CLASS C. James Buckwalter 1

Transcription

1 CLASS C James Buckwalter 1

2 Class-C Amplifier Like class-b, I quiescent = 0 Unlike class-b, conduction angle < 180 deg Gain is low since device is turned on for only a short period Iout Imax Vmin Vmax ut match match Harmonics are shorted RL Vce Ic Vrf time Idc time James Buckwalter 2

3 Class-C Amplifier I quiescent = 0 I dc < I rf / p Iout Imax P out (rf) < 1/4 V rf I rf P in (dc)= V o I dc h max =p/4*(v max -V min )/(V max +V min ) Vmin Vmax ut match match Harmonics are shorted RL Vce Ic Vrf time Idc time James Buckwalter 3

4 Load Impedance Is A Resonant Network Short at all harmonics here Representative Z values At fo -j2w j2w 10 W match Z fo RL Z=RL at fo Z=0 at 2fo, 3fo, 4fo, At 2fo -j1w j4w 10 W fo 2fo James Buckwalter 4

5 Class-C Waveform Ic Idc time I ( t) = ìï í îï i PK cosq - I DC when i PK cosq > I DC 0 otherwise cosf = I DC i PK I DC is the (negative) offset i PK is the amplitude I DC = 1 F ò I ( q ) dq = 1 2p 2p 0 I DC = 1 ( p i sinf - I F PK DD ) F ò 0 ( ) i PK cosq - I DD dq I FUND = 1 p I FUND = 2 p 2p ò 0 F ò 0 ( ) I q cosqdq = 2 p ( ) I q ( i PK cos 2 q - I DC cosq )dq James Buckwalter 5 F ò 0 cosqdq I FUND = 2 æ1 p 2 i PKF+ 1 4 i ö ç PK sin2f - I DC sinf è ø I FUND = 2 p i æ 1 PK 2 F+ 1 ö ç sin2f - cosfsinf è 4 ø I FUND = i æ PK ç p F - 1 è 2 sin2f ö ø

6 Class-C Conduction Φ is the conduction angle during which the current conducts through the transistor. What happens for Φ of pi or 2PI? I FUND = i æ PK ç p F - 1 è 2 sin2f ö ø James Buckwalter 6

7 Class-C Efficiency I DC = i PK ( sinf -FcosF) p h = P RF,MAX P DC = v PK i PK 2V DD I DC = æ çf - 1 sin 2F è 2 2 sin F ( ) ö ø ( ) - Fcos( F) ( ) As conduction angle approaches 0, the efficiency approaches 100%. What is the penalty? James Buckwalter 7

8 Class C Waveform Analysis Calculate efficiency h = i pk æ 1 v 2 PK i PK = 1 v pk p F sin2f ö ç è ø V DC I DC 2 V DC i pk sinf - FcosF p æ çf - 1 è 2 sin2f ö ø sinf - FcosF f ( q 0 ) = 1 2 ( ) h = V V FUND FUND,MAX f ( q 0 ) V FUND,MAX V DC ( ) = V pk V RF,MAX f ( q 0 ) V RF,MAX V DC h = P OUT V MAX -V MIN f ( q 0 ) P OUT,MAX V MAX +V MIN lim f q 0 ( q 0) =1 lim f ( q 0 ) = p 0 q 0 p 2 4 James Buckwalter 8

9 Power Amplifier Comparison James Buckwalter 9

10 Class-C Amplifier Efficiency Class C has very good efficiency because whenever the device has current, Vds is particularly low Vds ID Vrf time Idc time James Buckwalter 10

11 Class-C Waveform Analysis How about the loadline resistance? Iout Imax R L = V FUND I FUND Vmin Vmaxut V FUND = V MAX -V MIN 2 æ I FUND = I S q sin2q ö ç 0 è ø ( ) I MAX = I S - I O = I S 1- cosq 0 I FUND = I æ S p q sin2q ö ç 0 = I MAX è ø p R L = p V MAX -V MIN I MAX 1- cosq 0 q sin2q 0 q sin2q 0 1- cosq 0 RL / RL Class A Theta (degrees) James Buckwalter 11

12 Gain and Conduction Angle Gain (db) Class A Class AB "ideal" Class AB "real" 6dB Class B ideal Class B real Class C "real" Pin (dbm) James Buckwalter 12

13 Power Amplifier Comparison Maximum voltage swing is 2V o V min Gain is 6 db lower for class-b than class-a, expect it to be even lower than class-c. Power density is the same for class A and B but lower for class C. James Buckwalter 13

14 Other Classes of Amplifiers PA research is focused around getting highpower power at high-efficiency. Class D Amplifiers Push-pull style amplifier Class E/F Amplifiers Switching amplifiers which can allow 100% PAE but require care with harmonics Class J Amplifier Overdriven class-a James Buckwalter 14

15 CLASS F James Buckwalter 15

16 Class F Amplifier Add harmonic tuning to Class B amplifier Nominally open circuit at odd harmonics Short circuit at even harmonics (In reality, need to optimize for given transistor) V ds begins to look like a square wave match match Harmonic tuning 3fo fo RL RL James Buckwalter 16

17 Class F Amplifier With added 3rd harmonic V 3fo =1/9 V fo, V fo can reach the highest value without causing clipping I quiescent = 0 I dc =I ave = I rf / p P dc = V DD I ave h = 9 8 p 4 V max -V min V max +V min Iout Imax Vds Id Iave Vmin Vmax Vrf (?) time time James Buckwalter 17 ut

18 Class F Strategy Adding 3rd harmonic to voltage flattens its top and bottom, so it begins to approach a square wave With 3rd harmonic added, the fundamental can be increased at fixed signal swing (before clipping) Get even better results adding 5th harmonic James Buckwalter 18

19 Fourier Series Example /2 2/p = /2 Vfo / (/2) = 0.63/0.5 = 1.26: This is for perfect square wave (includes all odd harmonics) Vfo / (/2) ~ 9/8 = 1.125: This is just 3 rd harmonic time James Buckwalter 19

20 Class F Waveform Analysis Is there power delivered to load at 2f o? No, V 2fo =0 Is there power delivered to load at 3f o? No, I 3fo =0 Prf=1/2 Vfo Ifo= 1/4 Vfo Irf = 1/4 Irf*(Vmax-Vmin)/2 * 9/8 Pdc= Vdc Idc = Irf/p*(Vmax+Vmin)/2 Efficiency =p/4 *9/8*(Vmax-Vmin)/(Vmax+Vmin) 2 Waveforms of Transistor ltage(blue) and Current (black) Idc=Irf/p just as for Class B Ifo = Irf /2 just as for Class B Vfo= RL(fo) Ifo V, I angle (degrees) For Class F Vmax=Vdc+8/9 Vfo Vmin=Vdc- 8/9 Vfo Vfo=(Vmax-Vmin)/2*9/8 Vmax= Vdc+Vfo Vmin=Vdc-Vfo for class B James Buckwalter 20

21 Class F Amplifier Implementation: Accounting for Output Capacitance m1 freq= 1.096GHz S(1,1)=0.252 / impedance = Z0 * ( j0.509) m2 freq= 2.089GHz S(1,1)=1.000 / impedance = Z0 * (1.514E-7 - j7.444e-4) S(1,1) m2 m1 m3 m3 freq= 2.951GHz S(1,1)=0.998 / impedance = Z0 * ( j43.448) James Buckwalter 21 freq (100.0MHz to 10.00GHz)

22 Class F Amplifier Alternative Implementation Short at all harmonics here Zo=RL lo/4 match fo Z=RL at fo Z=0 at 2fo, 4fo Z=inf. at 3fo, 5fo,... RL James Buckwalter 22

23 Class F Example David Schmelzer and Stephen I. Long, CSICS 2006 GaN FETs at 2GHz Class F amplifier 86% PAE, 17W James Buckwalter 23

24 Harmonic Load Tuning X2=Im(Znet) at 2fo X3=Im(Znet) at 3fo Class F -1 Class F -1 Class F Class F Class B Znet Cds RL XL(f) James Buckwalter 24

25 Other Approaches for High-Efficiency Control the voltage and current waveforms to prevent conduction while the voltage Class D: Switch current and voltage Class E: ZVS and ZVS derivative switching Class F -1 James Buckwalter 25