Download Class B Amplifiers

Class-A and Class-B
Amplifiers
Operation and Circuits
Summary : Previous Lecture
• The ac load line tells us what the
maximum output voltage swing will
be for the given amplifier. The
incoming signal will cause a current
swing above and below the Q point.
• The efficiency of an amplifier is the
portion of the power drawn from the
dc power supply that actually
transferred to the load, given as a
percentage.
• Higher efficiency is always better.
High efficiency means that a smaller
percentage of the power drawn from
the supply is used by the amplifier
itself.
Maximum Efficiency in a SeriesFed Class-A Amplifier
• When an ac signal is applied to the amplifier, the output current and voltage will
vary about the operating point Q. In order to achieve the maximum symmetrical
swing of current and voltage (to achieve maximum output power), the Q point
should be located at the centre of the dc load line. In that case, operating point
is IC = VCC/2RC and VCE = VCC/2 .
Class-A: Power Delivered to The Load
Using Peak Output Voltage or Peak Output Current
The average power delivered to the load is;
I RC
VP I P V
PL =
=
=
2
2RC
2
2
p
2
p
Maximum efficiency occurs when
Note:Usually we use Rc
or RL to represent the
load resistance.
Remember! It is just a
variable.
Vcc
Ip =
; Vp = VCC (two dc sources)
RC
Vcc
(single dc source)
; V = VCC
Ip =
P
2RC
2
Class-A: Power Delivered to The Load
Using Peak-to-Peak Output Voltage and
Peak-to-Peak Output Current
The average power delivered to the load is;
PL =
Vpp I pp
8
Maximum efficiency occurs when
Vcc V = V
I pp =
cc (single dc source)
; pp
RC
Class A: Power Supplied by the
Source
The total dc power supplied by one source is;
PS = VCC IC
Since, the Q point is located at the centre of the dc load
VCC
line; hence
Ic =
2Rc
Class-A Amplifier: Exercise 1
• Calculate the (i) output power (ii) input power and (iii)
collector efficiency of the amplifier circuit shown in figure
below. It is given that input voltage results in a base current of
10 mA peak.
Class-A Amplifier: Exercise 1
Solution : First draw the dc load line by locating two end points as shown
in figure below.
Ic(sat)=Vcc/Rc=20 V/20 Ohm= 1 A = 1000 mA
VCE=VCC=20.
The Q point of the circuit can be located as under:
Class-A Amplifier: Exercise 1
Solution
Class-A: Power Dissipated
• The maximum power dissipated by a transistor
P(Q,max) = ICQVCEQ
Class-A Amplifier: Exercise 2
• For the transistor in the common-emitter
circuit in Figure Ex. 2, the parameters are:
β=80, VCC =30V, and IC,max =1.2A.
• (a) Design the values of RL and RB for VCC =
30 V. What is maximum power dissipated
in the transistor?
• (b) Using the value of RL in part (a), find
IC,max and VCC if PQ,max = 5 W.
• (c) Calculate the maximum undistorted ac
power that can be delivered to RL in parts
(a) and (b) for the assumption that iC ≥0
and 0≤vCE ≤VCC.
Figure Ex.2
Class-A Amplifier: Exercise 2
Solution (a)
• Given β=80, VCC =30V, and IC,max =1.2A.
• (a) Design the values of RL and RB for VCC = 30 V.
What is maximum power dissipated in the transistor?
Class-A Amplifier: Exercise 2
Solution (b)
• Given β=80, VCC =30V, and IC,max =1.2A
• (b) Using the value of RL in part (a), find IC,max and VCC if PD,max
= 5 W.
Class-A Amplifier: Exercise 2
Solution (c)
• Given β=80, VCC =30V, and IC,max =1.2A
• (c) Calculate the maximum undistorted ac power that can
be delivered to RL in parts (a) and (b) for the assumption
that iC ≥0 and 0≤vCE ≤VCC.
Class-A Amplifiers:
Disadvantage
• The primary disadvantage of the Class A amplifier is its
low efficiency. We have seen that the majority of the
power drawn from the power supply is used by the
amplifier itself, with only a small percentage being
delivered to the load.
• The Class A amplifier that we have been using, has ICQ set
for approximately the middle of the DC load line with no
input signal.
• This means that the Class A amplifier is using power even
though we have no input signal.
Class-B Amplifiers : Overview
• The Class B amplifier was developed to improve on this low
efficiency problem.
• The maximum theoretical efficiency rating for the Class B
amplifier is approximately 78.5%.
• The Class B amplifier consumes very little power when there is
no input signal. This is because ICQ is close to zero.
• The Class B amplifier requires two transistors, each conducting
for approximately 180O of the incoming waveform.
Class-B Amplifier : Biasing
• When the amplifier is in its
quiescent state, (it has no input
signal) both transistors are biased
at cutoff.
• When the input is positive,
- Q1 is biased above cutoff transistor conducts, producing a
replica of the positive input at the
output.
- Q2 remains in cutoff.
• When the input is negative,
- Q1 is biased in cutoff.
- Q2 is biased above cutoff, and
the transistor conducts, producing a
replica of the negative input at the
output.
Q1
Q2
Class-B Amplifier
Complementary-Symmetry
Q1 ON
Q1
Q2 OFF
Q1 OFF
Q2
The fact that both transistors are never fully on at the same time is
the key to high efficiency rating of this amplifier.
Q2 ON
Class-B : DC Operating Characteristics
• Matched transistors should be used in Class B
amplifiers because any difference in the
operating characteristics of individual
transistors will cause non-linearity and output
distortion.
• The voltage drop across the two transistors
(VCE) would be one half of VCC ---the
transistors are a matched pair.
• The only devices in the circuit are the two
transistors, and their resistance is the same.
Therefore, VCC will split evenly across both
devices.
• This will be true regardless of the “on” state
of the transistor. The transistors will present
the same resistance ratio to the circuit, and so
VCC will always split evenly across them.
Class-B : DC Operating Characteristics
The voltage across the two
transistors would be fairly
constant and the collector current
would be reasonably unrestricted.
This gives us the vertical load line
shown
Class-B : AC Operating Characteristics
Maximum Efficiency (Idealized Power Efficiency)
• To find the ac load line, find iC(sat). The voltage across RL is equal to
VCE of the transistors. This voltage is one half of VCC.
• When either transistor is conducting, its operating point moves up
the ac load line. The operating point of the other transistor remains
at cutoff. The voltage swing of the conducting transistor can go all
the way from cutoff to saturation. Since one half of VCC is across
each transistor, vce(off) can be found as.
peak a.c. output
voltage (Vp)
peak a.c. output
current( Ip)
= v CE(off )
VCE (off ) Vcc
= IC(sat ) =
=
RL
2RL
Class B : AC Operating Characteristics
Maximum Efficiency
Class B : Total DC Power
Since each power source supplies half sinewave of
current, the average value is;
IS =
Vp
p RL
=
Ip
p
The total power supplied by a single source is;
æ Vp ö VCC I p
PS = VCC I S = VCC ç
÷=
p
è p RL ø
Class-B : Dual Power Supply
(Split Supply)
For maximum efficiency:
What is the peak ac output voltage?
What is the peak ac output current?
What is the total DC power supplied?
Class B : Dual (Split) Power
Supply
For maximum efficiency:
peak a.c. output
voltage (Vp)
= VCE (off ) = Vcc
peak a.c. output
current( Ip)
= IC (sat ) =
Vcc
RL
The total power supplied by a dual
power supply is:
æ Vp ö
PS = 2VCC I S = 2VCC ç ÷
è p RL ø
Class B: Power Delivered to The Load
The average power delivered to the load is;
I RL
VP I P V
PL =
=
=
2
2RL
2
2
p
2
p
Maximum efficiency occurs when
Vcc
Ip =
; Vp = VCC (dual power supply)
RL
Vcc
(single power supply)
; V = VCC
Ip =
P
2RL
2
Class B: Power Supplied by
Dual Power Supply
The total power supplied by a single power supply is;
æ Vp ö
PS = VCC I S = VCC ç
÷
è p RL ø
The total power supplied by a dual power
supply is:
æ
ö
Vp
PS = 2VCC I S = 2VCC ç
÷
è p RL ø
Class B: Power Dissipated
• The power dissipated (as heat) by the transistors in class B
amplifier is the difference between the input power delivered
by DC source(s) and the output power delivered to the load.
P2T = Ps - PL
• Where P2T is the power dissipated by the two transistors.
Hence, the power dissipated by each transistor is
P2T Ps - PL
PT =
=
2
2
Class B Amplifier: Exercise 3
• For a class B amplifier using a supply of VCC = 12V and driving a
load of 8Ω, determine
(i) maximum load power
(ii) DC. input power
(iii) efficiency.
Class B Amplifier: Exercise 3
Solution
Maximum efficiency occurs when
Ip =
Vcc
2RL
PL(max) =
=
VP =
VCC
2
Vp I p Vcc Vcc
=
´
2
4 2RL
12 12
´
= 2.25W
4 2(8)
æ Vp ö Vcc Vcc
PS = VCC I S = VCC ç
´
÷=
è p RL ø p 2RL
=
12
p
´
12
= 2.87W
2(8)
(single power supply)
h=
=
PL
´100
PS
2.25
´100 = 78.4%
2.87
Class B: Crossover Distortion
Crossover distortion is an inherent problem
with the Class B amplifier, caused by the
biasing arrangement.
Note the flat spot in the waveform as it
“crosses over” the zero point.
During this short period, both transistors are
off and the output is zero volts.
The crossover from one transistor to the
other is not instantaneous. The “on
transistor” turns off before the “off
transistor” turns on.
This can be eliminated by biasing the
transistors slightly above cutoff.
Class B: Crossover Distortion
• The diagram shown is a magnified
view of the cutoff region. We bias the
transistor at soft cutoff to avoid
crossover distortion. At soft cutoff,
there is still a small amount of
collector current flowing
• If we bias the transistor at hard cutoff,
we will eliminate most of the collector
current but we introduce crossover
distortion.
• This is because it takes time for the
transistor to come out of hard cutoff
and begin to conduct. Biasing at soft
cutoff reduces this transition time and
thus reduces crossover distortion.
Class AB: Eliminating Crossover
Distortion
• Both transistors are conducting when
the signal level is at zero volts; hence,
the amplifier does not have the
crossover distortion problems inherent
with the Class B amplifier.
• Crossover distortion occurs only when
both transistors are in cutoff.
• That situation does not normally occur
with the Class AB Amplifier.
• In Class AB operation, the transistors
conduct for a portion of the input cycle
that is greater than 180O but less than
360O .
• It can be seen that both transistors will
be conducting at the same time for a
small portion of the wave.