Download Class-AB POWER AMP

DMT 231 / 3
ELECTRONICS II
Lecture II
Power Amplifiers
(Class AB & Class C)
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 product of iCn and iCp is constant,
therefore if iCn increases iCp decreases
but does not to zero
POWER AMPLIFIER
– Class-AB Operation
Example III
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 III – 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 III – 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 III – 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 III – 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 AMP
Class-AB
Qn & Qp are assumed
matched transistors
Small biasing voltage
to eliminate dead band
POWER AMP
Class-AB
Various techniques are used in
obtaining the bias voltage VBB in
class AB power amplifier circuit.
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)
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 iB1 and iE1 also
decrease.
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)
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
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
2vI

1   R
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
requrements.
POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE IV
(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.
POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE IV – 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.
POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE IV – 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 
61
POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE IV – 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 
61
POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE IV – 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 
61
POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE IV – 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
POWER AMP
Class-AB with Input Buffer Transistors
EXAMPLE IV – 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.
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.