1 emt 112/4 analogue electronics 1 power amplifiers syllabus power amplifier classification, class...
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EMT 112/4ANALOGUE ELECTRONICS 1
Power Amplifiers
SyllabusPower amplifier classification, class A, class B, class AB, amplifier distortion, class C and D, transistor power dissipation, thermal management.
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Part III (cont’d)Power amplifier circuits
– Class AB
POWER AMPLIFIERS
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Class-ABPOWER AMP
Small biasing voltage to eliminate dead band
Qn & Qp are assumed matched transistors
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Class-ABPOWER AMP
Various techniques are used in obtaining the bias voltage VBB in class AB power amplifier circuit.
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Class-AB – Diode BiasingPOWER AMP
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.
I D
V D
+V D-
I D
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Class-AB – Diode BiasingPOWER AMP
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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Class-AB – Diode BiasingPOWER AMP
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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17/03/08
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Class-AB – Diode BiasingPOWER AMPEXAMPLE 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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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 (cont’d)
For both Qn and Qp;
75 A; 10 13 SQI
For D1 and D2;
A 103 14SDI
Determine;
a) IBias and VCC;b) The quiescent collector
currentc) iCn and iCp when the
output voltage is at its peak positive value
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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution
LLo RPV rmsa)
V 32.685
rmspeak 2 oo VV
V 94.8
V 8.118.0
peak oCC
VV
Select VCC = 12 V
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Class-AB – Diode BiasingPOWER AMP
At the +ve peak of the output voltage;
L
oLEn R
Vii peak
maxpeak
1peak
peakEn
Bn
ii
EXAMPLE 3 – Solution (cont’d)
A 1.128
94.8
mA 14.776
12.1
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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)
To maintain a minimum of 5 mA through the diodes;
DBn IiI Bias
Select IBias = 20 mA
514.7
mA 19.7
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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)
b) Under quiescent
condition (vI = 0),
ID = 20 mA
(neglecting iBn);
SD
DTBB I
IVV ln2
14
3
103
1020ln026.02
V 416.1
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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)
Assuming Qn and Qp are matched transistors;
V 708.02
BB
BEpBEn
V
VV
Hence;
TBB VVSQCQ eII 2/
mA 67CQI
026.0/708.01310 e
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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)
c) At the peak +ve value of output voltage;
peakmax LEn ii
mA 14.7max Bni
mA 12.1
maxBias BnD iII
mA 5.314.720
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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)
14
3
103
103.5ln026.02BBV
maxmax 1 EnCn ii
V 347.1
12.1751
75
A 105.1
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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)
SQ
CnTBEn I
iVV
maxln
1310
105.1ln26
mV 781
BEnBBBEp VVV
V 566.0781.0347.1
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Class-AB – Diode BiasingPOWER AMPEXAMPLE 3 – Solution (cont’d)
TBEp VVSQCp eIi /
026.0/566.01310 e
mA 285.0
Hence, when the output voltage is at its peak positive value;
A 105.1Cni
and;mA 285.0Cpi
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Class-AB – Diode BiasingPOWER AMPEXERCISE 1
a) +5 V;
b) -5 V;
c) at its peak negative value.
For the same amplifier as in Example 3 above, find iCn and iCp when the output
voltage is;
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Class-AB – VBE Multiplier BiasingPOWER AMP
Thus, VBB can be set by selecting suitable values for R1 and R2.
2
1
R
VI BER
2
11
21
1 R
RV
RRIV
BE
RBB
Neglecting the base current of Q1;
The biasing circuit comprising Q1, R1, R2 and IBias, provides the biasing voltage VBB.
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Class-AB – VBE Multiplier BiasingPOWER AMP
A fraction of IBias flows through Q1, so that;
1
11 ln
S
CTBE I
IVV
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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Class-AB – VBE Multiplier BiasingPOWER AMP
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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Class-AB with Input Buffer TransistorsPOWER AMP
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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Class-AB with Input Buffer TransistorsPOWER AMP
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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Class-AB with Input Buffer TransistorsPOWER AMP
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.
12 BBI iii (Kirchhoff’s Current Law)
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Class-AB with Input Buffer TransistorsPOWER AMP
Neglecting and
we have;
2
2 1 R
VVvi
n
BEIB
,3Rv ,4Rv 3Bi 4Bi
and;
1
1 1 R
VvVi
p
EBIB
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Class-AB with Input Buffer TransistorsPOWER AMP
If;
RRRVVVV EBBE 21 , ,
and; pn
then;
RVvV
R
VVvi EBIBEII
11
RvI
1
2
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Class-AB with Input Buffer TransistorsPOWER AMP
Since the voltage gain is approximately unity, the output current is;
L
I
L
OO R
v
R
vi
The current gain is;
LI
Oi R
R
i
iA
2
1
which is quite substantial. A large current gain is desirable since the output stage must meet the power requirements.
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24/03/08
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Class-AB with Input Buffer TransistorsPOWER AMPEXAMPLE 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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Class-AB with Input Buffer TransistorsPOWER AMP
2121 EERR iiii
mA 2.72
6.015
EXAMPLE 4 – Solution
(a) For vI = 0 (quiescent currents);
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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Class-AB with Input Buffer TransistorsPOWER AMP
L
I
L
OO R
v
R
vi
EXAMPLE 4 – Solution (cont’d)
(b) For vI = 10;
Because the voltage gain is approx. unity;
mA 100100
10
mA 1003 OE ii
mA 64.161
100
13
3
EB
ii
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Class-AB with Input Buffer TransistorsPOWER AMP
mA 2.2
2
106.015
EXAMPLE 4 – Solution (cont’d)
1
1 R
vVVi IBER
311 BRE iii
mA 56.064.12.2
μA 18.961
56.0
11
1
EB
ii
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Class-AB with Input Buffer TransistorsPOWER AMPEXAMPLE 4 – Solution (cont’d)
Since Q4 tends to turn off
when vI increases, iB4 is
negligible. Therefore;
222 R
Vvvii EBIRE
2
156.010
mA 2.12
μA 20061
mA 2.12
12
2
EB
ii
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Class-AB with Input Buffer TransistorsPOWER AMPEXAMPLE 4 – Solution (cont’d)
The input current;
12 BBI iii
μA 19118.9200
The current gain;
524191.0
100
I
Oi i
iA
37
L
i R
RA
2
1
Class-AB with Input Buffer TransistorsPOWER AMPEXAMPLE 4 – Solution (cont’d)
If the previous expression i.e.
we have;
610
1.02
2601
iA
is used,
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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Class-AB with Input Buffer TransistorsPOWER AMPEXERCISES
Problems 8.36 through 8.38 in the text book