it is
VCC
IC (pp) =
RC

 Using the maximum voltage swing in the equation below yields

VCE (pp)IC (pp)

P0 (ac) =
8

VCC
VCC
⁄R
C
=
8
𝑉𝐶𝐶 2
=
8𝑅𝐶
 The maximum power input can be calculated using the dc bias current set to
one-half the maximum value:

VCC /RC
Maximum Pi (dc) = VCC (maximumIC ) = VCC 2

𝑉𝐶𝐶 2
=
2𝑅𝐶
 Then the maximum efficiency is given by,

P0 (ac)
%η = × 100%
Pi (dc)

𝑉𝐶𝐶 2⁄
8𝑅𝐶
= 2
𝑉𝐶𝐶 ⁄
2𝑅𝐶
= 25%
 The maximum efficiency of a class A series fed amplifier is thus seen to be
25%.

 The maximum efficiency occurs only for ideal conditions of both voltage and
current swing .thus practical circuits will have less than this percentage.
TRANSFORMER-COUPLED CLASS A AMPLIFIER

 A form of class A amplifier having maximum efficiency of 50% uses a

transformer to couple the output signal to the load as shown in Fig.

 The transformer can step up or step down a voltage applied to primary coil.

VOLTAGE TRANSFORMATION

 The transformer can step up or step down a voltage applied to one side directly
as the ratio of the turns (or number of windings) on each side.

 The voltage transformation is given by

𝑉2 𝑁2
=
𝑉1 𝑁1
  Equation shows that if the number of turns of wire on the secondary side is
larger than on the primary, the voltage at the secondary side is larger than the
voltage at the primary side.

CURRENT TRANSFORMATION

 The current in the secondary winding is inversely proportional to the number

of turns in the windings.

 The current transformation is given by

I2 N1
=
I1 N2

 If the number of turns of wire on the secondary is greater than that on the
primary, the secondary current will be less than the current in the primary.

IMPEDANCE TRANSFORMATION

 Since the voltage and current can be changed by a transformer, an impedance

“seen” from either side (primary or secondary) can also be changed.
  As shown in Fig, impedance 𝑅𝐿 is connected across the transformer
secondary.

 This impedance is changed by the transformer when viewed at the primary

side (𝑅𝐿 ′ ).

𝑉2
𝑅𝐿 𝑅2 ⁄𝐼 𝑉2 𝐼1 𝑉2 𝐼1 N2 N2 N2
2
′ = = = = = = ( )2
𝑅𝐿 𝑅1 𝑉1⁄ 𝐼2 𝑉1 𝑉1 𝐼2 N1 N1 N1
𝐼1

N
 If we define a = N1 , where a is the turns ratio of the transformer, the above
2
equation becomes
𝑅𝐿 ′ 𝑅1 N1
= = ( )2 = a2
𝑅𝐿 𝑅2 N2

 We can express the load resistance reflected to the primary side as

𝑅𝐿 ′ = a2 𝑅𝐿
Where 𝑅𝐿 ′ is the reflected impedance.

Signal Swing and Output Ac Power

 Figure shows the voltage and current signal swings from the circuit of
Transformer-coupled audio power amplifier.

 From the signal variations shown in Fig, the values of the peak-to-peak Signal
swings are

𝑉𝐶𝐸 (𝑝𝑝) = 𝑉𝐶𝐸𝑚𝑎𝑥 − 𝑉𝐶𝐸𝑚𝑖𝑛

 𝐼𝐶 (𝑝𝑝) = 𝐼𝐶𝑚𝑎𝑥 − 𝐼𝐶𝑚𝑖𝑛

 The ac power developed across the transformer primary can then be calculated
using

(𝑉𝐶𝐸𝑚𝑎𝑥 − 𝑉𝐶𝐸𝑚𝑖𝑛 )(𝐼𝐶𝑚𝑎𝑥 − 𝐼𝐶𝑚𝑖𝑛 )

𝑃0 (𝑎𝑐 ) =
8

 For the ideal transformer, the voltage delivered to the load can be calculated
using the following equation.
N2
𝑉𝐿 = 𝑉2 = V
N1 1
 The power across the load can then be expressed as
𝑉𝐿 2 (𝑟𝑚𝑠)
𝑃𝐿 =
𝑅𝐿

N
 Similarly the load current can be calculated as 𝐼𝐿 = 𝐼2 = N1 IC with the output
2
ac power
𝑃𝐿 = 𝐼𝐿 2 (𝑟𝑚𝑠)𝑅𝐿

Efficiency

 The input dc power obtained from the supply is calculated from the supply dc
voltage and thus average power drawn from the supply.

Pi (dc) = VCC ICQ

 For the transformer-coupled amplifier, the power dissipated by the transformer
is small (due to the small dc resistance of a coil) and is ignored.
  The only power loss considered here is that dissipated by the power transistor
and calculated by

𝑃𝑄 = 𝑃𝑖 (𝑑𝑐 ) − 𝑃0 (𝑎𝑐)

Where 𝑃𝑄 is the power dissipated as heat.

P0 (ac)
%η = × 100%
Pi (dc)

Maximum theoretical efficiency

 For a class A transformer-coupled amplifier, the maximum theoretical

efficiency goes up to 50%.

 Based on the signals obtained using the amplifier, the efficiency can be
expressed as

VCE − VCEmin 2
%η = 50( max ) %
VCEmax + VCEmin

 Larger the value of VCEmax and smaller the value of VCEmin , the closer the
efficiency approaches the theoretical limit of 50%.

Class B Amplifier operation

 Class B operation is provided when the dc bias leaves the transistor biased just
off, the transistor turning on when the ac signal is applied.

 This is essentially no bias and conducts for only one half cycle.

 To obtain output for full cycle, it is required to use two transistors and have
each conduct on opposite half-cycles, the combined operation providing a full
cycle of output on opposite half cycles of output signal.
  Since one part of the circuit pushes the signal high during one half cycle and
other part pulls the signal low during the other half cycle, the circuit is referred
to as push-pull circuit.

 Class B operation provides greater efficiency than was possible using single
transistor in class A operation.

Input (DC) Power

 The amount of input power can be calculated using

𝑃𝑖 (𝑑𝑐 ) = 𝑉𝐶𝐶 𝐼𝑑𝑐
Where 𝐼𝑑𝑐 is the average or dc current drawn from the power supplies.

 In class B operation, the current drawn from a single power supply has the
form of a full-wave rectified signal, while that drawn from two power supplies
has the form of a half wave rectified signal from each supply.

 In either case, the value of the average current drawn can be expressed as
2
𝐼𝑑𝑐 = 𝐼(𝑝)
𝜋
where I(p) is the peak value of the output current waveform.
 Hence input power is equal to
2
𝑃𝑖 (𝑑𝑐 ) = 𝑉𝐶𝐶 𝐼(𝑝)
𝜋

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.

 If one is using an rms meter to measure the voltage across the load, the output
power can be calculated as

𝑉𝐿 2 (𝑟𝑚𝑠)
𝑃0 (𝑎𝑐 ) =
𝑅𝐿
 If one is using an oscilloscope, the peak, or peak-to-peak, output voltage
measured can be used:

𝑉𝐿 2 (𝑝𝑝) 𝑉𝐿 2 (𝑝)
𝑃0 (𝑎𝑐 ) = =
8𝑅𝐿 2𝑅𝐿
 The larger the rms or peak output voltage, the larger the power delivered to the
load.

Efficiency

 The efficiency of the class B amplifier can be calculated using the basic
equation:

P0 (ac)
%η = × 100%
Pi (dc)

𝑉𝐿 2 (𝑝)⁄
2𝑅𝐿
=
2
𝑉𝐶𝐶 𝜋 𝐼 (𝑝)

𝑉𝐿 (𝑝)
using 𝐼 (𝑝 ) = ⁄𝑅
𝐿

π 𝑉𝐿 (𝑝)
%η = × 100%
4 𝑉𝐶𝐶
  Equation shows that the larger the peak voltage, the higher the circuit
efficiency, up to a maximum value when 𝑉𝐿 (𝑝) = 𝑉𝐶𝐶 , this maximum
efficiency then being

π
maximum efficiency = × 100% = 78.5%
4

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.
𝑃2𝑄 = 𝑃𝑖 (𝑑𝑐 ) − 𝑃0 (𝑎𝑐)
where𝑃2𝑄 is the power dissipated by the two output power transistors
 The dissipated power handled by each transistor is then
𝑃2𝑄
𝑃𝑄 =
2

Maximum Power Considerations

 For class B operation, the maximum output power is delivered to the load
when 𝑉𝐿 (𝑝) = 𝑉𝐶𝐶
𝑉𝐶𝐶 2
𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑃0 (𝑎𝑐 ) =
2𝑅𝐿
 The corresponding peak ac current I(p) is then
𝑉𝐶𝐶
𝐼 (𝑝 ) =
𝑅𝐿
 The maximum value of average current from the power supply is
2 2 𝑉𝐶𝐶
𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝐼𝑑𝑐 = 𝐼 (𝑝) =
𝜋 𝜋 𝑅𝐿
 Using this current to calculate the maximum value of input power results in
2 𝑉𝐶𝐶 2𝑉𝐶𝐶 2
𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑃𝑖 (𝑑𝑐 ) = 𝑉𝐶𝐶 (𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝐼𝑑𝑐 ) = 𝑉𝐶𝐶 ( )=
𝜋 𝑅𝐿 𝜋𝑅𝐿
 The maximum circuit efficiency for class B operation is then
P0 (ac)
%η = × 100%
Pi (dc)
 𝑉𝐶𝐶 2⁄
2𝑅𝐿
= 2 × 100%
2𝑉𝐶𝐶 ⁄
𝜋𝑅𝐿
𝜋
= × 100%
4
= 78.54%

Efficiency in another form

 The maximum efficiency of a class B amplifier can also be expressed as

follows:
𝑉𝐿 2 (𝑝)
𝑃0 (𝑎𝑐 ) =
2𝑅𝐿

2 𝑉𝐿 (𝑝)
𝑃𝑖 (𝑑𝑐 ) = 𝑉𝐶𝐶 𝐼𝑑𝑐 = 𝑉𝐶𝐶 [ ]
𝜋 𝑅𝐿

𝑉𝐿 2 (𝑝)⁄
P0 (ac) 2𝑅𝐿
%η = × 100% = × 100%
Pi (dc) 2 𝑉 (𝑝)
VCC [ 𝐿 ]
𝜋 𝑅𝐿

𝑉𝐿 (𝑝)
= 78.54 × 100%
𝑉𝐶𝐶

Class B amplifier circuits

 A number of circuit arrangements for obtaining class B operation are possible.

 The input signals to the amplifier could be a single signal, the circuit then
providing two different output stages, each operating for one-half the cycle.

 If the input is in the form of two opposite polarity signals, two similar stages
could be used, each operating on the alternate cycle because of the input
signal.
 One means of obtaining polarity or phase inversion is using a transformer, the
transformer coupled amplifier having been very popular for a long time.

 Figure shows a center-tapped transformer to provide opposite phase signals.

 If the transformer is exactly center-tapped, the two signals are exactly opposite
in phase and of the same magnitude.

 Opposite polarity inputs can easily be obtained using an op-amp having two
opposite outputs or using a few op-amp stages to obtain two opposite polarity
signals.

 The circuit of Fig. uses a BJT stage with in-phase output from the emitter and
opposite phase output from the collector.
Transformer coupled push-pull amplifier

 The circuit of Fig. 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.

 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 half-cycle 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 half-cycle to
the load.

 The overall signal developed across the load then varies over the full cycle of
signal operation.
Complementary symmetry circuits

 Every transistor will conduct for half cycle.

 Single input signal is applied to the base of both transistors.
 npn transistor will be biased in conduction for positive half cycle of the input.
 During negative half cycle pnp transistor is biased into conduction when input
goes to negative.

Disadvantages

 One disadvantage is that the need of two separate voltage supplies.

  Cross over distortion in the output signal.
 This cross over distortion is referred to as the nonlinearity in the output signal
during cross over from positive to negative or vice-versa.

Amplifier distortion

 Any signal varying over less than the full 3600 cycle is considered to have
distortion.

 An ideal amplifier is capable of amplifying a pure sinusoidal signal to provide

a larger version, the resulting waveform being a pure sinusoidal frequency
sinusoidal signal.

 When distortion occurs, output will not be an exact duplicate of input signal
(except for magnitude).

 Distortion can occur because the device characteristic is not linear. In this case
non linear or amplitude distortion occurs.

 Distortion can also occur because the circuit elements and devices respond to
the input signal differently at various frequencies, this being frequency
distortion.

 One technique for describing distorted but period waveforms uses Fourier
analysis, a method that describes any periodic waveform in terms of its
fundamental frequency component and frequency components at integer
multiples- these components are called harmonic components or harmonics.

Harmonic Distortion

 A signal is considered to have harmonic distortion when there are harmonic

frequency components.

 If fundamental frequency has amplitude A1, and nth frequency component has
an amplitude of An.

 Harmonic distortion can be defined as

 |𝐴 𝑛 |
% 𝑛𝑡ℎ ℎ𝑎𝑟𝑚𝑜𝑛𝑖𝑐 𝑑𝑖𝑠𝑡𝑜𝑟𝑡𝑖𝑜𝑛 = %𝐷 = × 100%
|𝐴1 |

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 relation

%THD = √(D2 2 + D3 2 + D4 2 + ⋯ . . ) × 100%

Second harmonic distortion

1
((ICmax + ICmin ) − ICQ
D2 = |2 | × 100%
(ICmax − ICmin )

In voltage terms,

1
(VCEmax + VCEmin ) − VCEQ
D2 = |2 | × 100%
(VCEmax − VCEmin )

Power of signal having distortion

 Power delivered to the load resistor 𝑅𝐶 due to the fundamental component of

the distorted signal is
𝐼1 2 𝑅𝐶
𝑃1 =
2
 Total power due to all the harmonic components of the distorted signal is,
𝑅𝐶
𝑃 = (𝐼1 2 + 𝐼2 2 + 𝐼3 2 + ⋯ )
2

 In terms of Total harmonic distortion

2 𝑅𝐶
𝑃 = (1 + 𝐷2 2 + 𝐷3 2 + ⋯ )𝐼1
2

𝑃 = (1 + 𝑇𝐻𝐷2 )𝑃1