CLASS F - University of California, Santa Barbara...Class-D Amplifier • Supercharged “inverter” or push-pull amplifier can operate as a class-D amplifier. • Output network

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


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


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


-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


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)


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


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


86% PAE, 17W

Harmonic Load Tuning


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


SWITCHING PAS


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).


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


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.


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


driver

Vdd

Vhfo

time

Vdd

Vh

time

time

IL

Iave

Isw1

Irf

timeIave

Irf

Isw2

Isw1

Isw2IL

Class-D Amplifier Analysis


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)


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!


Class-D Amplifier Duty Cycle


Eliminating Crowbar Current

• Change duty cycle of pull-up and pull-down

• Slow rise/fall times


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


Cgs

Cgd

Cin

vin vout

Class-D Amplifier Implementation

• Inverter is “easy” to understand (NMOS and PMOS).

• Other push-pull amplifiers are possible


NOTE: NFET ONLY

Class-D Audio Amplifier


Switched Capacitor PA

• Segmented inverters can be used for digital control


Yoo, Allstot et al.U of Washington

Current Mode Class-D Amplifier


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


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”


Current-Mode Class-D Amplifier• Achieves Zero Voltage Switching (ZVS)

• Potentially more efficient at high frequency


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.


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


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

CLASS F - University of California, Santa Barbara...Class-D Amplifier • Supercharged “inverter” or push-pull amplifier can operate as a class-D amplifier. • Output network

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