its power amplifier on edc slides

1. Power Amplifier

2. Series Fed Class A Amplifier
• The β of a power transistor is
generally less than 100, the
overall amplifier circuit using
power transistor that are
capable of handling power or
current while not providing
much voltage gain.
• DC bias operation:
• VCC = IBRB +VBE
• IC = βIB
• VCC = VCE +ICRC

3. Series Fed Class A Amplifier
• The largest collector
swing will be possible if
Q point is set at mid of
the dc load line

4. General efficiency of Class A amplifier
• Pin (dc) = ICQVCC =
Vcc
2Rc
VCC
• (when Q-point is in centre of dc
load line)
• Pin (dc) =
V2
cc
2Rc
=
VCEmax+VCEmin 2
2Rc
Po(ac) =
V2
o(p−p)
2 2 2
Rc
=
VCE
2
(p−p)
2 2 2
Rc
=
VCEmax
−VCEmin
2
8Rc

5. General efficiency of Class A amplifier
• The general efficiency of Class A
Amplifier is
• % η =
Po(ac)
Pin(dc)
100%
= 25
VCEmax
−VCEmin
VCEmax
+VCEmin
2
%
For ideal case
VCEmax =VCC
VCEmin = 0
Then
% ηmax = 25 %
Then
% ηmax = 25 %
• Practically efficiency is 10 to 20
%.

6. Transformer coupled Class A amplifier
• Here R’L =(
𝑁1
𝑁2
)2RL = α2 RL
• α is called turn ratio of transformer
• Since VCE can’t be negative the
maximum permissible decrease in
VCE below its Q point value is VCEQ =
VCC.
• Thus the maximum possible peak
value of VCE form Q-point is VCC
volt.
• To achieve maximum peak to peak
o/p variation, the intercept of the
ac load line on the VCE axis should
therefore 2 VCC volts

7. Transformer coupled Class A amplifier
• The ICQ is selected so that the ac
load line a line having slope -
1/R’L, intersects the VCE axis at
2VCC volts.
• When ICQ is set for maximum
signal swing so that VCEmax =
2VCC, ICQ is one half of ICmax. That
is ICmax =2 ICQ

8. Maximum efficiency :
• Pin (dc) = VCCICQ
• Po(ac)max =
VCEp−pmax
2 2
ICp−pmax
2 2
• =
2VCC 2ICQ
8
• =
VCC ICQ
2
• % ηmax
= Po ac max
Pin(dc)
100%
• = 50 %
• The actual efficiency rating of a
transformer coupled class A
amplifier will generally be less
than 40%. Because
• VCEQ will always be some value
that is less than VCC
• Transformer is subject to various
power losses, couple losses and
hysteresis losses.

9. General Efficiency :
• VCEQ = VCC =
VCEmax+VCEmin
2
• Pin (dc) = VCCICQ= VCC
VCEQ
R
• =
VCEmax
+VCEmin
2
4R
• Po(ac)=
Vo(p−p)
2 2
2
1
𝑅
• =
VCEmax
−VCEmin
2
8𝑅
• General Efficiency
• % η =
Po(ac)
Pin(dc)
100%
• % η = 50
VCEmax
−VCEmin
VCEmax
+VCEmin
2
%

10. Class B Amplifier:

11. Class B Amplifier: General Efficiency
• Here,
• Idc/ Iavg =
𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒
𝑏𝑎𝑠𝑒
• =
1
π 0
π
𝐼𝑃𝑠𝑖𝑛ω𝑡 𝑑ω𝑡 =
2IP
π
• Pin(dc) = VccIdc = Vcc
2IP
π
• Po(ac) =
𝑉2
𝐿
(
𝑃
)
2𝑅𝐿
• % η =
Po(ac)
Pin(dc)
100%
• Taking IP =
𝑉𝐿
(
𝑃
)
𝑅𝐿
• % η =
VL
(
P
)
Vcc
78.5%
• For maximum efficiency
• VL(P) = VCC
• % ηmax = 78.5%

12. Condition for maximum Power dissipation:
• Power loss by class B amplifier/ Power
dissipation by transistors:
• PD(2Q) = Pi(dc) –Po(ac)
• =
2VccVLp
πRL
-
V2
LP
2RL
• Differentiating with respect to VLP
•
dPD
(
2Q
)
dVLP
=
2Vcc
πRL
-
VLP
RL
• for maximum loss
•
dPD
(
2Q
)
dVLP
= 0 =
2Vcc
πRL
-
VLP
RL
• VLP =
2Vcc
π
• Maximum loss /maximum power
dissipation: when VLP =
2Vcc
π
• Ploss max= PD(2Q)max =
2V2cc
π2𝑅𝐿
• Thus QN and QP must be capable of safely
dissipating
1
2
PD(2Q)max
Efficiency of class B Amp when maximum
power dissipation at transistors
% η =
VL
(
P
)
Vcc
78.5%
% ηmin =
0.6366VCC
Vcc
78.5% = 50%
when
VLP =
2Vcc
π

13. Condition for maximum Power dissipation:
• Power loss by class B amplifier/ Power
dissipation by transistors:
• PD(2Q) = Pi(dc) –Po(ac)
• =
2VccVLp
πRL
-
V2
LP
2RL
• Differentiating with respect to VLP
•
dPD
(
2Q
)
dVLP
=
2Vcc
πRL
-
VLP
RL
• for maximum loss
•
dPD
(
2Q
)
dVLP
= 0 =
2Vcc
πRL
-
VLP
RL
• VLP =
2Vcc
π
• Maximum loss /maximum power
dissipation: when VLP =
2Vcc
π
• Ploss max= PD(2Q)max =
2V2cc
π2𝑅𝐿
• Thus QN and QP must be capable of safely
dissipating
1
2
PD(2Q)max
Efficiency of class B Amp when maximum
power dissipation at transistors
% η =
VL
(
P
)
Vcc
78.5%
% η=
0.6366VCC
Vcc
78.5% = 50%
when
VLP =
2Vcc
π

14. Phase splitter Circuit:

15. Transformer coupled Push-pull class B amplifier
• The bases of 2 CE connected identical
transistor Q1 and Q2 have been
connected to the opposite ends of the
secondary of the input transformer
• Load RL is coupled to the collector
terminals of the 2 transistor through
output transformer T2
• During +ve half of i/p only Q1
conducts and o/p appears across the
load
• Similarly during –ve half of i/p only Q2
conducts and o/p appears across the
load.

16. Complementary –Symmetry Class B Push Pull
amplifier
• Requires match transistors
( ex : BC547, BC 557 ; BD 139, BD 140, etc)
Requires two separate voltage supplies.
Using kVL,
Vo = VCC- VCEN
Vo is maximum when VCEN = VCENsat
So, Vomax = VCC- VCENsat
Again using KVL,
Vin =VBEN+ Vo
For maximum Vo , its corresponding Vin is
=VCC-VCENsat+VBEN
In a similar manner, for Q2
Vomax= -VCC+VECPsat
And its corresponding Vin is
= -VCC+VECPsat-VEBP

17. Crossover distortion:
• Refers to the fact that during the
signal cross over from the +ve to –
ve (or vice versa) there is some
non-linearity in the o/p signal.
• This results from the fact that the
circuit does not provide exact
switching of one transistor OFF and
other ON at the zero- voltage
condition
• Both transistor may be partially OFF
so that the output voltage does not
follow the input around the zero
voltage condition.
• Biasing the transistor in Class AB
improves this operation by biasing
both transistor to be ON for more
than half a cycle.

18. Transfer characteristic curve of Class B
complementary symmetry Push Pull Amp

19. Numerical:
• Calculate the efficiency of
transformer coupled push pull
power amplifier for a supply
voltage of 20 V and output of (i)
VP= 20 V and (ii) VP = 16 V
• Solution:
• First case:
• % η = 50
VCEmax
−VCEmin
VCEmax+VCEmin
2
%
• % η = 50%
• Second case :
• % η = 50
VCEmax−VCEmin
VCEmax+VCEmin
2
%
• % η = 32%

20. Numerical:
• For a class B amplifier providing
a 20V peak signal to a 16 Ω load
(speaker) and a power supply of
VCC = 30 V, determine the input
power, output power and circuit
efficiency
• Solution:
• Pin(dc) = Vcc
2𝐼𝑃
π
=23.9 W
• Ip =
𝑉𝐿𝑃
𝑅𝐿
=1.25 A
• Po(ac) =
𝑉2
𝐿𝑃
2𝑅𝐿
= 12.5 W
• %η =
𝑃𝑜(𝑎𝑐)
𝑃𝑖𝑛 𝑑𝑐
% =52.3%

21. Numerical:
• For a class B amplifier using a
supply of VCC = 30 V and driving a
load of 16 Ω, determine the
maximum input power, maximum
output power and maximum
transistor dissipation.
• Solution:
• For maximum case VP = VCC
• Pin(dc)max = Vcc
2𝐼𝑃
π
= 35.81 W
• Ip =
𝑉𝑃
𝑅𝐿
=1.8 A
• Po(ac)max =
𝑉2𝐿
2𝑅𝐿
= 28.11 W
• %ηmax =
𝑃𝑜(𝑎𝑐)
𝑃𝑖𝑛 𝑑𝑐
% = 78.5 %
• PD(2Q) = Pin(dc) – Po(ac) = 7.6 W
• Maximum transistor dissipation
• PD(2Q) max =
2V2cc
π2𝑅𝐿
= 11.3 W

22. Numerical:
• It is required to design a class B
output stage to deliver an average
power of 20 W to an 8 Ω load. The
power supply is to be selected such
that VCC is about 5 V greater than the
peak output voltage. This avoids
transistor saturation and associated
non linear distortion, and allows for
including short circuit protection
circuitry. Determine the supply
voltage required , the peak current
drawn from each supply , the total
supply power, and the power
conversion efficiency. Also determine
the maximum power that each
transistor must be able to dissipate
safely.
• Solution:
• Po(ac) =
𝑉2𝐿
2𝑅𝐿
• VL = 17.8 V
• Vcc = 23 V ( selected)
• Ip =
𝑉𝑃𝐿
𝑅𝐿
= 2.22 𝐴
• Pin (dc) = Vcc
2𝐼𝑃
π
= 32.5 W
• Po(ac) = 20 W
• %η =
𝑃𝑜(𝑎𝑐)
𝑃𝑖𝑛 𝑑𝑐
% = 61.5 %
• PD(2Q) max =
2V2cc
π2𝑅𝐿
= 13.4 W
• PD(Q) max =13.4/2 = 6.7 W

23. Class AB Push Pull Amplifier:
• The basic circuit of class AB push
pull amplifier is the same as that
of class B push pull amplifier
except that the voltage drop
across base –emitter junction is
approximately equal to
threshold voltage.
• The distortion introduced in
class AB amplifier is less than
that in class B amplifier but
more than in class A amplifier.
• Drawbacks:
• Low conversion efficiency than
class B amplifier
• Wastage of standby power

24. Complementary –symmetry class AB Push pull
amplifier
• Resistor R1 and R2 provide voltage
divider bias to forward bias the base
emitter junction of transistor Q1
and Q2
• Drawbacks:
• Requires two separate voltage
supplies.
• Difficult to obtain matched
complementary transistors
• If there is an unbalance in the
characteristics of the two
transistors, even harmonics will no
longer be cancelled and
considerable distortion will be
introduced.

25. Complementary symmetry class AB Push Pull
amplifier using diodes

27. Darlington complementary-symmetry class
AB push –pull Amplifier
• Darlington –connected transistor
provides higher output current and
lower output resistance
• During +ve half cycle npn transistors
will be biased into conduction
whereas pnp transistors at cut off. So
current ICQ1 flows from +Vcc to
ground through load
• During –ve half cycle pnp transistors
will be biased into conduction
whereas npn transistor are at cut off.
So current ICQ1 = ICQ2 flows from
ground to –Vcc through load.

28. Darlington complementary symmetry class AB
Push pull amplifier using diodes

29. Quasi complementary Push Pull amplifier
• The push pull operation is achieved by
using complementary transistor (Q1
and Q2) before the matched npn
output transistor (Q3 and Q4)
• Q1 and Q3 forms Darlington
connection that provides o/p from a
low impedance emitter follower
• Similarly the connection of Q2 and Q4
forms a feedback pair which provides
low impedance drive to the load.
• Resistor R2 can be adjusted to
minimize cross-over distortion by
adjusting the dc bias condition
• Most popular form of power amplifier

30. Quasi complementary Push Pull amplifier
using diodes:
• Draw yourself

31. Tuned Power amplifier:
• Tuned amplifiers amplify selective
frequency only using LC network
and hence are useful in
communication receiver.
• The response of tuned amplifier is
similar to band pass filter with
center frequency ωo.
• Tuned amplifier finds application in
the radio frequency (RF) and
intermediate frequency (IF)
sections of communication receiver
and in variety of other system.

32. Basic Principle: Single Tuned Amplifier:
• The basic principle
underlying the design
of tuned amplifier is
the use of a parallel
LCR circuit as load or
at the input of BJT or
FET amplifier.

33. • From the ac equivalent network,
• Vo = -gmVi
1
𝑌𝐿
• Vo = −
gmVi
sC+
1
R
+
1
sL
•
Vo
Vi
= −
gm
C
S
S2+
S
RC
+
1
LC
………(i)
• The transfer function of a band
pass filter is given by
• T(s) =
ns
s2+s
ωo
Q
+ωo2
• Comparing eqn (i) with the
transfer function
• ωo =
1
𝐿𝐶
• B =
ωo
Q
=
1
RC
• And Q = RCωo
• At centre frequency (ωo), gain is
•
Vo
Vi
= -gmR

34. Numerical:
• It is required to design a tuned
amplifier of the type shown
having fo = 1 MHz , 3-dB
Bandwidth = 10 KHz and centre
frequency gain = -10. The FET
available has at bias point gm =
5mA/V and rd = 10 KΩ.
• ωo =
1
𝐿𝐶
• 2πfo =
1
𝐿𝐶
• L= 3.18 μH
• Solution :
• Centre freq gain = -gmR’
• R’= 2 K
• R’ = rd//R
• R=2.5 K
• ∆f= 10KHz (given)
• B =
1
R′C
• 2π∆f =
1
R′C
• C=7958 pF

35. Stagger-tuned voltage amplifier:
• If the different tuned circuit
which are cascaded are tuned to
slightly different frequency, it is
possible to obtain an increased
bandwidth with maximal
flatness around the centre
frequency.
• in figure L1C1 and L2C2 are
tuned to different frequency f1
and f2.

36. Stagger-tuned voltage amplifier:
• If an optimum stagger tuning is
employed the response curve of
the amplifier is very close to a
rectangular response curve.
Such a response curve is called
Butterworth response.

37. Synchronous Tuning:
• Synchronous tuning is used to
get sharp curve having least BW.
• The 3-dB BW of overall amplifier
is related to that of individual
tuned circuit
• BW =
ωo
Q
2
1
𝑁 − 1
• Where N – no of stage
• 2
1
𝑁 − 1 is BW shrinkage factor