OUTPHASING PA. James Buckwalter 1

Transcription

1 OUTPHASING PA James Buckwalter 1

2 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

3 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

4 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

5 Outphasing Amplifiers It s fair to say this is an OLD idea. Chireix, H. High Power Outphasing Modulation, Proc. IRE, Vol. 3. No. 11, Nov. 1935, pp 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 cartesian coordinates ( ) = m I t s t where ( )cos( wt)+ m Q ( t)sin wt A( t) = m I ( t)+ m Q ( t) q t ( ) = A t æ m ( ) = tan -1 I ( t) ç ( t) è m Q ( )cos wt +q t ö ø ( ) ( ) James Buckwalter 5

6 Outphasing Concept Outphasing exploits cos( A)+cos B ( ) = cosç A+ B We can define an outphasing angle, φ. s 1 t Note that s1 and s are constant envelope signals. æ è ö æ cos A - B ö ç ø è ø ( ) = cos( wt +q +f) s ( t) = cos( wt +q -f) James Buckwalter 6

7 Outphasing Concept How is the signal reconstructed? s 1 t Since, ( ) = A MAX ( ) s ( t) = A MAX cos wt +q +f cos( A)+cos B ( ) = cosç A+ B Then, determine the outphasing angle. s( t) = s 1 ( t)+ s ( t) = A MAX cos( wt +q)cos( f) æ è ö æ cos A - B ö ç ø è ø cos( wt +q -f) æ A( t) A( t) = A MAX cos( f) f = cos -1 ç è A MAX ö ø James Buckwalter 7

8 Alternative Derivation Consider a slightly different representation. s 1 t Now, ( ) = A MAX ( ) s ( t) = - A MAX sin wt +q -f sin ( A) -sin( B) = cosç A+ B Then, determine the outphasing angle. s( t) = s 1 ( t)+ s ( t) = A MAX cos( wt +q)sin( f) æ è ö æ sin A - B ö ç ø è ø ( ) sin wt +q +f æ A( t) A( t) = A MAX sin( f) f = sin -1 ç è A MAX ö ø James Buckwalter 8

9 Approaches for Load Pulling James Buckwalter 9

10 Outphasing Isolated Combining Recall how an isolated combiner works. ( t) = A MAX s 1 + s = A MAX cos( wt +q)cos f When the voltage across Port and Port 3 is unbalanced, where does the power go? Voltage across resistor is i dis = s 1 - s Z o ( ) s 1 s = - A MAX sin( wt +q)sinf Z o James Buckwalter 10 ( t) = A MAX ( ) cos wt +q +f cos( wt +q -f)

11 Outphasing Isolated Combining Find the output power as a function of the outphasing angle s 1 + s = A MAX cos( wt +q )cos( f) A s 1 + s = MAX 4 A s 1 + s = MAX ( 1+ cosf) ( 1+ cosf) Normalized Output Power Outphasing Angle (deg) James Buckwalter 11

12 Outphasing Isolated Combining How much power is dissipated? A s 1 - s = MAX sin f 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. Normalized Output Power Outphasing Angle (deg) James Buckwalter 1

13 Drain Efficiency for Outphasing with Isolated Power Combining 100 Normalized Efficiency (%) This is no better than the backoff behavior of a class-a amplifier Power Backoff James Buckwalter 13

14 Outphasing Load Modulation N o t e t h i s i s p r e d i c a t e Load modulation V 1 =Ve jf V =Ve - jf Current through the load I = V 1 -V R ( ) = V R e jf - e - jf Therefore, the impedance is Z 1 = V 1 I = R Z = - V I = -R e jf ( e jf - e - jf ) e - jf ( e jf - e - jf ) = R cosf + j sinf j sinf = R = -R cosf - j sinf j sinf = R ( 1- j cotf) ( 1+ j cotf) James Buckwalter 14

15 Outphasing Load Modulation ( ) Z = R Z 1 = R 1- j cotf 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. ( 1+ j cotf) James Buckwalter 15

16 Series Compensation Z 1 = R ( 1- j cotf) Z = R ( 1+ j cotf) Consider adding a series compensation element to each PA. Z 1 = R ( 1- j cotf) + jx C Z = R ( 1+ j cotf) - jx C Compensated with Xc = jr/. This is still not ideal behavior. James Buckwalter 16

17 Shunt Compensation Transform the series network to a parallel network. Z 1 = R The quality factor Q is ( 1- j cotf) = R S + jx S Q = X S R S = cotf Therefore the parallel network becomes R P = R S 1+Q X P = X S 1+Q ( ) = R ( S 1+ cot f) = Q The quality factor is a function of the outphasing angle. R sin f = - R cotf tan f sin f = - R 1 sinf cosf = -R sinf James Buckwalter 17

18 Parallel Load Shunt compensation is introduced through the opposite sign reactance. Compensated with Xc = R/0.75. Y 1 = 1 - sinf jx C jr + sin f R Y = sinf jx C jr + sin f R James Buckwalter 18

19 Output Power The output power has already been determined as a function of the outphasing angle ( ) P o = A MAX 1+ cosf R The dc power consumption depends on the loadline P DC = A ( MAX p i 1 + i ) Assuming class-b æ 1 ö æ 1 ö i 1 = Y 1 V 1 = A MAX e jf ç + jb c i = Y V = A MAX e - jf ç - jb c è Z 1 ø è Z ø Z 1 = R e jf jf e- Z = R cosf cosf James Buckwalter 19

20 Efficiency The efficiency can now be written as a function of the outphasing angle A MAX ( 1+ cosf) ( 1+ cosf) h = P o P DC = A MAX R æ ç p è 1 Z 1 + jb c + 1 Z - jb c = p ö 4 ø R Z 1 + jrb c + R Z - jrb c James Buckwalter 0

21 Implementation James Buckwalter 1