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OUTPHASING PA
©James Buckwalter 1
Average Efficiency
• We recognize the importance of average efficiency.
• However, PA design –to now- has focused on peak efficiency.
• Other techniques should be developed to provide peak efficiency over a range of power levels.
©James Buckwalter 2
Insight into Loadline for High Efficiency
• High efficiency can be realized for lower power conditions that peak saturated output power.
• There is no fundamental reason why efficiency cannot be optimized over different output powers
• The trick is to vary the loadline.
©James Buckwalter 3
Three Techniques Exploit Average Efficiency
• Doherty Amplifiers: Loadline modulation through turning on PAs at different power conditions
• Outphasing Amplifiers: Loadline modulation through keeping PAs at peak power and producing a phase shift.
• Envelope Tracking: Loadline remains constant but VDD of transistor changes with signal amplitude
©James Buckwalter 4
Outphasing Amplifiers
• It’s fair to say this is an OLD idea.Chireix, H. “High Power Outphasing Modulation,” Proc. IRE, Vol. 23. No. 11, Nov. 1935, pp. 1370-1392.
• Sometimes called Chireix, LINC (linear in nonlinear components)
• Basic outphasing uses two amplifiers that are controlled purely through signal phase.
• The polar coordinates can be transformed to the cartesiancoordinates
where
©James Buckwalter 5
s t( ) =mI t( )cos wt( )+mQ t( )sin wt( ) = A t( )cos wt +q t( )( )
A t( ) = mI2 t( )+mQ
2 t( ) q t( ) = tan-1 mI t( )mQ t( )
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èçç
ö
ø÷÷
Outphasing Concept
• Outphasing exploits
• We can define an outphasing angle, φ.
• Note that s1 and s2 are constant envelope signals.
©James Buckwalter 6
cos A( )+ cos B( ) = 2cosA+B
2
æ
èç
ö
ø÷cos
A-B
2
æ
èç
ö
ø÷
s1 t( ) = cos wt +q +f( ) s2 t( ) = cos wt +q -f( )
Outphasing Concept
• How is the signal reconstructed?
• Since,
• Then, determine the outphasing angle.
©James Buckwalter 7
s t( ) = s1 t( )+ s2 t( ) = AMAX cos wt +q( )cos f( )
s1 t( ) =AMAX
2cos wt +q +f( ) s2 t( ) =
AMAX
2cos wt +q -f( )
A t( ) = AMAX cos f( ) ®f = cos-1 A t( )AMAX
æ
èç
ö
ø÷
cos A( )+ cos B( ) = 2cosA+B
2
æ
èç
ö
ø÷cos
A-B
2
æ
èç
ö
ø÷
Alternative Derivation
• Consider a slightly different representation.
• Now,
• Then, determine the outphasing angle.
©James Buckwalter 8
s t( ) = s1 t( )+ s2 t( ) = AMAX cos wt +q( )sin f( )
s1 t( ) =AMAX
2sin wt +q -f( ) s2 t( ) = -
AMAX
2sin wt +q +f( )
A t( ) = AMAX sin f( ) ®f = sin-1 A t( )AMAX
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èç
ö
ø÷
sin A( ) -sin B( ) = 2cosA+B
2
æ
èç
ö
ø÷sin
A-B
2
æ
èç
ö
ø÷
Approaches for Load Pulling
©James Buckwalter 9
Outphasing “Isolated” Combining
• Recall how an isolated combiner works.
• When the voltage across Port 2 and Port 3 is unbalanced, where does the power go?
• Voltage across resistor is
©James Buckwalter 10
idis =s1 - s2
2Zo= -AMAX
Zosin wt +q( )sinf
s1 + s2 = AMAX cos wt +q( )cos f( )
s1 t( ) =AMAX
2cos wt +q +f( )
s2 t( ) =AMAX
2cos wt +q -f( )
Outphasing “Isolated” Combining
• Find the output power as a function of the outphasing angle
©James Buckwalter 11
s1 + s2 = AMAX cos wt +q( )cos f( )
s1 + s22
=AMAX
2
42 1+ cos2f( )
s1 + s22
=AMAX
2
21+ cos2f( )
Outphasing Angle (deg)0 10 20 30 40 50 60 70 80 90
No
rmalized
Ou
tpu
t P
ow
er
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Outphasing “Isolated” Combining
• How much power is dissipated?
• The result is that the efficiency drops rapidly.
• This is undesirable from the standpoint of average efficiency.
• Need to allow amplifiers to modulate the load.
©James Buckwalter 12
s1 - s22
=AMAX
2
2sin2 f
Outphasing Angle (deg)0 10 20 30 40 50 60 70 80 90
No
rmalized
Ou
tpu
t P
ow
er
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Drain Efficiency for Outphasing with Isolated Power Combining
Power Backoff-10 -8 -6 -4 -2 0
No
rmalized
Eff
icie
nc
y (
%)
0
10
20
30
40
50
60
70
80
90
100
©James Buckwalter 13
This is no better than the backoff behavior of a class-A amplifier
Outphasing Load Modulation
• Load modulation
• Current through the load
• Therefore, the impedance is
©James Buckwalter 14
V1 =Ve jf V2 =Ve- jf
I =V1 -V2
R=V
Re jf - e- jf( )
Z1 =V1
I= R
e jf
e jf - e- jf( )= R
cosf + j sinf
2 j sinf=R
21- j cotf( )
Z2 = -V2
I= -R
e- jf
e jf - e- jf( )= -R
cosf - j sinf
2 j sinf=R
21+ j cotf( )
Note this is predicated
Outphasing Load Modulation
• The load modulation results in a reactive component which shows up on each PA.
• This is NOT good because it forces the PA to operate away from the ideal loadline.
• Note the ideal loadline would lie along the x-axis of the Smith chart.
©James Buckwalter 15
Z1 =R
21- j cotf( ) Z2 =
R
21+ j cotf( )
Series Compensation
• Consider adding a “series” compensation element to each PA.
• Compensated with Xc = jR/2.
• This is still not “ideal” behavior.
©James Buckwalter 16
Z1 =R
21- j cotf( ) Z2 =
R
21+ j cotf( )
Z1 =R
21- j cotf( )+ jXC
Z2 =R
21+ j cotf( ) - jXC
Shunt Compensation
• Transform the series network to a parallel network.
• The quality factor Q is
• Therefore the parallel network becomes
©James Buckwalter 17
Z1 =R
21- j cotf( ) = RS + jXS
Q =XS
RS= cotf The quality factor is a function
of the outphasing angle.
RP = RS 1+Q2( ) = RS 1+ cot2 f( ) =R
2sin2 f
XP = XS1+Q2
Q2= -R
2cotf
tan2 f
sin2 f= -R
2
1
sinf cosf=
-R
sin2f
Parallel Load
• Shunt compensation is introduced through the opposite sign reactance.
• Compensated with Xc = R/0.75.
©James Buckwalter 18
Y1 =1
jXC-
sin2f
jR+
2sin2 f
R
Y2 = -1
jXC+
sin2f
jR+
2sin2 f
R
Output Power
• The output power has already been determined as a function of the outphasing angle
• The dc power consumption depends on the loadline
©James Buckwalter 19
Po =AMAX
2
2R1+ cos2f( )
PDC = AMAX2
pi1 + i2( ) Assuming class-B
i1 =Y1V1 = AMAXejf 1
Z1
+ jBcæ
èç
ö
ø÷ i2 =Y2V2 = AMAXe
- jf 1
Z2
- jBcæ
èç
ö
ø÷
Z1 = Re jf
2cosfZ2 = R
e- jf
2cosf
Efficiency
• The efficiency can now be written as a function of the outphasing angle
©James Buckwalter 20
h =Po
PDC=
AMAX2
2R1+ cos2f( )
AMAX2 2
p
1
Z1
+ jBc +1
Z2
- jBc
æ
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ö
ø÷
=p
4
1+ cos2f( )R
Z1
+ jRBc +R
Z2
- jRBc
Implementation
©James Buckwalter 21
