Alexandria High Institute for Engineering and Technology
Power Amplifiers
Sameh M. Selim
Introduction:
Power amplifiers are large-signal amplifiers. This generally means that
much larger portion of the load line is used during signal operation than
in small-signal amplifier. In this chapter, we will cover 4 classes of power
amplifiers: class A, class B, class AB, and class C.
Class A:The o/p signal varies for as full 360o of the cycle. Fig. (1-a)
shows that this requires the Q-point to be biased at a level so that at
least half the signal swing of the o/p may vary up and down without
going to a high enough voltage to be limited by the supply voltage or
too low to approach the lower supply level, or 0 V in this description.
Class B:The o/p signal varies over one –half the i/p signal cycle, or for
180o of the signal. The dc bias point for class B is therefore at 0 V, with
the o/p then varying from this bias point for a half cycle. Obviously, the
o/p is not a faithful reproduction of the i/p if only one half- cycle is
present. Two class B operations: one to provide o/p on the +ve halfcycle and another to provide operation on the –ve o/p half-cycle - are
necessary. The combined half-cycles then provide an o/p for a full 360o
of operation. This type of operation is referred to as
push-pull operation.
Class AB:An amplifier may be biased at a dc level above the zero-basecurrent level of class B and above one-half the supply voltage level of
class A. Class AB operation still requires a push-pull connection to
achieve a full o/p cycle. The o/p signal swing occurs between 180o and
360o.
The following Figure shows an ac load line with the Q-point moved away
from center toward cutoff. The collector current can only swing down to
near zero and an equal amount above ICQ. The VCE can only swing up to
its cutoff value and an equal amount below VCEQ. If the amplifier is
driven any further than this, it will "clip" at cutoff, as shown in the
following Figure.
The following Figure shows an ac load line with the Q-point moved away
from center towards saturation. In this case, the O/P variation is limited
by saturation. The collector current can only swing up to near saturation
and an equal amount below ICQ. The collector-to-emitter voltage can
only swing down to its saturation value and an equal amount above
VCEQ. If the amplifier is driven any further, it will "clip" at saturation.
Class C:The o/p of a class C amplifier is biased for operation at less than
180o of the cycle and will operate only with a tuned circuit, which
provides a full cycle of operation for the tuned or resonant frequency.
(Used for radio or communications)
Amplifier Efficiency: defined as the ratio of o/p power to i/p power.
The maximum efficiency of a class A is 25% with a direct or series –fed
load connection and 50 % with a transformer connection to the load.
Class B operation, with no dc bias power for noi/p signal, can be shown
to provide a maximum efficiency up to 78.5%.
The following table summarizes the operation of the amplifier classes.
Amplifier Classs
A
AB
Operating Cycle
360o
180o-360o
B
C
180o < 180o
Power Efficiency 25%-50% 25%(50%) to 78.5% 78.5%
Class A, series fed Amplifier: The transistor used in this circuit is a
power transistor that is capable of operatingin the range of a few to
tens of watts.
DC Bias Operation: VCC and RB fix the dc base-bias current at
𝐼𝐵 =
𝑉𝐶𝐶 −0.7 𝑉
𝑅𝐵
IC =  IB
VCE = VCC - ICRC
(1)
(2)
(3)
AC Operation: When an i/p ac signal is applied to the amplifier, the o/p
will vary from its dc bias operating voltage and current. Fig.(a) shows a
small i/p signal. As the i/p signal is made larger, the o/p will vary further
around the established dc bias untileither 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/Rc) at the high end of its swing.
For the collector-emitter voltage, the limit is either 0 V or the supply
voltage, VCC.
Power considerations
The power input of the amplifier is provided by the supply. With no i/p
signal, the dc current drawn is the collector bias current ICQ. The power
drawn from the supply is:
(4)
Even with an ac signal applied, the average current drawn from the
supply remains the same.
The power output is provided by the o/p voltage and current varying
around the bias point. This ac power is delivered to the load RC, and can
be expressed in many ways.
Using RMS signals
(5)
Using Peak Signals
(6)
Using Peak-to-Peak signals
(7)
Efficiency %the efficiency of the amplifier is calculated using
(8)
Maximum Efficiencyfor class A series fed amplifier, the maximum
efficiency can be determined using the maximum voltage and current
swings. For the voltage swing it is
Maximum VCE(p-p) = VCC
For the current swing it is
Maximum IC(p-p) = VCC/RC
Using thepeak-to-peakform of the power output from (7) yields
The maximum power i/p can be calculated using the dc bias current set
to one-half the maximum value:
Using equation (8) to calculate the maximum efficiency:
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%.
Example1
Calculate the i/p power, o/p power, and efficiency of the amplifier circuit
shown in Figure for an i/p voltage that results in a base current of 10 mA
peak.
Solution:using equations (1) to (3), determine the Q-point to be:
IBQ = 19.3 mA
ICQ = 0.483 A
VCEQ = 10.34 V
This bias point is marked on the transistor collector characteristic shown
in Figure. The ac variation of the o/p signal can be obtained graphically
using the dc load line (by connecting VCE=VCC = 20 V with IC = VCC/RC =
1000 mA = 1 A).
When the i/p ac base current increases from its dc bias level, the
collector current rises by
IC(p) = IB(p) = 25 (10 mA peak) = 250 mA peak
Using equation (6) yields
𝑃𝑜 (𝑎𝑐 ) =
𝐼𝐶2 (𝑝)
2
𝑅𝐶 =
(250×10−3 𝐴)
2
2
(20 Ω) =0.625 W
Using equation (4) results in
Pi(dc) = VCCICQ = (20 V)(0.48A) = 9.6 W
Using equation (8), the amplifier power efficiency can be calculated
%𝜂 =
𝑃𝑜 (𝑎𝑐)
𝑃𝑖 (𝑑𝑐)
× 100%, check = 6.5%
Transformer-Coupled Class A Amplifier
Transformer Action:
 Voltage Transformation
𝑽𝟐
𝑽𝟏
=
𝑵𝟐
(9)
𝑵𝟏
 Current Transformation
𝑰𝟐
𝑰𝟏
=
𝑵𝟏
(10)
𝑵𝟐

𝑹𝑳
𝑹′𝑳
=
Impedance Transformation
𝑹𝟐
𝑹𝟏
=
𝑽𝟐 /𝑰𝟐
𝑽𝟏 /𝑰𝟏
=
𝑽 𝟐 𝑰𝟏
𝑽 𝟏 𝑰𝟐
If we define a = N1/N2
𝑹′𝑳
𝑹𝑳
=
𝑹𝟏
𝑹𝟐
𝑵𝟏 𝟐
= (𝑵 ) = 𝒂𝟐 (11)
𝟐
The load resistance reflected to the primaryside is:
𝑹𝟏 = 𝒂𝟐 𝑹𝟐 or𝑹′𝑳 = 𝒂𝟐 𝑹𝑳 (12)
Example2
Calculate the effective resistance seen looking into the primary of a 15:1
transformer connected to an 8-Ω load.
Solution:Using eq.(12), get 𝑅𝐿′ = (15)2 (8Ω) = 1800 Ω
Example3
What transformer turns ratio is required to match a 16- speaker load
so that the effective load resistance seen at the primary is 10 k?
𝑁1 2
𝑅𝐿,
Solution: Using eq. (11), check(𝑁 ) = 𝑅 = 625,
2
So,
𝑁1
𝑁2
𝐿
= 25: 1
Operation of Amplifier Stage
DC Load Line:The transformer (dc) winding resistance determines the dc
load line for the above circuit. Typically, this dc resistance is very small
(ideally 0 ), so a 0  dc load line is a straight vertical line, drawn from
the point VCEQ = VCC.
Quiescent operating Point: The operating point can be obtained
graphically at the point of intersection of the dc load line and the base
current set by the circuit. The collector quiescent current can then be
obtained from the operating point.
Keep in mind that, in class A operation, the dc bias point sets the
conditions for the maximum undistorted signal swing for both collector
current and collector-emitter voltage. If the i/p signal produces a
voltage swing less than the maximum possible, the efficiency of the
circuit at that time will be less than 25%.
AC Load Line: It is necessary to calculatethe ac load resistance "seen"
looking into the primary side of the transformer, then draw the ac load
line on the collector characteristic. The reflected load resistance (R'L) is
calculated using Eq. (12).
Draw the ac load line so that it passes thro' the operating point and has
a slope equal to (-1/R'L). Notice that the ac load line shows that the o/p
signal swing can exceed the value of VCC. It is therefore necessary after
obtaining the ac load line to check that the possible voltage swing does
not exceed transistor maximum ratings.
Signal Swing and Output AC Power:The Figure shows the voltage
and current signal swings from the above circuit. The values of the peakto-peak signal swings are
The ac power developed across the transformer primary can be
calculated as
(13)
The ac power calculated is that developed across the primary of the
transformer. Assuming an ideal transformer, we find that the power
delivered by the secondary to the load is approximately that calculated
using Eq. (13).
The o/p ac power can also be determined using the voltage delivered to
the load.
For the ideal transformer, the voltage delivered to the load can be
calculated using Eq.(9):
The power across the load can then be expressed as Eq. (5)
Using Eq. (10) to calculate the load current yields
With the o/p power then calculated using
Example 4
Calculate the ac power delivered to the 8- speaker for the following
circuit. The component values result in a dc base current of 6 mA, and
the i/p signal (Vi)results in a peak base current swing of 4 mA.
Solution: The dc load line is drawn vertically fro the voltage point:
VCEQ = VCC = 10 V
For IB = 6 mA, the operating point on the following graph is
VCEQ = 10 V and ICQ = 140 mA
The effective ac resistance seen at the primary is
R'L = (3)2(8) = 72 
To help draw the load line, consider the following procedure:
For a current swing of𝐼𝐶 =
𝑉𝐶𝐸
𝑅𝐿′
=
10 𝑉
72 Ω
= 139 𝑚𝐴
Mark point A: ICEQ + IC = 140 mA + 139 mA = 279 mA along the y-axis
Connect point A thro' 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 collector-emitter voltage obtained fro the graph
are, respectively,
VCEmin = 1.7 V
ICmin = 25 mA
VCEmax = 18.3 V
ICmax = 255 mA
Using Eq. (13), the ac power delivered to the load can be calculated
𝑃𝑜 (𝑎𝑐 ) =
(18.3𝑉 − 1.7𝑉 )(255𝑚𝐴 − 25𝑚𝐴)
= 𝟎. 𝟒𝟕𝟕 𝑾
8
Efficiency: The average i/p (dc) power drawn from the supply:
𝑃𝑖 (𝑑𝑐 ) = 𝑉𝐶𝐶 𝐼𝐶𝑄
(14)
The dissipated power, PQ by the power transistor, ignoring the
dissipated power by the transformer,
𝑃𝑄 = 𝑃𝑖 (𝑑𝑐 ) − 𝑃𝑜 (𝑎𝑐 )
(15)
Example 5
For the above circuit and results of example 4, calculate the dc i/p
power, power dissipated by the transistor, and efficiency of the circuit
for i/p signal of example 4.
Solution:From Eq. (14), Pi(dc) = (10 V)(140 mA) = 1.4 W
From Eq. (15), PQ = 1.4 W – 0.477 W = 0.92 W
%𝜼=
𝑷𝒐 (𝒂𝒄)
𝑷𝒊 (𝒅𝒄)
× 𝟏𝟎𝟎%, 𝒄𝒉𝒆𝒄𝒌 = 𝟑4.1%
Maximum Theoretical Efficiency: For a class A transformer-coupled
amplifier, themaximum theoretical efficiency goes up to 50%. Based on
the signals obtained using the amplifier, the efficiency can be expressed
as
(16)
The larger VCEmax and the smaller of VCEmin, the closer the efficiency
approaches the theoretical limit of 50%.
Example 6
Calculate the efficiency of a transformer-coupled class A amplifier for a
supply of 12 V and outputs of:
(a) V(p) = 12 V.
(b) V(p) = 6 V.
Solution:
(a)since VCEQ = VCC = 12 V,
VCEmax = VCEQ + V(p) = 12 + 12 = 24 V
VCEmin = VCEQ – V(p) = 12 V – 12 V = 0 V
using Eq. (16), % η = 50%
(b) check VCEmax = 18 V, VCEmin = 6 V
% η = 12.5%
(c)
VCEmax = 14 V,
% η = 1.39%
VCEmin = 10 V
(c) V(p) = 2 V.
Class B Amplifier Operation: A 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 the
transistor conducts current for only one-half of the signal cycle.
To obtain o/p for the full cycle of signal, it is necessary to use 2
transistors, each conduct on opposite half cycles, and the combined
operation providing a full cycle of o/p signal.
Since one part of the circuit pushes the signal high during one half-cycle
and the other part pulls the signal low during the other half-cycle, the
circuit is referred as a push-pull circuit.
Input (DC) Power: The i/p dc power
provided by the power supply can
calculated using:
be
(17)
Where Idc is the average, or dc
current drawn from the power supplies.
(18)
Using Eq.(18) in Eq. (17) results in
(19)
Output (AC) Power: Using rms meter to measure the voltage across the
load, Po(ac) is
(20)
If one uses an oscilloscope, the peak or peak-to-peak o/p voltage can
be used:
(21)
The larger the rms or peak o/p voltage, the larger is the power
delivered to the load.
Efficiency: The efficiency of the class B amplifier can be calculated using
the basic equation
𝑷𝒐 (𝒂𝒄)
%𝜼=
× 𝟏𝟎𝟎%
𝑷𝒊 (𝒅𝒄)
Using Eqs. (19) and (21) in the efficiency equation results in
%𝜼 =
𝑽𝟐𝑳 (𝒑)/𝟐𝑹𝑳
𝟐
𝝅
𝑽𝑪𝑪 [( )𝑰(𝒑)]
× 100% =
𝝅 𝑽𝑳 (𝒑)
𝟒 𝑽𝑪𝑪
× 100% (22)
[ Using I(p)=VL(p)/RL] . Equation (22) shows that the larger the peak
voltage, the higher is the circuit efficiency, up to maximum value when
VL(p) = VCC, this maximum efficiency then being
Maximum efficiency =
𝝅
𝟒
× 𝟏𝟎𝟎 = 𝟕𝟖. 𝟓%
Power Dissipation by output Transistors: This is the difference between
the i/p power delivered by the supplies and the o/p power delivered to
the load,
P2Q = Pi(dc) – Po(ac)
(23)
P2Q is the power dissipated by the 2 o/p transistors. The dissipated
power handled by each transistor is
PQ =
𝑷𝟐𝑸
𝟐
(24)
Maximum Power Considerations For class B operation, the maximum
o/p power is delivered to the load when VL(p) = VCC:
(25)
The corresponding peak ac current I(p) is then
𝐼 (𝑝) =
𝑉𝐶𝐶
𝑅𝐿
So the maximum value of the average current from the power supply is
𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝐼𝑑𝑐 =
2
2𝑉𝐶𝐶
𝐼 (𝑝) =
𝜋
𝜋𝑅𝐿
Using this current to calculate the maximum value of i/p power results
(26)
Using (25) and (26), the maximum circuit efficiency for class B operation
is then
𝜋
𝑚𝑎𝑥𝑖𝑚𝑢𝑚 % 𝜂 = 4 × 100% = 78.54%
(27)
For class B operation, the maximum power dissipated by the o/p
transistors does not occur at the maximum power i/p or o/p condition.
The maximum power dissipated by the two o/p transistors occurs when
the o/p voltage across the load is
𝑉𝐿 (𝑝) = 0.636𝑉𝐶𝐶
(=
2
𝑉 )
𝜋 𝐶𝐶
For a maximum transistor power dissipation of
(28)
Example 7
For a class B amplifier providing a 20-V peak signal to a 16- load
(speaker) and a power supply of VCC = 30 V, determine the i/p power, o/p
power, and circuit efficiency.
Solution:
𝑉𝐿 (𝑝) 20𝑉
𝐼𝐿 (𝑝) =
=
= 1.25 𝐴
𝑅𝐿
16Ω
The dc value of the current drawn from the power supply is then
𝐼𝑑𝑐 =
2
𝐼 (𝑝), 𝑐ℎ𝑒𝑐𝑘 = 0.796 𝐴
𝜋 𝐿
The i/p power delivered by the supply voltage is
𝑃𝑖 (𝑑𝑐 ) = 𝑉𝐶𝐶 𝐼𝑑𝑐 , 𝑐ℎ𝑒𝑐𝑘 = 23.9 𝑊
The i/p power delivered to the load is
𝑉𝐿2 (𝑝)
𝑃𝑜 (𝑎𝑐 ) =
, 𝑐ℎ𝑒𝑐𝑘 = 12.5 𝑊
2𝑅𝐿
The circuit efficiency is calculated, check
% = 52.3%
Example 8
For a class B amplifier using a supply of VCC = 30 V and driving a load of
16 , determine the maximum i/p power, o/p power, and transistor
dissipation.
Solution:
Using Eq. (25), maximum Po(ac) = 28.125 W (check)
Using Eq. (26), maximum Pi(dc) = 35.81 W (check)
The maximum %  = 78.54%
(as expected)
Using Eq. (28), maximum PQ = 0.5(maximum P2Q) = 5.7 W
The maximum efficiency of a class B amplifier can also be expressed as
follows:
𝑉𝐿2 (𝑝)
𝑃𝑜 (𝑎𝑐 ) =
2𝑅𝐿
𝑃𝑖 (𝑑𝑐 ) = 𝑉𝐶𝐶 𝐼𝑑𝑐
2𝑉𝐿 (𝑝)
= 𝑉𝐶𝐶 [
]
𝜋𝑅𝐿
From which, we can get:
% 𝜂𝑚𝑎𝑥 = 78.54
𝑉 𝐿 (𝑝 )
𝑉𝐶𝐶
%
(29)
Example 9
Calculate the efficiency of a class B amplifier for a supply voltage of VCC
= 24 V with peak o/p voltages of:
a) VL(p) = 22 V.
Solution:
Using Eq. (29) gives
b) VL(p) = 6 V.
Class B Amplifier circuits: The i/p signal s to the amplifier could be
single signal, the circuit then providing 2 different o/p stages, each
operating for one-half the cycle. If the i/p is in the form of 2 oppositepolarity signals, 2 similar stages could be used, each operating on the
alternate cycle because of the i/p signal.
The Figure shows different ways to obtain phase-inverted signals from a
single i/p signal. Fig. (a) shows a center-tapped transformer to provide
opposite-phase signals. Fig. (b) uses a BJT stage with in-phase o/p from
the emitter and opposite-phase o/p from the collector. If the gain is
made nearly 1 for each o/p, the same magnitude results.
A- Transformer-Coupled Push-Pull Circuits
During the first half-cycle of operation, transistor Q1 is driven ON,
whereas transistor Q2 is driven Off. The current I1 thro' the transformer
results in the first half-cycle of signal to the load.
During the second half-cycle of the i/p signal, Q2 conducts, whereas Q1
stays Off, the current I2 thro' 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.
B- Complementary-Symmetry Circuits
Using complementary transistors (npn and pnp) it is possible to obtain a
full cycle o/p across a load using half-cycles of operation from each
transistor.
One disadvantage of the circuit is the need for 2 separate voltage
supplies. Another disadvantage is Crossover distortion which refers to
the fact that during the signal crossover from +ve to –ve. there is some
nonlinearity in he o/p signal. This results from the fact that the circuit
does not provide exact switching of one transistor off and the other ON
at zero voltage condition.
A more practical version of a push-pull circuit using complementary
transistors is shown in the following Figure. The load is driven as the o/p
of an emitter-follower so that the load resistance of the load is matched
by the low o/p resistance of the driving source. The circuit uses
complementary Darlington-connected transistors to provide higher o/p
current and lower o/p resistance.
C- Quasi-Complementary Push-Pull Amplifier
In practical power amplifier circuits, it is preferable to use npn
transistors for both high current-o/p devices. Since the push=pull
connection requires complementary devices, a pnp high power
transistor must be used. A practical means of obtaining complementary
operation while using the same matched npn transistors for the o/p is
provided by a quasi-complementary circuit shown in the following
Figure. The push-pull operation is achieved by using complementary
transistors (Q1 and Q2) before the matched npn o/p transistors (Q3 and
Q4). Notice that Q1 and Q3 form a Darlington connection that provides
o/p from a low-impedance emitter follower. The connection Q2 and Q4
forms a feedback pair, which similarly provides a low-impedance drive
to the load. Resistor R2 can be adjusted to minimize crossover distortion
by adjusting the dc bias condition.
Example 10
For the circuit shown, calculate the i/p power, o/p power, and power
handled by each o/p transistor and the circuit efficiency for an i/p of
12 V rms.
Solution
Given: VCC = 25 V, RL = 4 , Vi (rms)=12 V
Vi(p)=√2𝑉𝑖 (𝑟𝑚𝑠)= 16.97V, check
= VL(p) (the amplifier has ideally a voltage gain = 1)
𝑃𝑜 (𝑎𝑐 ) =
𝑉𝐿2 (𝑝)
2𝑅𝐿
= 𝟑𝟔 𝑾, check
𝑉𝐿 (𝑝)
= 4.24 𝐴
𝑅𝐿
𝐼𝐿 (𝑝) =
2
𝐼 (𝑝) = 2.7 𝐴
𝜋 𝐿
𝑃𝑖 (𝑑𝑐 ) = 𝑉𝐶𝐶 𝐼𝑑𝑐 = 𝟔𝟕. 𝟓 𝑾, check
𝐼𝑑𝑐 =
𝑃𝑄 =
𝑃2𝑄
%𝜂 =
2
𝑃𝑜
𝑃𝑖
=
𝑃𝑖 −𝑃𝑜
2
= 𝟏𝟓. 𝟕𝟓 𝑾, check
= 𝟓𝟑. 𝟑 %, check
Example 11
For the circuit of example 10, calculate the maximum i/p power,
maximum o/p power, i/p voltage for maximum power operation, and
power dissipated by the o/p transistors at this voltage.
Determine the max. power dissipated by the o/p transistors, and the i/p
voltage at which it occurs.
Solution
VCC = 25 V, RL = 4 ,

𝑚𝑎𝑥. 𝑃𝑖 (𝑑𝑐 ) =

𝑚𝑎𝑥. 𝑃𝑜 (𝑎𝑐 ) =
% 𝜂 𝑚𝑎𝑥. =
2
2𝑉𝐶𝐶
𝜋𝑅𝐿
2
𝑉𝐶𝐶
2𝑅𝐿
= 99.47 W, check
= 78.125 W, check
𝑃𝑜
× 100 = 𝟕𝟖. 𝟓𝟒 %
𝑃𝑖
To achieve max. power operation, the o/p voltage must be:
 VL(p) = VCC = 25 V
 P2Q = Pi – Po = 21.3 W
2
2𝑉𝐶𝐶
 𝑚𝑎𝑥. 𝑃2𝑄 = 𝜋2 𝑅 = 𝟑𝟏. 𝟔𝟔 𝑾
𝐿
This max. dissipation occurs at
 VL = 0.636 VL(p) = 15.9 V
Note that at VL = 15.9 V, the o/p transistors dissipate 31.66 W,
Whereas at VL = 25 V they only dissipate 21.3 W.
Amplifier Distortion: A pure sinusoidal signal has a single frequency
at which the voltage varies positive and negative by equal amounts. Any
signal varying over less than the full 360o cycle is considered to have
distortion. Nonlinear distortion occurs when the device characteristic is
not linear. This can occur with all classes of amplifier operation.
Frequency distortion occurs because the circuit elements and devices
respond to the i/p signal differently at various frequencies.
One technique for describing distortion is by using Fourier analysis, a
method that describes any periodic waveform in terms of its
fundamental frequency component and its integer multiples
(harmonics). For example a signal that is originally 1 kHz could result,
after distortion, in a frequency component at 1kHz and harmonic
components at 2 kHz, 3 kHz, 4 kHz, and so on. The original signal of 1
kHz is called fundamental frequency; those at integer multiples are the
harmonics. The 2 kHz is called a second harmonic, that at 3 kHz is the
third harmonic, and so on.
Harmonic Distortion % Dn
If the fundamental frequency has amplitude A1, and the nth frequency
component has amplitude An, the harmonic distortion can be defined
as:
(30)
Total Harmonic Distortion % THD
(31)
Example 12
 Calculate the harmonic distortion components for an o/p signal
having fundamental amplitude of 2.5 V, second harmonic amplitude
0.25 V, third harmonic of 0.1 V, and fourth harmonic of 0.05 V.
 Calculate the total harmonic distortion.
Solution
Using Eq. (30) yields
Using Eq. (31), we obtain
Second Harmonic Distortion the Figure shows a collector current
waveform IC with the quiescent ICQ, minimum ICmin, and maximum ICmax
signal levels. The signal shown indicates that some distortion is present.
An equation that approximately describes the distorted waveform is
𝑖𝐶 ≈ 𝐼𝐶𝑄 + 𝐼𝑜 + 𝐼1 𝑐𝑜𝑠𝜔𝑡 + 𝐼2 𝑐𝑜𝑠2𝜔𝑡
(32)
ICQ the quiescent current which occurs with zero i/p signal; Io is the
nonzero average (dc) current of the distorted signal; I1 is the
fundamental component of the distorted ac signal; and I2 is the second
harmonic component at twice the fundamental frequency.
a- At point 1 (t = 0):
𝑖𝐶 = 𝐼𝐶𝑚𝑎𝑥 = 𝐼𝐶𝑄 + 𝐼𝑜 + 𝐼1 𝑐𝑜𝑠0 + 𝐼2 𝑐𝑜𝑠0
𝑖𝐶𝑚𝑎𝑥 = 𝐼𝐶𝑄 + 𝐼𝑜 + 𝐼1 + 𝐼2
(a)
b- At point 2 (t = /2):
𝑖𝐶 = 𝐼𝐶𝑄 = 𝐼𝐶𝑄 + 𝐼𝑜 + 𝐼1 𝑐𝑜𝑠
𝐼𝐶𝑄 = 𝐼𝐶𝑄 + 𝐼𝑜 − 𝐼2
𝜋
2𝜋
+ 𝐼2 𝑐𝑜𝑠
2
2
(b)
c- At point 3 (t = ):
𝑖𝐶 = 𝐼𝐶𝑚𝑖𝑛 = 𝐼𝐶𝑄 + 𝐼𝑜 + 𝐼1 cos 𝜋 + 𝐼2 cos 2𝜋
𝐼𝐶𝑚𝑖𝑛 = 𝐼𝐶𝑄 + 𝐼𝑜 − 𝐼1 + 𝐼2
(c)
Adding (a), (c), substituting Io = I2 from (b), we get
𝐼𝐶𝑚𝑎𝑥 + 𝐼𝐶𝑚𝑖𝑛 − 2𝐼𝐶𝑄
𝐼𝑜 = 𝐼2 =
4
Subtracting (c) from (a), we get
𝐼1 =
𝐼𝐶𝑚𝑎𝑥 − 𝐼𝐶𝑚𝑖𝑛
2
Using the definition of the second harmonic distortion as
𝐼2
𝐷2 = | | × 100%
𝐼1
1
(𝐼
)−𝐼𝐶𝑄
+𝐼
2 𝐶𝑚𝑎𝑥 𝐶𝑚𝑖𝑛
𝐷2 = |
𝐼𝐶𝑚𝑎𝑥 −𝐼𝐶𝑚𝑖𝑛
| × 100%
(33)
In a similar manner, D2 can be expressed in terms of measured
collector-emitter voltages:
1
(𝑉
+𝑉𝐶𝐸𝑚𝑖𝑛 )−𝑉𝐶𝐸𝑄
2 𝐶𝐸𝑚𝑎𝑥
𝐷2 = |
𝑉𝐶𝐸𝑚𝑎𝑥 −𝑉𝐶𝐸𝑚𝑖𝑛
| × 100%
(34)
Power of a signal Having Distortion the o/p power delivered to the
load resistor RC due to the fundamental component of the distorted
signal is
𝑃1 =
𝐼12 𝑅𝐶
(35)
2
The total power due to all the harmonic components of the distorted
signal can be calculated using
𝑃 = (𝐼12 + 𝐼22 + 𝐼32 + … . )
𝑅𝐶
(36)
2
Taking 𝐼12 as a common factor, the total power can be expressed in
terms of the total harmonic distortion "THD",
𝑃 = (1 +
𝐷22
+ 𝐷32
+
= (1 + 𝑇𝐻𝐷 2 )𝑃1
… . )𝐼12
𝑅𝐶
2
(37)
Example 13
Calculate the second harmonic distortion if an o/p waveform displayed
on an oscilloscope provides the following measurements:
a) VCEmin = 1 V, VCEmax = 22 V, VCEQ = 12 V.
b) VCEmin = 4 V, VCEmax = 20 V, VCEQ = 12 V.
Solution
Using Eq. (34), check that
a) D2 = 2.38 %
b) D2 = 0 %
Example 14
For a harmonic distortion reading of D2 = 0.1, D3 = 0.02, and D4 = 0.01,
with I1 = 4 mA and RC = 8 , calculate the total harmonic distortion
"THD", fundamental power component, and total power.
Solution
Using Eq. (31), check that THD  0.1
Using Eq. (35), check that 𝑃1 =
𝐼12 𝑅𝐶
2
= 64 W
Using Eq. (37), check that P = 64.64 W