An amplifier receives a signal from some pick up 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 mill volts from a cassette or CD input, or a few micro–volts 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. 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
amplification primarily to increase the voltage of the input signal. Large–signal or power amplifiers,
on the other hand, primarily provide sufficient power to an output load to drive a speaker or other
power device, typically a few watts to tens of watts. 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.

Classification of Output Stages

An amplifier can be classified
a) According to frequency range
• DC amplifier (0 freqency)
• Audio amplifier (20 Hz to 20 KHz)
• Video amplifier (up to few MHz)
• Radio frequency amplifier (a few KHz to l00’s of MHz)
• Ultra high frequency amplifier (l00’s or 1000 of MHz)
b) According to use
• Current amplifier
• Voltage amplifier
• Power amplifier
c) According to the type of load
• Untuned amplifier
• Tuned amplifier
d) According to the method of operation
Class A: Transistor is so biased that output (or collector) current flows for full 360º of the ac
cycle. This operation 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, or too low to approach the lower supply level, or 0V as shown
in below figure.
 V0

Power supply
level
Full 360º output swing
Class A dc
bias level

OV t

 Class A operation means that transistor operates in active (or linear) region of its load
line. So output waveform is exactly similar to the input waveform. (High fidelity)
 It has a maximum efficiency of 25% (without transformer) and 50% (with transformer)
and minimum efficiency of 0%.
 C > B > A (denotes efficiency)
• Class B: A class B circuits provides an output signal varying over one-half the input signal
cycle, or for 180º of signal as shown in figure. The dc bias point for class B is therefore at 0V,
with the output then varying from this bias point for a half cycle. 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 360º of operation. This type of connection is referred to as push-
pull operation. The class B operation by itself creates a much distorted output signal since
reproduction of the input takes place for only 180º of the output signal swing.

V0

180º output swing

Class B dc
bias level
OV

• Class AB: 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, but the dc
bias level is usually closer to the zero-base-current level for better power efficiency. For class
AB operation, the output signal swing occurs between 180º and 360º and is neither class A
nor class B operation.
• Class C: The output of a class C amplifier is biased for operation at less than 180º of the cycle
and will operation 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.
• Class D: 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
make 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.

Amplifier efficiency:
The power amplifier efficiency of an amplifier defined as the ratio of output power to input
power, improves (gets higher) going from class A to class D. In general terms we see that 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, occurring for the largest output voltage and current swing, is only
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 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%.

Comparisons of Amplifier Classes

Class A AB B C D
Operating cycle 360º 180ºto 360º 180º less than 180º pulse operation

Power efficiency 25% to 50% 25% (50%) and 78.5% 78.5% Greater
than B over 90%

Class A Output Stage

Series– Fed class A Amplifier

 Vcc

Ic Load
RB Rc

IB
C1

Power Transistor

Vi

Fig.4.2(a) Series – Fed Class A Amplifier

• It is called series fed as load RC is connected in series with transistor output.

• The difference between series fed (large signal) and small signal version is that signals
handled by large signal circuit are in the range of volts and transistor used is a power
transistor that is capable of operating in the range of few to tens of watts.
• Series fed circuit is not the best to use as large signal amplifier because of its poor power
efficiency.
•  is less than 100 (high current or power gain but low 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 = IB
With the collector– emitter voltage then
VCE = VCC – ICRC
Ic
VCC
Rc
dc load line
Q-point
ICQ IBQ

VCE
0 VCEQ = VCC VCC
2
Fig: 4.2(b) Transistor characteristics showing load line and Q point
• First DC load line is drawn using valve of VCC and RC since they are constant. (VCC & RC are
constant).
• The quiescent point is calculated using above equations.
• If the dc bias collector current is set at one-half the possible signal swing (between 0 and
VCC/RC) the largest collector current swing will be possible. Additionally, if the quiescent
collector is set at one half the supply voltages (0 and V CC) the largest voltage swing will be
possible.
• With the Q point set at this optimum bias point, the power considerations for the given circuit
(series fed circuit) are determined.
• AC operation:
When an input ac signal is applied to the amplifier for the given circuit, the output will vary
from its dc bias operating voltage and current. 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 0V or the supply voltage VCC.
Ic
Input signal
Vcc
Rc

Output
current swing
O VCE
Vcc

Output voltage swing

Fig: 4.2(c) Amplifier input and output signal variation

Maximum efficiency for Ideal case:
V02
Output power (minimum), Po(ac) = IoVo = R
C

But,
Vopp VCC Vopp
Vo = = [Vrms = ]
2 2 2 2 2 2
[VOPP means peak to peak output voltage= VCC for ideal case]

(VCC / 2 2)2 VCC2

 Po(ac) = RC = 8R .......... (i)
C

Input power from collector supply VCC,

Pi(dc) = VCC Idc(avg)
= VCC. ICQ
 VCEQ
= VCC. R
C

VCC VCC
= VCC. 2R [ VCEQ = 2 ]
C

VCC2
or, Pi(dc) = 2R ....... (ii)
C

Po(ac)
 Maximum efficiency () = P (dc) × 100%
i

VCC2 2RC
= 8R V 2 × 100%
C CC

= 25%
Therefore, max = 25% for Series Fed Class A amplifier.
General efficiency of class A amplifier: (Series fed class A amplifier):
VCC
• DC biasing is set at VCEQ = 2 and ICQ is chosen such that it doesn't exceed the maximum
collector dissipation for the transistor Q1.
VCE or V0

VCC

VCEQ = Vcc
2
VCE max

VCE min
t
0

Fig.4.2(d) Graphical analysis of Class A amplifier with resistive load

Here, from figure

VCE max + VCE min VCC
VCEQ = 2 = 2 ...... (i)

The power supplied by dc source

VCEQ
Pi(dc) = VCC. R
C
 VCC
= VCC. 2R
C

(VCE max + VCE min)2

= 2RC .......... (ii)

Similarly, output power obtained from amplifier as,

VCE max – VCE min2 1 V2rms
Po(ac) =  . RC [Po(ac) = R ]
 2 2  C

(VCE max – VCE min)2

or, Po(ac) = 8RC .......... (iii)

Thus, efficiency of amplifier

Po(ac)
gen = P (dc) × 100%
i

(VCEmax – VCE min)2 100 %

= (V +V )2 × 4
CEmax CEmin

 VCEmax – VCEmin 2
 gen =   × 25%
VCEmax + VCEmin
[Note: From above equation we can see that most series–fed circuits will provide efficiencies
of much less than 25%.]
Transformer coupled class A amplifier: or // single ended power amplifier (Ideal case):
• It has better efficiency of 50% because no power is wasted in collector resistor RC as is series
fed.
• A matching transformer (step-down) is provided to couple the high impedance collector
circuit to low impedance load.
+Vcc

N2 RL
R N1

C1

Vi
 Fig.4.2(e) Transformer Coupled Class A amplifier

The transformer (dc) winding resistance determines the dc load line for the above circuit.
Typically, this dc resistor is small (ideally 0). Thus, a 0 dc load line is a straight vertical line.
Ic

dc load line
Ic max

IOP
Q-point

ac load line
V
O 2Vcc CE
VCEQ = Vcc
VOP

VOP (P-P) = 2Vcc

Fig. 4.2(f)

Here,
Input power,
Pi(dc) = VCC. ICQ
VCC VCC2
= VCC. R = R [From figure 4.2(f)]
C C

V o2
Po(ac) = R
C

Vopp
But, Vo =
2 2
2 VCC
= [From figure 4.2(f)]
2 2
VCC
=
2
VCC2
 Po(ac) = 2R
C

Po(ac)
 Efficiency () = P (dc) × 100%
i

VCC2 RC
= 2R × V 2 × 100 %
C CC

VCC 2
= 50 V 
 
CC

max = 50%
 Hence the efficiency of transformer coupled amplifier is 50% for ideal case i.e. when Vopp = 2VCC

General efficiency of Transformer Coupled Class A amplifier: add figure

We know, for transformer coupled class A amplifier
VCE (max) + VCE (min)
VCEQ = VCC = 2
Then,
VCEQ
Pi(dc) = VCC. R
C

VCC VCC2
= VCC. R = R
C C

[VCE (max) + VCE (min)]2

or, Pi(dc) = 4RC
Also,
VCE (max) – VCE (min)2 1
Po(ac) =   . RC
 2 2 
(VCE max – VCE min)2
= 8RC
General Efficiency
Po(ac)
 gen = P (dc) × 100%
i

 VCEmax – VCEmin2
 gen = 50   %
VCEmax + VCEmin

Q. Calculate the efficiency of a transformer coupled class A amplifier for a supply of 12VDC and
output of (i) VP = 12V and (ii) VP = 6V. [065 Chaitra Regular]
Solution
We have,
For class A transformer coupled amplifier:
VCE max – VCE min 2
gen = 50 × V %
 +V
CE max CE min 
VPP 2
= 50 × 2V  %
 CEQ 
VPP 2 2VP 2 VP 2
= 50 × 2V  %= 50 × 2V  % = 50 × V  %
 CC   
CC  
CC

(i) When VP = 12 V
12 2
 = 50 ×12 = 50%
 
 (ii) When VP = 6V
6 2 1
 = 50 × 12 = 50 × 4 = 12.5 %
 

Class B Output Stage:

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 i.e. IC flows for only 180º of ac cycle. Operating point
is set at cut-off. To obtain output for the full cycle of signal, it is necessary to use two transistors
and have each conduct on opposite half-cycles, the combined operation providing a full cycle of
output signal. Since one part of the circuit pushed the signal high during one half-cycle and the
other parts pulls the signal low during the other half cycle, the circuit is referred to as a push-pull
circuit, with each half operating on alternate half-cycles, the load then receiving a signal for the
full ac cycle. The power transistor used in the push-pull circuit are capable of delivering the
desired power to load, and the class B operation of these transistors provides greater efficiency
then was possible using a single transistor.

One - half circuit

Load

One - half circuit

Fig. 4.3(a) Block diagram of Class B

• Class B amplifier has maximum efficiency of 78.5% and minimum of 50%

Advantage:

The main advantage of class B/AB amplifier over the class A amplifier is that there is very little
current in the transistor when there is no input signal. This results in low power dissipation when
there is no signal.

Efficiency of Class B amplifier:

Let Vop be the peak output voltage produced by class B amplifier. Let RL (or rp) be the equivalent
load resistance in the output circuit.

Then,
 V2r m s (VOP/ 2)2 VOP2
PO(ac) = rp = rp = 2r ....... (i)
p

Let VCC be the dc source voltage and the equivalent dc current supplied by the dc source is the
average dc value of the ac output current which may be expressed as,
2 2 VOP 2 VOP
Idc = I = = r
 OP  rp  p
Thus, dc power supplied by source is,
2VOP
Pi (dc) = VCC. Idc = VCC. ......(ii)
rp
Po (ac)
 Efficiency () = Pi (dc) × 100%

VOP2 rp
= × 100%
rp 2VOP. VCC
VOP
 = . 25  %  general formula
VCC

When VOP  VCC

max = 25  % = 78.5%

Condition for minimum efficiency:

Power loss by class-B amplifier,
PLoss = Pi(dc) – Po(ac)
2VCC. VOP VOP2
or, PLoss = – 2 r ............ (iii)
 rp p

Differentiating (iii) with respect to output voltage VOP,

dP Loss 2VCC VOP
dVOP = rp – rp
For maximum loss,
dP Loss 2VCC VOP
= 0 = – r
dVOP rp p

VOP 2VCC
or, rp = rp
2
or, VOP = V ....... (iv)
 CC
or VOP = 0.6366VCC

Thus, when output voltage is 63.66% of maximum possible output voltage the efficiency of class
B amplifier becomes minimum which may be expressed as,
 VOP
min = V 25% [From general formula]
CC

0.6366VCC
or, min = VCC 25% [From (iv)]

min = 50%

Class B amplifier circuits:

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

+VCC
180º out of phase
RC
R1

Vin C1
R2
C2
} Push-pull
i/p signals

RE C3 Same phase

Fig.4.3(b) Phase splitter using BJT

• Same output of input through emitter & 180º out of (opposite output of input) phase through
collector.
Inverting

Ampl
V1 ifier Out of phase
EF V1

EF V2
same phase
Transformer- Coupled Push-Pull Stages

Transformer coupled class B push pull amplifier:

VCC

ICQ1
Q1 IL

Vin
R1

R2

Q2
RL
ICQ2

Fig. 4.7(a) Push Pull Circuit

• When Vin = 0, V1 = V2 = 0 then ICQ1  ICQ2 = 0

• During first half cycle, transistor Q1 conducts whereas Q2 is at cut-off mode. So ICQ1 current
flows through transformer which results in first half cycle of signal to the load.
• During next half cycle, Q2 conducts whereas Q1 is at cutoff mode. So ICQ2 current flows
through transformer resulting in second half cycle.
• The output transformer will combine the two outputs of Q1 & Q2.
* Cross-over distortion:

V0

0.5V
Q1
t
Q2

cross over
distortion
Fig.: Cross over distortion
In the result of small currents for V < VThreshold the output is non-linear and is much smaller that
it would be if response were linear. This effect is called cross-over distortion. When unbiased, a
class B push pull amplifier has no output until the input voltage exceeds approx 0.7V or 0.5V.
This results in clipping between positive and negative half cycles. This effect is called crossover
distortion.

Elimination:
We need to apply slight forward bias to each diode of transistor i.e. the Q-point should be slightly
above cut-off to avoid this distortion. In order to minimize crossover distortion, transistor must
operate in a class AB mode where a small stand by current flows at zero excitation which is
described in section 4.4.

Efficiency:
IC1(mA)

IO P
Q point
0 VCE1 VCE 2

VCC

VCC

IC2(mA)
Fig. 4.7(b) Composite Characteristics for Class B push pull amplifier

Here,
PO(ac) = IoVo

Vo2 (VCC / 2)2 VCC2

= Z = ZL = 2Z .......... (i)
L L

= maximum possible output power

Pi(dc) = VCCIdc
 2 2 VCC 2 VCC2
= VCC. IOP = VCC Z = Z ........ (ii)
  L  L
 Efficiency,
PO(ac)
 = P (dc) × 100%
i

VCC2  ZL
= 2Z . 2 . V 2 × 100% =  25%
L CC

Therefore,
= 78.5%
= maximum efficiency
Transformer less complementary symmetry push-pull amplifier using complementary
matched transistor Q1 and Q2. (Class B).

+V CC

Q1(npn)
emitter follower
Vin

Q2(pnp) RL

–VCC

Fig. 4.7(c) Complementary Symmetry push pull amplifier

VRL

crossover
distortion
(both Q1 and Q2 off)

• A single input signal is applied to base of both transistors

• During first half cycle Q1 (npn) transistor will be forward biased as it operates in active mode
while Q2 is at cut-off. So current IC flows from transistor Q1 to load to ground.
• During negative half cycle, Q2 (pnp) will be biased into conduction and Q1 at cut-off. So,
current flows from ground to –VCC.

Class AB Amplifier
An amplifier may be biased at a dc level above the zero base current level of class B amplifiers
and above one half the supply voltage level of class a amplifier. This mode of operation is class
AB mode of operation of amplifier, where a small standby current flows at zero excitation
resulting in less distortion than class B.

Class AB operation still needs a push-pull connection to achieve a full output cycle but the dc bias
level is usually closer to zero base current level for better power efficiency. For class AB operation
the output signal swing occurs between 180º and 360º and is neither class A nor class B operation.

+VCC

QN
VBB
2

vi v0
V BB
2
RL

QP

–VCC

Fig.4.4 (a) Class AB operation

v0

slope = 1

vi
0

Fig. 4.4(b) Transfer Characteristics of Class AB stage

Here a bias voltage VBB is applied between the bases of QN and QP, giving rise to bias current IQ.
(VBB/2VT)
IN = IP = IQ = ISe

For small Vin, both transistors conducts and crossover distortion is almost completely eliminated.
The value of VBB is selected to yield the required quiescent current IQ.
Operation
When Vin goes positive by certain amount the base voltage of QN increases by same amount Vin.
This increases IN according to the corresponding increase in VBEN. However since the voltage
between two base is constant i.e. VBB, the increase in VBEN will result in an equal decrease in VBEP
and hence in IP. So QP will be conducting a current that decreases as Vo increases; for V0 the
current in QP can be ignored. For negative input voltage Vin the opposite occurs. So far small Vin
both transistors conduct as Vin increased or decreased one of the two transistors take over the
operation.

Tuned Amplifier:
The amplifier used for amplifying a signal of specific frequency or narrow band of frequency is
known as tuned amplifier. Tuned amplifiers amplify selective frequency only using LC network
and hence are useful in receivers. It works only at resonance frequency. Resonant LC circuit as a
load provides high impedance and so tuned amplifiers can provide high gain.
VCC

RB C R L
y=admittance

V0
C1

Vi

Fig.4.8 (a) Simple Class A tuned amplifier