Class A amplifier analysis and efficiency calculation

INTERNATIONAL SCHOOL OF TECHNOLOGY &
SCIENCES (FOR WOMEN)
DEPARTMENT OF ELECTRONICS &
COMMUNICATION ENGINEERING
Prepared By:
Mr. K V VENKATARAMANA,
Assistant Professor.
K V VENKATARAMANA‖ ECA ‖ Dept of ECE ‖ ISTS
II BTECH II SEMESTER
ELECTRONIC CIRCUIT ANALYSIS
UNIT IV

Contents:
Classification of Amplifier
Class A Power amplifier and itsAnalysis
Example
Harmonic Distortions
Class B push-pull amplifiers and itsAnalysis
Example
Complementary Symmetry push-pull Amplifier
Class AB PowerAmplifier
Class C PowerAmplifier
Thermal Stability and Heatsinks
Distortion in Amplifiers

Introduction
Power amplifiers are used to deliver a relatively high
amount ofpower, usually to a low resistance load.
Typical load values range from 300W (for
transmission antennas) to 8W (for audio speaker).
Although these load values do not cover every
possibility, they doillustrate the fact that power
amplifiers usually drive low- resistance loads.
Typical output power rating of a power amplifier will
be 1W orhigher.
Ideal power amplifier will deliver 100% of the
power it draws from the supply to load. In practice,
this can never occur.
 The reason for this is the fact that the components in
the amplifier will all dissipate some of the power that
is being drawn form the supply.K V VENKATARAMANA‖ ECA ‖ Dept of ECE ‖ ISTS

Concept of Power Amplifier
Provide sufficient power to an output load to
drive other power device.
 To deliver a large current to a small loadresistance
e.g. audio speaker;
 To deliver a large voltage to alarge load resistance
e.g. switching power supply;
 To provide a low output resistance inorder to avoid
loss of gain and to maintain linearity (to minimize
harmonic distortion)
 To deliver power to the loadefficiently

6 K V VENKATARAMANA‖ ECA ‖ Dept of ECE ‖ ISTS

Power Amplifier Power Dissipation
P1 = I 2R
1 1
P2 = I 2R
2 2
ICQ
RC
RE
R1
R2
VCC
I1
I2
ICC
P = I2
R
C CQ C
PT = I2
R
TQ T
P = I2
R
E EQ E
IEQ
The total amount of power
being dissipated by the
amplifier, Ptot , is
Ptot = P1 + P2 + PC + PT + P E
The difference between this
total value and the total
power being drawn from the
supply is the power that
actually goes to the load –
i.e. output power.

Power Amplifier Efficiency 
 A figure of merit for the power amplifier is its efficiency,
 .
Efficiency (   of an amplifier is defined as the
ratio of ac output power (power delivered to load) to dc
input power . B y
formula:
 A s we will see, certain amplifier configurations have
much higher efficiency ratings than others.
This is primary consideration when deciding which
type of power amplifier to use for a specific application.
dc input power P(dc)

ac output
i
o

power
 100% 
P (ac)
100%

“AUDIO AMPLIFIERS” – Unit-5
“RADIO FREQUENCY AMPLIFIERS” –
Unit-6
 D C POWER AMPLIFIERS
9
Power Amplifiers Classification
(Depending on Type of output load)

Class-A Power Amplifier: Class-B Power Amplifier:
Class-C Power Amplifier:

Class A - The transistor
conducts during the whole
cycle of sinusoidal input
signal
Class B - The transistor
conducts during one-half cycle
of input signal
Class AB - The transistor
conducts for slightly more
than half a cycle of input
signal
Class C - The transistor
conducts for less than half a
cycle of input signal

Efficiency Ratings
 The maximum theoretical efficiency ratings of
class-A, B, and C amplifiers are:
Amplifier
ClassA
Class B
Class C
Maximum Theoretical
Efficiency, max
25%
78.5%
99%

Steps for Analysis of Power
Amplifiers:Draw the Circuit Diagram of relevant type
of Power Amplifer
Draw the Graphical operating point
representation
Make DC analysis (AC components as 0) &
Calculate ‘Pdc’
Make AC analysis (DC components as 0) &
Calculate ‘Pac’
Measure the Percentage Efficiency of the
amplifier (% n)
Show the Power Dissipation
Advantages and Disadvantages14 K V VENKATARAMANA‖ ECA ‖ Dept of ECE ‖ ISTS

Class A Amplifier
o u tp ut waveform  same shape  input waveform + 
phase shift.
 The collector current is nonzero 100% of thetime.
 inefficient, since even with zero input signal, ICQ is
nonzero (i.e. transistor dissipates power in
the rest, or quiescent, condition)
vin vout
Av

Class A Amplifier Classification
Series FED/Direct Coupled Transformer Coupled
RC
RE
R1
R2
RL
v in
ICQ
I1
+VCC
ICC
i C

 Series FED/Direct Coupled Class A Power
Amplifier Analysis
Common-emitter (voltage-divider) configuration (RC-
coupled amplifier)
RC
RE
R1
R2
RL
vin
ICQ
I1
+VCC
ICC

 Typical Characteristic Curves for Class A
Operation
Configuration : No inductor @ transformer are used
(a) Common-emitter amplifier,
(b) dc load line (the Q point is at centre of the load line)
(c)instantaneous power dissipation versus time in the
transistor

 DC Input Power
RC
+VCC
RE
R1
R2
RL
vin
ICQ
I1
ICC
The total dc power, Pi(dc) , that an
amplifier draws from the power supply :
CC CCi
P(dc) V I
I  I  ICC CQ 1
I  ICC CQ
(I  I )CQ 1
i CC CQ
P(dc)  V I
Note : That this equation is valid for most amplifier power analyses. We can
rewrite for the above equation for the ideal amplifier as
Pi(dc) VCC ICQ

 AC Output Power
R1//R2
vcevin
vo
ic
RC//RLrC
Po(ac) ic(rms)vo(rms)
2
Vm
Vo(rms)  2
Im
Io(rms) 
V m

V p p
2 2
I m

I p p
2
Vpp
V maxV min
 2
I pp

Im ax  I m in
22 2
o
P (ac) 
Vm
*
Im

Vm *Im
2 8
o
Vpp
*
Ipp
P (ac)    2 2 
Vpp * Ipp
8
o
P(ac) 
(VmaxVmin)*(Im axImin)

Efficiency
:
o
P(ac) 
8
Pi(dc) VCC ICQ
%
Pac
*100
Pdc
8*Vcc*Icq
%
2*Vcc* Icq
*100
Imax=2Icq & Imin= 0
8*Vcc*Icq
20
Vmax=VCC & Vmin= 0
%
(Vmax Vmin)(Im ax Im in)
*100
%25%

 Advantages:
1) Circuit is Simple to Design
2) Less no. of Components
 Dis-Advantages:
1) Load Resistor is directly connected to collector. This causes
considerable wastage of Power
2) Power Dissipation is More
3) Efficiency is very poor, due to large power Dissipation
21

Transformer Coupled Amplifier
•
•
•
•
• The theoretical maximum efficiency of a basic
RC-coupled class-A amplifier is limited to 25%.
In practical circuit, the efficiency is less
than 25%. Used for output power of
about 1 W only.
Transformer coupling can increase the
maximum efficiency to 50%
Disadvantage of transformer coupling –
expensive & bulky.K V VENKATARAMANA‖ ECA ‖ Dept of ECE ‖ ISTS

Neglecting transformer resistance and assuming RE is small;
VCEQ VCC

 DC Input Power
The total dc power, Pi(dc) , that an
amplifier draws from the power supply :
CC CCi
P(dc) V I
I  I (I  I )
CC CQ CQ 1
CC CQi
P(dc)  V I
Note : That this equation is valid for most amplifier power analyses. We can
rewrite for the above equation for the ideal amplifier as
Pi(dc) VCC ICQ
i C

 AC Output Power
2
V pp

V maxV min
i C
Po(ac) i1(rms)v1(rms) Po(ac) i2(rms)v2(rms)
At Primary side A.C power generated:
L
L
R'
V2
1rmsR' 1rmsPo(ac)i1(rms)v1(rms)I2
22 2
1111 Im* mVIm mV
o * P(ac) 
2
V p p
V m 
2
Ip p
I m 
2
I pp

Im ax  I m in

2 8
o
8
o
P(ac) 
 We consider an idea Transformer, means Power delivered on primary side isequal
to the secondary side load, and the values of currents and voltages are similar.
Vpp
*
Ipp
P (ac)    2 2 
Vpp * Ipp
Pi ( d c )  V C C I C Q

Efficiency
:
o
P(ac) 
8
Pi(dc) VCC ICQ
%
Pac
*100
Pdc
8*Vcc*Icq
%
4*Vcc* Icq
*100
Vmax=2VCC & Vmin= 0
Imax=2Icq & Imin= 0
8*Vcc*Icq
%
(Vmax Vmin)(Im ax Im in)
*100
%50%
28

Power
Dissipation:
29
pdpDCpAC
 In case of Class A amplifiers, Power Dissipation is maximum when no input
signal is applied.

31
 Advantages:
1) Efficiency be somewhat high compared to series fed Class A
2)Impedance matching is required for maximum power transfer is
possible
 Dis-Advantages:
1) Due to Transformer, the circuit becomes bulkier, heavier and
costlier compared to series fed type
2) Frequency response of the circuit is poor
3) Circuit is complicated to design and implementation compared to
series fed type

Class A
advantages:•No cross over distortion
•No switching distortion
•Lower harmonic distortion in the voltage amplifier
•Lower harmonic distortion in the current amplifier
•No signal dependent distortion from the power supply
•Constant and low output impedance
•Simpler design
Class A dis-advantages:
• Low Efficiency
• Power Dissipation more
• Wastage of Power

Example
Calculate the input power [Pi(dc)], output power
[Po(ac)], and efficiency [] of the amplifier circuit for
an input voltage that results in a base current of
10mA peak.
+VCC = 20V
IC
Vi
  25
RC
20
RB
1k
Vo
2 2
100%  6.5%

C
C
c(sat)
Pi(dc)
Pi(dc)  VCC ICQ  (20V )(0.48A) 9.6W
 (20)  0.625W
250103
A2
C ( peak )
R
I2
Po(ac) 

20V
 1000mA  1A
R 20

VCC
I
19.3mA
20V  0.7V
1kRB
VCC VBE
IBQ 
 
Po(ac)
VCE (cutoff )  VCC  20V
IC( peak )  Ib( peak )  25(10mA peak)  250mA peak
ICQ  IB  25(19.3mA)  482.5mA  0.48A
VCEQ  VCC  ICRC  20V  (0.48A)(20)  10.4V

Class B Power
Amplifier Consists of Push-Pull & complementary pair electronic devices
 One conducts for one half cycle of the input signal and the other
conducts for another half of the input signal
 When the input is zero, both devices are off, the bias currents are
zero and the output is zero.
 Ideal voltage gain is unity

 For input larger than zero, A turn ON and supplies current to the
load.
 For input less than zero, B turn ON and sinks current from the
load

Class-B transformer coupled push
pull amplifier
36
Driver Transformer
Load Transformer

 DC Input Power
CC dc CC m

P(dc) V I 
2
V I
•DC biasing point is on the X-axis
so the Vceq=VCC. And Iceq
becomes 0. hence Q-point (Vcc,0).
• Each Transformer output is an Half-rectified
Waveform. Then peak value of the output current is the average d.c current
I1 (dc) Im
I 2 (dc) Im
 
  
Im
 2Im
I (dc)  Im

 AC Output Power
Po(ac) i1(rms)v1(rms) Po(ac) i2(rms)v2(rms)
At Primary side A.C power generated:
L
L
R'
V2
1rmsR' 1rmsPo(ac)i1(rms)v1(rms)I2
R
VI R
L
m L
o '
2
m
2 '
222
Vm Im Vm*Im
2 2
* P(ac) 

Efficiency
:
%
Pac
*100
Pdc
• Vm = VCC
4 VCC
%

* Vm *100
%78.5%
2
oP(ac) 
Vm *Im
CC m

P(dc) 
2
V I
4
38
VCC
%

* Vcc*100

Power
Dissipation:pdpDCpAC
pd 2
VccImVm*Im
 2
 In case of Class A amplifiers, Power Dissipation is maximum when no input
signal is applied. But in Case of Class B, when input is 0 then Vm=0. Hence
Power dissipation is very less compared to Class A.
39

41
 Advantages:
1) Efficiency is much higher than Class A operation.
2) When there is no input signal, the power dissipation is zero.
3) Harmonics get cancelled. This reduces the harmonic distortion.
4) Ripples present in supply voltage also get eliminated.
5) Due to the Transformer impedence Matching is possible.
Dis-Advantages:
1) Frequency response of the circuit is poor
2) Center Tap transformers are necessary.
3) Transformers make the circuit bulky and hence costlier

Complimentary Symmetry Class B Circuit

43
Conducts
Non-Conduct
Non-Conducts
Conducts

44
 Advantages:
1) As the circuit is transformer less, its weight, size and cost are less.
2)Due to the common collector configuration, impedance matching
is possible.
3)Frequency response of the circuit is improves due to transformer
less Class B amplifier circuit
Dis-Advantages:
1) Circuit needs two separate voltage supplies
2) Output is distorted to cross-over distortion

DEAD BAND
CROSSOVER DISTORTION

 Cross-over distortion free
circuits• Push-Pull Distortion Less circuit:
45

• Complementary Push-Pull Distortion Less circuit:
46

Class AB Power Amplifier

Class AB Voltage Transfer Curve

Class C Power Amplifier

Load line
Operation:
51

Collector Currents & Output Current
 iCpiCn  iL

Comparison of Amplifiers
Class A B C AB
Operating 360˚ 180˚ Less that 180˚ 180˚ to
360˚ Cycle
Position of Q- Center of Load On X-axis Below X-axis In between
X-
Point Line axis to
center
of Load line
Efficiency Poor, Better, High In
between 25% to 50% 78.5% 78.5% to
50%
Distortion No Present Highest Present

Amplifier Distortion :
Amplifier Distortion can take on many forms
such as Amplitude, Frequency and Phase
Distortion due to Clipping
Distortion in Amplifiers

Amplitude Distortion
:
Frequency Distortion
:

Phase Distortion
:

Thank you

Class A amplifier analysis and efficiency calculation

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