Transcript power amp - class-ab

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.
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POWER AMPLIFIERS
Part III (cont’d)
Power amplifier circuits
– Class AB
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POWER AMP
Class-AB
Qn & Qp are assumed
matched transistors
Small biasing voltage
to eliminate dead band
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POWER AMP
Class-AB
Various techniques are used in
obtaining the bias voltage VBB in
class AB power amplifier circuit.
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POWER AMP
Class-AB – Diode Biasing
The base-emitter of a BJT is basically a p-n junction and
hence exhibits similar characteristics as that of a diode –
voltage across a diode in a forward mode is almost
constant over a range of the diode current.
VD
ID
+
VD
-
ID
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POWER AMP
Class-AB – Diode Biasing
The above diode characteristic
is utilised in biasing of the
push-pull power amplifier
In the absence of
input signal (vI = 0),
most of IBias flows
through D1 and D2
establishing a small
bias voltage VBB for
the base-emitter
junctions of Qn and Qp.
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POWER AMP
Class-AB – Diode Biasing
When vI is at its peak
+ve, iBn may be large.
IBias shall be sufficient
to supply both iBn and
the current through D1
and D2 in order to
maintain the bias
voltage VBB.
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POWER AMP
Class-AB – Diode Biasing
EXAMPLE 3
A diode biasing class AB
power amplifier is to meet
the following specifications;
• RL = 8 ;
• output power to PL = 5
W;
• peak output voltage to
be not more than 80%
of VCC;
• minimum value of ID to
be no less than 5 mA
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POWER AMP
Class-AB – Diode Biasing
EXAMPLE 3 (cont’d)
For both Qn and Qp;
I SQ  1013 A;   75
For D1 and D2;
I SD  3  10 14 A
Determine;
a) IBias and VCC;
b) The quiescent collector
current
c) iCn and iCp when the
output voltage is at its
peak positive value
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Class-AB – Diode Biasing
POWER AMP
EXAMPLE 3 – Solution
a)
Vo rms   PL RL
 5  8  6.32 V
Vopeak   2Vorms 
 8.94 V
VCC 
Vo peak 
0.8
 11.8 V
Select VCC = 12 V
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POWER AMP
Class-AB – Diode Biasing
EXAMPLE 3 – Solution (cont’d)
At the +ve peak of the
output voltage;
iEnpeak   iL max  
Vo peak 
RL
8.94

 1.12 A
8
iBnpeak  
iEnpeak 
1 
1.12

 14.7 mA
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POWER AMP
Class-AB – Diode Biasing
EXAMPLE 3 – Solution (cont’d)
To maintain a minimum of
5 mA through the diodes;
I Bias  iBn  I D
 14.7  5
 19.7 mA
Select IBias = 20 mA
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POWER AMP
Class-AB – Diode Biasing
EXAMPLE 3 – Solution (cont’d)
b)
Under quiescent
condition (vI = 0),
ID = 20 mA
(neglecting iBn);
VBB
 ID 

 2VT ln 
 I SD 
 20 103 

 2  0.026 ln 
14 
 3 10 
 1.416 V
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POWER AMP
Class-AB – Diode Biasing
EXAMPLE 3 – Solution (cont’d)
Assuming Qn and Qp are
matched transistors;
VBEn  VBEp

VBB
 0.708 V
2
Hence;
I CQ  I SQeVBB / 2VT
 10 13 e 0.708/ 0.026
I CQ  67 mA
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POWER AMP
Class-AB – Diode Biasing
EXAMPLE 3 – Solution (cont’d)
c)
At the peak +ve value
of output voltage;
iEnmax   iL peak 
 1.12 mA
iBnmax   14.7 mA
I D  I Bias  iBnmax 
 20 14.7  5.3 mA
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POWER AMP
Class-AB – Diode Biasing
EXAMPLE 3 – Solution (cont’d)
VBB
 5.3 103 

 2  0.026 ln 
14 
 3 10 
 1.347 V
iCn max  

1 
iEnmax 
75

1.12
1  75
 1.105 A
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POWER AMP
Class-AB – Diode Biasing
EXAMPLE 3 – Solution (cont’d)
 iCn max  

VBEn  VT ln 
 I

SQ


 1.105 
 26 ln  13 
 10 
 781 mV
VBEp  VBB  VBEn
 1.347  0.781  0.566 V
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POWER AMP
Class-AB – Diode Biasing
EXAMPLE 3 – Solution (cont’d)
iCp  I SQe
VBEp / VT
 10 13 e 0.566/ 0.026
 0.285 mA
Hence, when the output
voltage is at its peak positive
value;
iCn  1.105 A
and;
iCp  0.285 mA
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POWER AMP
Class-AB – Diode Biasing
EXERCISE 1
For the same amplifier as in
Example 3 above, find iCn
and iCp when the output
voltage is;
a) +5 V;
b) -5 V;
c) at its peak negative value.
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POWER AMP
Class-AB – VBE Multiplier Biasing
The biasing circuit comprising
Q1, R1, R2 and IBias, provides the
biasing voltage VBB.
Neglecting the base current
of Q1;
VBE1
IR 
R2
VBB  I R R1  R2 
 R1 
 VBE1 1  
 R2 
Thus, VBB can be set by selecting suitable
values for R1 and R2.
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POWER AMP
Class-AB – VBE Multiplier Biasing
A fraction of IBias flows through
Q1, so that;
 I C1 

VBE1  VT ln 
 I S1 
Under quiescent
condition, iCn and iCp are
small and hence iBn and
iBp are negligible.
As vI increases, iCn increases followed by
an increase of iBn. IC1 decreases but VBE1
is almost constant.
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POWER AMP
Class-AB – VBE Multiplier Biasing
To facilitate adjustment
of the ratio R1/R2 and
hence, the value of VBB,
a third resistor Rv is
included in the circuit.
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POWER AMP
Class-AB with Input Buffer Transistors
R1, R2 and the emitter-followers
Q1 and Q2 establish the required
quiescent bias.
R3 and R4 (usually of low
values) are incorporated to
provide thermal stability.
The output voltage is
approximately equal to the input
voltage (emitter-follower)
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POWER AMP
Class-AB with Input Buffer Transistors
When the input voltage vI
increases, the base voltage of
Q3 increases and the output
voltage vO increases. The
emitter current of Q3 increase to
supply the load current iO. The
base current of Q3 increases.
The increase in base voltage of
Q3 reduces the voltage across,
and the current through R1. This
means that iB1 and iE1 also
decrease.
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POWER AMP
Class-AB with Input Buffer Transistors
Also when the input voltage vI
increases, the voltage across R2
increases and iE2 and iE2
increase. The input current iI
accounts for the reduction in iB1
and the increase in iB2 i.e.
iI  iB 2  iB1 (Kirchhoff’s
Current Law)
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POWER AMP
Class-AB with Input Buffer Transistors
Neglecting vR 3 , vR 4 , iB 3 and iB 4
we have;

vI  VBE   V 
iB 2 
1   n R2
and;
V   vI  VEB 
iB1 
1   p R1
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POWER AMP
Class-AB with Input Buffer Transistors
If;
V   V  , VBE  VEB , R1  R2  R
and;  n   p  
then;
vI  VBE  V  V   vI  VEB
iI 

1   R
1   R
2v I

1   R
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POWER AMP
Class-AB with Input Buffer Transistors
Since the voltage gain is approximately
unity, the output current is;
vO vI
iO 

RL RL
The current gain is;
iO 1   R
Ai  
iI
2 RL
which is quite substantial. A large
current gain is desirable since the
output stage must meet the power
requirements.
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POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE 4
(a) Determine the quiescent
bias currents in all
transistors;
(b) Calculate all the currents
labeled in the figure and the
current gain when vI = 10 V.
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POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE 4 – Solution
(a) For vI = 0 (quiescent currents);
iR1  iR 2  iE1  iE 2
15  0.6

 7.2 mA
2
Assuming all transistors are
matched, the bias currents in Q3 and
Q4 are also approximately 7.2 mA
since the base-emitter voltages of Q1
and Q3 are equal and those of Q2
and Q4 are equal.
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POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE 4 – Solution (cont’d)
(b) For vI = 10;
Because the voltage gain is
approx. unity;
vO vI
iO 

RL RL
10

 100 mA
100
iE 3  iO  100 mA
iB 3
iE 3
100


 1.64 mA
1 
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POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE 4 – Solution (cont’d)
V   VBE  vI 
iR1 
R1
15  0.6  10

 2.2 mA
2
iE1  iR1  iB3
 2.2 1.64  0.56 mA
iE1
0.56
iB1 

 9.18 μA
1 
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POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE 4 – Solution (cont’d)
Since Q4 tends to turn off
when vI increases, iB4 is
negligible. Therefore;
iE 2  iR 2
vI  vEB  V 

R2
10  0.6   15

2
 12.2 mA
iB 2
iE 2
12.2 mA


 200 μA
1 
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POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE 4 – Solution (cont’d)
The input current;
iI  iB 2  iB1
 200  9.18  191 μA
The current gain;
iO
100
Ai  
 524
iI 0.191
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POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE 4 – Solution (cont’d)
If the previous expression i.e. Ai
we have;
Ai

1   R

2 RL
is used,

1  60 2 

 610
2  0.1
The higher gain is due the fact that the base currents of
Q3 and Q4 are neglected in deriving the expression.
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POWER AMP
Class-AB with Input Buffer Transistors
EXERCISES
Problems 8.36 through 8.38 in the text book
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