Transcript power amp - classification

EMT 112/4
ANALOGUE ELECTRONICS 1
Power Amplifiers
Syllabus
Power amplifier classification, class A, class B, class
AB, amplifier distortion, class C and D, transistor
power dissipation, thermal management.
POWER AMPLIFIERS
Part II
Classification of Power
Amplifiers
POWER AMPLIFIER – Classification
They are grouped together based on their Q-points on the
DC load line.
POWER AMPLIFIER – Classification
In class-A; the transistor conducts during the
whole cycle of sinusoidal input signal
POWER AMPLIFIER – Classification
In class-B; the transistor conducts during
one-half cycle of input signal
POWER AMPLIFIER – Classification
In class-AB; the transistor conducts for slightly
more than half a cycle of input signal
POWER AMPLIFIER – Classification
In class-C; the transistor conducts for less
than half a cycle of input signal
POWER AMPLIFIER – Class-A Operation
For maximum swing ( +ve and –ve), transistor is biased such
that the Q point is at centre of the load line.
The transistor conducts for a full cycle of the input signal
POWER AMPLIFIER – Class-A Operation
Instantaneous power dissipation in transistor is;
pQ  vCE iC
For sinusoidal input signal;
iC  I CQ  I p sin t
And;
vCE
VCC

 V p sin t
2
POWER AMPLIFIER – Class-A Operation
For maximum possible
swing;
I p  I CQ
And;
VCC
Vp 
2
Therefore;
pQ 
VCC I CQ
2
1  sin
2
t 
POWER AMPLIFIER – Class-A Operation
When the input signal = 0, the transistor must be capable of
handling a continuous power of;
VCC I CQ
2
Efficiency;
PL

PS
PL = average ac power to the load
PS = average power supplied by the source (VCC)
POWER AMPLIFIER – Class-A Operation
For maximum possible swing;
1
1  VCC
PL  V p I p  
2
2 2
VCC I CQ

 I CQ 
4

Power supplied by the source;
PS  VCC I CQ
The efficiency;

VCC I CQ
4
Maximum theoretical
efficiency of class A
amplifier is therefore
25%
 VCC I CQ  0.25
POWER AMPLIFIER – Class-B Operation
Consists of 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
Both devices are off when the input is zero
(See Figure)
POWER AMPLIFIER – Class-B Operation
When vI = 0, both A and B are
OFF and therefore vO = 0.
The complementary pair
POWER AMPLIFIER – Class-B Operation
When vI > 0, A is ON and B is
OFF and vO > 0.
POWER AMPLIFIER – Class-B Operation
When vI < 0, A is OFF and B is
ON and vO < 0.
POWER AMPLIFIER – Class-B Operation
The transfer characteristic of the
complementary pair
POWER AMPLIFIER – Class-B Operation
Approximate Class-B : Complementary push-pull circuit
POWER AMPLIFIER – Class-B Operation
Approximate Class-B : Complementary push-pull circuit
Assuming ideal transistor;
when vI = 0;
both Qn & Qp are off and
vO = 0
POWER AMPLIFIER – Class-B Operation
Approximate Class-B : Complementary push-pull circuit
vO = vI  vBEn
When vI > 0; Qp is off & Qn
conducts and becomes an emitter
follower, sourcing the current IL
to RL.
POWER AMPLIFIER – Class-B Operation
Approximate Class-B : Complementary push-pull circuit
When vI < 0; Qn is off & Qp conducts and becomes an
emitter follower, sinking the current IL from RL.
POWER AMPLIFIER – Class-B Operation
Approximate Class-B : Complementary push-pull circuit
The problem with ClassB: transistors are not
ideal.
POWER AMPLIFIER – Class-B Operation
Crossover Distortion
Assuming cut-in
voltage of transistor is
0.6 V, vO = 0 for a
range 0.6 V < vI < 0.6
V.
The transfer
characteristic becomes
non-linear (See
Figure).
The range where both
transistors are
simultaneously off
known as the dead
band.
POWER AMPLIFIER – Class-B Operation
Crossover Distortion
The output will be distorted – crossover distortion (Figure)
POWER AMPLIFIER – Class-B Operation
Crossover Distortion
Crossover distortion can be eliminated by biasing the
transistor with small quiescent current – class-AB
POWER AMPLIFIER – Class-B Operation
Idealized Power Efficiency
VCC  V   V 
POWER AMPLIFIER – Class-B Operation
Idealized Power Efficiency
POWER AMPLIFIER – Class-B Operation
Idealized Power Efficiency
The Q-point is at
the cutoff point of
both transistors
(zero collector
current)
POWER AMPLIFIER – Class-B Operation
Idealized Power Efficiency
The output voltage of
the idealized class-B is;
vO  V p sin t
The maximum possible
value of Vp is VCC
The instantaneous
power dissipation in Qn
is;
pQn  vCEniCn
POWER AMPLIFIER – Class-B Operation
Idealized Power Efficiency
For 0  t   , the collector
current is;
iCn 
Vp
RL
sin t
For   t  2 the collector
current is;
iCn  0
POWER AMPLIFIER – Class-B Operation
Idealized Power Efficiency
The collector-emitter
voltage is;
vCEn  VCC  V p sin t
Therefore the instantaneous
power dissipation in Qn is;
pQn  vCEniCn
 VCC
 Vp

 V p sin t  sin t 
 RL

for 0  t  
POWER AMPLIFIER – Class-B Operation
Idealized Power Efficiency
And;
pQn  0 for   t  2
Therefore, the average
power dissipation in Qn
is;
PQn 
VCCV p
RL

Vp
2
4 RL
POWER AMPLIFIER – Class-B Operation
Idealized Power Efficiency
Plotting PQn versus Vp as given by the equation;
PQn 
VCCV p
RL

Vp
2
4 RL
gives us the following curve:
POWER AMPLIFIER – Class-B Operation
Idealized Power Efficiency
Because of symmetry;
Differentiating
PQn 
PQn  PQp
VCCV p
RL

Vp
2
4 RL
with respect to Vp, for maximum PQn gives us;
2
VCC
PQn max   2
 RL
which occurs when
Vp 
2VCC

POWER AMPLIFIER – Class-B Operation
Idealized Power Efficiency
Since each power source supplies half sinewave of
current, the average value is;
IS 
Vp
RL
The total power supplied by the two sources is;
 Vp 

PS  2VCC I S  2VCC 
 RL 
POWER AMPLIFIER – Class-B Operation
Idealized Power Efficiency
The power delivered to the load is;
PL 
2
O  rms 
V
RL
The efficiency is;

V

p
/ 2
RL

2

V
2
p
2 RL
PL V p
 
PS 4VCC
Maximum efficiency occurs when
V p  VCC
POWER AMPLIFIER – Class-B Operation
Idealized Power Efficiency
Under this condition;


4
 0.785
Maximum theoretical efficiency of class B
amplifier is therefore 78.5%
The efficiency obtained in practice is less than the maximum
value because of other circuit losses and because the peak
output voltage must remain less than VCC to avoid transistor
saturation which can cause distortion in the output signal.
POWER AMPLIFIER – Class-B Operation
Idealized Power Efficiency
The maximum transistor power dissipation occurs at;
Vp 
2VCC

Substituting in the expression for efficiency;

V p
4VCC
  2VCC  1



4VCC    2
50%
POWER AMPLIFIER
Crossover distortion can be
virtually eliminated by applying
small quiescent bias on each
transistor (See Figure)
If Qn and Qp are matched,
each emitter-base junction is
biased with VBB/2 when vI is
zero. Hence vO is also zero.
The quiescent collector
currents are;
iCn  iCp  I S eVBB / 2VT
– Class-AB Operation
POWER AMPLIFIER
– Class-AB Operation
As vI increases, the voltage at
the base of Qn increases and
vO increases. Qn operates as
an emitter follower supplying
current to RL. The output
voltage is given by;
VBB
vO  vI 
 vBEn
2
The collector current of Qn is;
iCn  iL  iCp (Neglecting the base
currents)
POWER AMPLIFIER
Since iCn must to supply the
load current, vBEn increases
which causes vBEp to decrease
because VBB is constant. The
decrease in vBEp results in a
decrease in iCp.
– Class-AB Operation
POWER AMPLIFIER
When vI goes negative, the
base voltage of Qp decreases
followed by a decrease in vO.
Qp operates as emitter
follower, sinking the load
current.
As iCp increases vEBp
increases causing a decrease
in vBEn and iCn.
– Class-AB Operation
POWER AMPLIFIER
– Class-AB Operation
Transfer characteristics (vO versus vI)
POWER AMPLIFIER
– Class-AB Operation
POWER AMPLIFIER
– Class-AB Operation
– Class-AB Operation
POWER AMPLIFIER
iCn  iCp relationship
vBEn  vEBp  vBB
Using the relationship I C  I S e
can be written as;
VBE / VT
, the above expression
 iCn 
 iCp 
 I CQ 

VT ln    VT ln    2VT ln 
 IS 
 IS 
 IS 
Hence;
iCniCp  I
2
CQ
The produc of iCn and iCp is constant,
therefore if iCn increases iCp decreases
but does not to zero
POWER AMPLIFIER
– Class-AB Operation
Example 8.8
Mn and Mp are matched transistors
with the following parameters;
VT  1 V;
K  0.20 A/V 2
If VDD = 10 V and RL = 8 ,
find the bias voltage VBB/2 for
IDQ = 0.05 A. Find also VGSn,
VSGp andvI if vO = 5 V.
POWER AMPLIFIER
– Class-AB Operation
Example 8.8 – Solution
iD  K vGS  VT

2
Since the MOSFETs are matched,
at quiescent point;
iD  I DQ
and
vGS  VGSQ  VSGQ
Hence;
I DQ
VBB

2
 VBB

 K
 VT 
 2

2
POWER AMPLIFIER
– Class-AB Operation
Example 8.8 – Solution (cont’d)
Substituting values;
 VBB 
0.05  0.2
 1
 2

which yields;
2
VBB
 1 .5 V
2
From the expression
iDn  K vGSn  VT
we have;

2
iDn
vGSn 
 VT
K
POWER AMPLIFIER
– Class-AB Operation
Example 8.8 – Solution (cont’d)
When vO  5 V;
iDn
vO
 iL 
RL
5

 0.25 A;
20
and
iDn
0.25
vGSn 
 VT 
 1  2.12 V
K
0.2
POWER AMPLIFIER
– Class-AB Operation
Example 8.8 – Solution (cont’d)
Since;
VBB  vGSn  vSGp
then;
vSGp  VBB  vGSn
 3  2.12  0.88 V
And;
VBB
3
vI  vO  vGSn 
 5  2.12   5.62 V
2
2
POWER AMPLIFIER
– Class-C Operation
Transistor conducts
for less than half a
cycle of input signal
•
•
•
Tuned circuit is required.
Used for RF amplifier.
Efficiency > 78.5%
B – E junction is reverse-biased to
obtain Q-point beyond cut-off.