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CLASS F
©James Buckwalter 1
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)
• Vds begins to look like a square wave
©James Buckwalter 2
RL
match
Vo
Harmonic tuning
RL
match
Vo
fo
3fo
Class-F Amplifier
• With added 3rd harmonic V3,pk=1/9 Vpk,
• Vpk can reach the highest value without causing clipping
• Iquiescent= 0
• Idc=Iave = Irf / p
• Pdc= VDD Iave
©James Buckwalter 3
Iout
VoutVoVmin
Imax
Vmax
timeIave
Id time
Vdd
Vds
Vrf (?)
h =9
8
p
4
Vmax -Vmin
Vmax +Vmin
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 4
-1.5
-1
-0.5
0
0.5
1
1.5
0 2 4 6 8 10 12
-1.5
-1
-0.5
0
0.5
1
1.5
0 2 4 6 8 10 12
Fourier Series Example
• Vpk / (Vo/2) = 0.63/0.5 = 1.26: This is for perfect square wave (includes all odd harmonics)
• Vpk / (Vo/2) ~ 9/8 = 1.125: This is just 3rd harmonic
©James Buckwalter 5
time
0
Vo/2
-Vo/2
2/p Vo=0.63 Vo
Class-F Waveform Analysis
• Is there power delivered to load at 2fo? No, V2,pk=0• Is there power delivered to load at 3fo? No, I3,pk=0• PRF=1/2 VFUND IFUND= 1/4 VFUND 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)
©James Buckwalter 6
Waveforms of Transistor Voltage(blue) and Current (black)
0
0.5
1
1.5
2
0 45 90 135 180 225 270 315 360 405 450 495 540 585 630 675 720 765
angle (degrees)
V, I
IDC=IRF/p just as for Class B
IFUND = IRF /2 just as for Class B
Vpk= RL(fo) Ipk
For Class F
Vmax=VDC+8/9 VFUND
Vmin=VDC- 8/9 VFUND
VFUND=(Vmax-Vmin)/2*9/8
Vmax= VDC+VFUND
Vmin=VDC-VFUND for class B
Class-F Amplifier Implementation:Accounting for Output Capacitance
©James Buckwalter 7freq (100.0MHz to 10.00GHz)
S(1
,1)
1.096E90.826 / 141.441
m1
2.089E90.793 / 99.795
m2
2.951E90.998 / -2.634
m3
m1freq=S(1,1)=0.252 / -79.574impedance = Z0 * (0.963 - j0.509)
1.096GHz
m2freq=S(1,1)=1.000 / -179.915impedance = Z0 * (1.514E-7 - j7.444E-4)
2.089GHz
m3freq=S(1,1)=0.998 / -2.634impedance = Z0 * (1.517 - j43.448)
2.951GHz
Class-F Amplifier
• Alternative Implementation
©James Buckwalter 8
RLmatch
Vdd
fo
lo/4Zo=RL
Z=RL at fo
Z=0 at 2fo, 4fo
Z=inf. at 3fo, 5fo,...
Short at all harmonics here
Class-F Example
• David Schmelzer and Stephen I. Long, CSICS 2006
• GaN FETs at 2GHz
• Class F amplifier
©James Buckwalter 9
86% PAE, 17W
Harmonic Load Tuning
©James Buckwalter 10
X2=Im(Znet) at 2fo
X3=Im(Znet) at 3fo
XL(f)
RLCds
Znet
Class FClass F
Class B
Other Approaches for High-Efficiency
• Control the voltage and current waveforms to prevent conduction while the voltage
• Class D: Switch current and voltage
• Class F-1
• Class E: ZVS and ZVS derivative switching
©James Buckwalter 11
SWITCHING PAS
©James Buckwalter 12
Switching-Mode Amplifiers• We have severe constraints from bias-point amplifiers in
terms of gain and efficiency.
• Switching-mode minimizes power dissipation in transistor:
– when voltage is high, current is zero
– when current is high, voltage is minimum
• Examples: Class E, Class D amplifiers
• But they require special considerations to operate linearly.
©James Buckwalter
13
Iout
Vout
Pdissdriver
Vh
time
Vsw
time
Isw
Transistor Model for PA
• Transistor acts like current source with Iout a linear replica of vin except in cutoff when vin<vTH
• To operate at high efficiency, we want to operate where the transistor DOES NOT want to behave like a current source.
• When Vout gets low enough, transistor acts like voltage source (in triode).
©James Buckwalter 14
Iout
Vout
Imax
Transistors in Triode
• Transistors are not perfect switches
– Finite rise/fall time
– Finite on-resistance
– Finite output capacitance
– Finite input capacitance
• Switch current is proportional to voltage across switch with conductance that changes between 0 and 1 with control voltage
©James Buckwalter 15
t = RONCOFF CDS
CGS
CGD
Class-D Amplifier
• Supercharged “inverter” or push-pull amplifier can operate as a class-D amplifier.
• Output network is constructed to provide a short at fundamental frequency
• Switches provide square wave voltage source
• Choose load for the combined switches to be nominally open circuit at odd and even harmonics
• In reality, need to optimize for given transistor.
©James Buckwalter 16
driver
Vdd
Vhfo
Class-D Amplifier Waveforms
• Switches alternately provide half-wave sine current
• Net output current is purely sinusoidal
• Sometimes called voltage-mode amplifier
©James Buckwalter 17
driver
Vdd
Vhfo
time
Vdd
Vh
time
time
IL
Iave
Isw1
Irf
timeIave
Irf
Isw2
Isw1
Isw2IL
Class-D Amplifier Analysis
©James Buckwalter 18
time
Vdd
Vh
time
time
IL
Iave
Isw1
Irf
timeIave
Irf
Isw2
PRF
=v
pk
2
2RL
Since the voltage is a square wave at the drain, the fundamental component that gets through output node is
v
pk=
2
pV
dd
PRF
=2V
dd
2
p 2RL
P
DC= I
DCV
DD
The average dc current is found from
IDC
=1
T
2VDD
p Rsin w
ot( ) dt £
2VDD
p 2R0
T /2
ò
PDC
= IDC
VDD
=2V
DD
2
p 2R
h =P
RF
PDC
=100%!!If this seems too good to be true…
Sources of Power Loss
1. On resistance of switches
2. Capacitance charge and discharge
3. Transient Power Loss (crowbar current)
©James Buckwalter 19
v
on= i
onr
on Assume the on current is constant P
diss= I
DC
2 ron
Ton
T
P
diss=
1
2CV 2 f
Pdiss
= i t( )v t( ) dt0
T
ò
Pdiss
=1
6I
DCV
DDf
Charge Dissipation
• Energy is lost each transition
• Energy loss is independent of on resistance.
• Want small output capacitance or small voltage swing!
©James Buckwalter 20
Class-D Amplifier Duty Cycle
©James Buckwalter 21
Eliminating Crowbar Current
• Change duty cycle of pull-up and pull-down
• Slow rise/fall times
©James Buckwalter 22
Non-overlapping clock generator
Input Capacitance
• As for most PAs, voltage gain is large.
• Power switch: Vin small, Vout large
• Input Capacitance Cin ~ Cgs + (1+Av) Cgd
• Cin is often dominated by Cgd
• This is different than in logic gates where Miller effect is not assumed to be significant
©James Buckwalter 23
Cgs
Cgd
Cin
vin vout
Class-D Amplifier Implementation
• Inverter is “easy” to understand (NMOS and PMOS).
• Other push-pull amplifiers are possible
©James Buckwalter 24
NOTE: NFET ONLY
Class-D Audio Amplifier
©James Buckwalter 25
Switched Capacitor PA
• Segmented inverters can be used for digital control
©James Buckwalter 26
Yoo, Allstot et al.U of Washington
Current Mode Class-D Amplifier
©James Buckwalter 27
Voltage
[V]
time[A]
timeCurrent
1dsI2dsI
1dsV 2dsV
M2on
M1on
0
0
M2on
M1on
Rload
Vdd
2
2p
Vddp
M2
VDD
M1
VDD
Rload
Waveforms are dual of VMCD
Entire current gets routed through M1 then M2Vload must be sinewave because harmonics are shorted
Current Mode Class-D Amplifier
©James Buckwalter 28
Zero Voltage Switching
• We recognize from the analysis of class-D amplifier operation.
• Loss due to device capacitance Cds rapid discharge when transistor switches, ½ C V 2.
• CMCD amplifier avoids this because V=0 when transistor switches from open to short ! “ZVS”
©James Buckwalter 29
Current-Mode Class-D Amplifier• Achieves Zero Voltage Switching (ZVS)
• Potentially more efficient at high frequency
©James Buckwalter 30
Device: Infineon CLY5
Balun : 50ohm coaxial cable
Rload = 50ohm
C out = 8 pF
RloadVin(-)Vin(+)
4p
4p
BalunC filter
L filter
Drain Efficiency = 79%
PAE = 72.5%
(includes Balun Loss)
Pout= 730 mW
H. Kobayashi et al (Fuji
Electric & UCSD)
10
15
20
25
30
0
25
50
75
100
-5 0 5 10 15 20
Pin vs. Pout and PAE
Pout (dBm)
PAE
Pout
[dB
m]
PA
E
Pin [dBm]
Inverse Class-F Switching Amplifier
• With different harmonic matching can get different waveforms, still with 100% efficiency. Class F-1 is dual of class F.
©James Buckwalter 31
Iout
VoutVddVmin
Imax
Vmax
Vdc=Vave =Vrf / p
Output current waveform has
fundamental, 3rd harmonic, 5th, etc
=> square wavetime
Vdc
Vce
time
Iave
Ic
match
Vdd
fo
lo/4Zo=RL
Z=RL at fo
Z=0 at 3fo, 5fo,…
Z=inf. at 2fo, 4fo,...
Still must worry about Cds to get Z
correct
Class F and Inverse Class-F
• Class F• Tends to minimize peak voltage excursion
• Good for devices with limited BV
• Requires open at 3rd harmonic, sometimes difficult with high Cds
• Can have CoutV2 losses
• Class F-1
• Has high peak voltage excursion
• Bad for devices with limited BV
• Requires open at 2nd harmonic, often easier for devices with high Cds
• Does not have CoutV2 losses
©James Buckwalter 32
timeIave
Idtime
Vo
Vds
time
Iave
Id
time
Vo
Vds
Class F-1
Class F
Harmonic Load Tuning
©James Buckwalter 33X1=0
X2=Im(Znet) at 2fo
X3=Im(Znet) at 3fo
XL(f)
RLCds
Znet
Class F-1
Class F-1
Class FClass F
Class B