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POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F ,...

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POWER AMPLIFIERS ©James Buckwalter 1
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Page 1: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

POWER AMPLIFIERS

©James Buckwalter 1

Page 2: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Power Amplifiers

• Purpose of a power amplifier– Generate high output power

– Efficient conversion of DC power to RF power

– Linear amplification

• Generally PAs will be – Common source

– Cascode

• Inductor is a “choke” to provide DC

• Capacitor is a “ac coupling” path to output

©James Buckwalter 2

Page 3: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Definition of Power

• Instantaneous Power

• Average Power

• There is a component that averages to zero that corresponds to energy stored and dissipated each cycle.

©James Buckwalter 3

P = v t( ) × i t( )

Pav =1

Tv t( ) × i t( )dt

t=0

T

ò

Page 4: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Definition of Power

• If the voltage and current is

• The average power becomes

• In phasor notation

©James Buckwalter 4

v t( ) =V cos wt( )

i t( ) = I cos wt +q( )

Pav =1

TVI cosq cos2 wt( )dt

t=0

T

ò =1

2VI cosq

v t( ) =V Re e jwt{ }

P =1

2Re vi*{ } =

1

2V

2Re Y{ } =

1

2I

2Re Z{ } Preac =

1

2Im vi*{ }

Page 5: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Example:

©James Buckwalter 5

Page 6: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Time Scales for Power Amplifiers

• Carrier T ~1ns, Period Ts ~100ns, Control ~ 1ms

©James Buckwalter 6

Page 7: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Power Scales for Power Amplifiers

©James Buckwalter 7

Page 8: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Large Signal Analysis

• Consider harmonic analysis

• Power

• In phasor

©James Buckwalter 8

P =V0I0 +1

2Vk Ik cos qk( )

k³1

å

v t( ) = Vk cos kwt( )k³0

å

i t( ) = Ik cos kwt +qk( )k³0

å

P =1

2Vk

2

k³1

å Re Yk{ } =1

2Ik

2

k³1

å Re Zk{ }

Page 9: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Harmonic Engineering

• Power is not simply V2

• May see strong harmonic content but power delivered at harmonics might be small due to impedance at harmonics

• Keep load reactive at harmonics.

©James Buckwalter 9

P =1

2Vk

2

k³1

å Re Yk{ } =1

2Ik

2

k³1

å Re Zk{ }

Page 10: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

PA Characteristics

©James Buckwalter 10

Voltage versus time Current versus time Loadline Impedance

Output versus Input Power Gain versus Power Efficiency versus Power

Harmonics

Page 11: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Notation

• Be careful about notation of voltage, current, and power.

• Average power

• For small signal, RMS voltage and current is common

• Therefore,

©James Buckwalter 11

vrms =v

2

P =1

2Re vi*{ }

P = Re vrmsi*

rms{ }

Page 12: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Units

• Peak Units

• RMS Units

v

RF= v

pksin w

RFt( )

vrms

=1

Tv

RFt( )( )

2

dt-T /2

T /2

ò

© James Buckwalter 12

Page 13: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

What does it mean to be “matched”?

• The antenna will appear to be 50 at the RF frequency so the input impedance of the receiver should be 50 .

• Why 50 ?

– It’s historical but an important (IEEE) standard.

– 30 Ohms offers the best power handling.

– 75 Ohms offers the lowest attenuation.

– All test interfaces are 50 .

©James Buckwalter 13

Page 14: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Power Amplifier Matching

• Loadline should be matched to 50 Ohm antenna output (or something like that).

• How do we do this?

©James Buckwalter 14

Page 15: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Power Delivered to a Load

Pdel ,L

=1

2Â v

LiL

*{ } =1

2Â i

LZ

LiL

*{ } =iL

2

2R

L

Pdel ,L

=v

S

2

2

RL

RS+ R

L( )2

+ XS+ X

L( )2

©James Buckwalter 15

Page 16: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Power Delivered to a Load

©James Buckwalter 16

Pdel ,L

=v

S

2

2

RL

RS+ R

L( )2

+ XS+ X

L( )2

• How do you maximize the power delivered?

• Vary the load while keeping the source impedance fixed.

Page 17: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Two Important Conclusions:Conjugate (Power) Matching

I)XS

= -XL

II)¶P

DL

¶RL

= 0® RS

= RL

Pdel ,L,max

=v

S

2

8RS

=v

L

2

2RL

©James Buckwalter 17

since vL

=R

L

RS+ R

L

vS

=1

2v

S

Page 18: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

What is the Penalty for Mismatch?

©James Buckwalter 18

Pdel ,L

=v

S

2

8RS

1

1+d X

2RS

æ

èç

ö

ø÷

Pdel ,L

Pav ,S

=1

1+d X

2RS

æ

èç

ö

ø÷

2

X

S= -X

L+dX

• The PA is very nonlinear at high power. This explains how the gain and power change under different conditions.

Page 19: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Characteristics of Transistor Amplifiers

©James Buckwalter 19

Page 20: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Ideal Assumptions

• The transistor acts like a current source

• Current controlled by input voltage as long as threshold condition is satisfied.

• Current is independent of output voltage as long as minimum/maximum output voltage is satisfied.

©James Buckwalter 20

Page 21: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Difference between Current Models

©James Buckwalter 21

Page 22: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Classes of Power Amplifier

• Bias point amplifiers

– Class-A

– Class-B

– Class-C

• Switching amplifier

– Class-E

– Class-F

©James Buckwalter 22

Page 23: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

PA Waveform

• Drain current consists of DC current and AC current

©James Buckwalter

iD = ID + id

id = ipk sin wot( )

vo = -id R

23

Page 24: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

PA Quiescent Conditions

• DC Conditions

VDS = VDD

IDS = IDD

• DC and AC Conditions

VDS = VDD+VAC

VAC = VOUT

IDS = IDD+IAC

IAC=-ILOAD=-VOUT/RL

©James Buckwalter 24

Page 25: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Transistors as Power Devices

• At high-power, the transistor is operating as a large signal device.

• Large voltage swing across drain-source junction

• Maximum voltage is VMAX~2Vddof transistor

• Minimum voltage is VMIN ~Vkwhere transistor begins to turn off

• Maximum current of transistor at VMIN is Imax

©James Buckwalter 25

Page 26: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Time Waveforms

©James Buckwalter 26

vin =VGG + vpk cos wot( )

iD = IDD - ipk cos wot( ) = IDD - iL

iL = ipk cos wot( )

vDS =VDD + iL RL

Page 27: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class A Operation

• Bias such that device never turns “off” (Vgs – Vt > 0)

• Note that the peak voltage across transistor is 2 Vdd!

©James Buckwalter

iD

= IDD

- ipk

cos wRF

t( )

v

DS=V

DD+ i

pkR

Lcos w

RFt( )

P

RF=

ipk

2 RL

2

27

Page 28: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-A Efficiency

• How much power consumption is required to keep the amplifier from turning off?

• Device power

• Maximum efficiency

©James Buckwalter

h =P

RF

PDC

=i

pk

2 RL

2IDD

VDD

£V

DD

2VDD

= 50%

IDD

³ ipk

VDD

³ ipk

RL+V

MIN

28

iD

= IDD

- ipk

cos wRF

t( )v

DS=V

DD+ i

pkR

Lcos w

RFt( )

P

dev= P

devdc( )+ P

devw( ) = I

DDV

DD-

1

2i

pk

2 RL

Page 29: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Voltage Swing Limitations

• Efficiency

• RF Voltage/Current

• This defines the limitation in delivering power

©James Buckwalter 29

h =P

o

PDC

=P

o

Pmax

Pmax

PDC

VRF ,MAX

=1

2V

MAX-V

MIN( )

VDD

=1

2V

MAX+V

MIN( )

VMAX

= 2VDD

-VMIN

IRF ,MAX

= IDD

PRF ,MAX

=1

2V

RF ,MAXI

RF ,MAX=

1

4V

MAX-V

MIN( ) IDD

PDC

=VDD

IDD

Page 30: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-A Efficiency vs RF Power

• Construct efficiency in terms of minimum device output voltage and d.c. voltage

©James Buckwalter 30

h =P

RF

PDC

=P

RF

PRF ,MAX

PRF ,MAX

PDC

h =P

RF

PRF ,MAX

1

2

VMAX

-VMIN( ) I

DD

VMAX

+VMIN( ) I

DD

h =P

RF

PRF ,MAX

1

2

2VDD

- 2VMIN

2VDD

=P

RF

PRF ,MAX

1

21-

VMIN

VDD

æ

èçç

ö

ø÷÷

Operating Power

Maximum Class-A Efficiency

Device FactorBack-off of amplifier

results in linear efficiency degradation with power

Page 31: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-A Output PowerDON’T THINK YOU NEED THIS SLIDE

• The peak output power can be re-written as a product of the maximum current for the device (where Imax = 2*Ipk) and the voltage swing (where ipk*R = Vdd-Vk)

• So called power triangle

©James Buckwalter 31

PRF

=i

pk

2 R

2=

1

2I

DD

VMAX

-VMIN

2

PRF

=1

4I

DDV

MAX-V

MIN( )

Page 32: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Power Triangle

• Another way to look at this is that the area under the IV curve should give the total power

©James Buckwalter 32

PRF

=1

2

IMAX

2

VMAX

-VMIN( )

2

PRF

=1

8I

MAX2V

DD-V

MIN( )Breakdown voltage (BV) may be larger than twice the VDD. More generally, we should take 2 VDD

Page 33: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class A Load-line Resistance

• Must increase current handling or voltage handling of device.

• Load line is defined by

• For Silicon CMOS, voltage is extremely limited. Large current means wide transistors and low output impedances.

©James Buckwalter

RLL

=V

MAX-V

MIN( )I

MAX

33

Page 34: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Example: 1 W Power Amplifier in CMOS

• You are asked to design a 1 W PA in a 1 V CMOS process.

• What is the size of the transistor assuming that the transistor can handle 1 mA per micron of DC current; i.e. 2mA per micron of Idd? The knee voltage is 0.3 V.

• What is the loadline resistance?

©James Buckwalter 34

Page 35: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Solution: 1 W Power Amplifier in CMOS

• Power delivered from PA:

• CMOS PAs handle high current!!!

©James Buckwalter

PDEL

=I

DD2V

DD-V

K( )4

IDD

=4P

DEL

2VDD

-VK

=4W

1.7V= 2.35A

35

Page 36: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Solution: 1 W Power Amplifier in CMOS

• The loadline resistance is

• This is an extremely small resistance. This makes matching difficult as we will discuss today!

©James Buckwalter

RLL

=V

MAX-V

K( )I

DC

=2.7V

2.35A=1.14W

36

Page 37: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

More Examples

©James Buckwalter 37

Page 38: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class A Load-line Resistance

• Load line can be defined under different conditions

• Different load lines selected for fixed Vdd

©James Buckwalter 38

P

RF=

1

8I

MAXV

MAX-V

MIN( )

h =P

RF

PRF ,MAX

1

21-

VMIN

VDD

æ

èçç

ö

ø÷÷

Page 39: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Power/Efficiency Trade-off

• Changing RL changes efficiency and output power.

• Larger loadlineresistance achieves higher efficiency at expense of power.

©James Buckwalter 39

Page 40: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Changing Loadline Resistance for Efficiency

• For fixed Vdd, changing the loadline, the output power reduces while the efficiency increases.

• This is an important observation used for Doherty, Chereix, and Envelope Tracking

©James Buckwalter 40

Page 41: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

CLOSER LOOK AT LOADLINE

©James Buckwalter 41

Page 42: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Effect of Output Capacitance

• Loadline is not static when transistor model is no longer quasi static

©James Buckwalter 42

iD

= IDD

- ipk

cos wRF

t( )+CDS

d

dtv

OUT

id

= -ipk

cos wRF

t( )+ jwRF

CDS

vOUT

Page 43: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Matching Conditions

• 1) DC loadline considerations dictate choice of RL to optimize Pout or PAE combination. (not necessarily gain)

• 2) Imaginary part of load admittance is established to match the imaginary part of device

©James Buckwalter 43

BL = -Á YO{ }

Page 44: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Simulating the Reactive Loadline

• De-embed the Cds and Cdg values and add these as “negative” capacitors to get a quasi-static loadline.

©James Buckwalter 44

Page 45: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Simulation of Waveforms

• Now we can look inside transistor to get loadline

©James Buckwalter 45

Page 46: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Load Pull Contour

• Load pull measurement (simulation) consists of systematically varying load impedance and measuring amplifier characteristics

• Load pull contours describe how Pout (efficiency, linearity, etc) vary with ZL

©James Buckwalter 46

Page 47: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Load Pull Contours

©James Buckwalter 47

Page 48: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Differences in Load Pull for Power and PAE

©James Buckwalter 48

Page 49: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

PA Design Methodology

1. Choose Vdd based on transistor technology with sufficient power handling capability and breakdown voltage

2. Using DC characteristics choose load line and verify the Pout can be obtained.

3. Determine the input impedance matching network using the bias condition on the average DC current corresponding to average output power

4. Determine the load susceptance and match the output to obtain RL and BL

5. Provide output match at harmonic frequencies6. Set-up bias network7. Optimize using simulator.

©James Buckwalter 49

Page 50: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Overdriven Class A

• You might wonder what happens when we turn a class A amplifier to “11”

• In general, you get higher output power and efficiency in the overdriven regime but this is difficult to analyze

• Vin determine Idrain through the nonlinear transistor characteristics

• Vdrain depends on the load impedance at different harmonics

• If Vdrain is too low, the transistor is forced into triodand becomes a voltage source causing current collapse.

©James Buckwalter 50

Page 51: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

©James Buckwalter 51

Page 52: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Power Added Efficiency

• Power added efficiency incorporates the RF power lost into driving the device.

• The PAE is never more than the drain efficiency.

• High gain is required from the PAE to make the amplifier efficient.

©James Buckwalter

, , , ,

,

1

11

rf out rf in rf out rf in

DC DC rf out

P P P PPAE

P P P

PAEG

52

Page 53: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

PAE

• Gain can be changing with output power as device is pushed into compression

©James Buckwalter 53

Po

=i

pk

2 R

2=

gmv

in( )2

2R

G =P

o

Pin

µ gm

2 R

Page 54: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

CLASS B

©James Buckwalter 54

Page 55: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-B Operation

• Drain current on for half the cycle

©James Buckwalter

iD

=i

pkcos w

ot( ) i

d> 0

0 id

£ 0

ì

íï

îï

vpk

= -iD

2Rcos w

RFt( )

P

RF=

vpk

2

2R£

VDD

2

2R

55

Note that the current is half-wave rectified by the voltage is a sine wave!

Page 56: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Fourier Components for Half-Wave

• Even harmonics result from half wave current

©James Buckwalter 56

iD

=i

pkcos w

ot( ) i

d> 0

0 id

£ 0

ì

íï

îï

iD

f( ) =i

pk

p+

ipk

2sin 2p ft( ) -

2ipk

p

cos 4pkft( )4k 2 -1k³1

å

Page 57: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-B Loadline

©James Buckwalter 57

What does this say about the possible output power?

Page 58: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Harmonic Matching for Class-B

©James Buckwalter 58

• At fundamental, generate loadline match

• At higher harmonics, ZL = 0

Page 59: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Power Dissipation in Class A and Class B

©James Buckwalter 59

Page 60: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-B Current Analysis

• IQ should be zero for DC (no power consumption with no input).

• However, quiescent current is obviously not the “average” current.

• To calculate the d.c. power consumption, the d.c. average power

• Looking at the half-wave drain current, we could also calculate

©James Buckwalter 60

IDC

=1

2piPK

sin q( ) dq £iPK

p0

p

ò

IFUND

=1

piPK

sin2 q( ) dq £iPK

20

p

ò

Page 61: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-B Efficiency

• The output power and DC power are

• Therefore, the maximum drain efficiency is

©James Buckwalter

h =P

RF

PDC

=

vPK

2

2RV

DDiPK

p

=p

2

vPK

2

RVDD

iPK

=p

4

vPK

VDD

®hMAX

= 78%

61

PRF

=1

2v

PKiPK

=1

2

vPK

2

R=

1

2R

iPK

2

æ

èçç

ö

ø÷÷

2

PDC

=VDD

IDC

=VDD

iPK

p

Page 62: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-A versus Class-B

• Comparing power of Class-A and B

• Comparing efficiency of Class-A and B

• What is the cost of class-B amplifier?

©James Buckwalter

PRF ,A

=V

DD

2

2RP

RF ,B=

VDD

2

2R

50% 78%A B

62

Page 63: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-B Efficiency Revisited• How does efficiency depend on knee voltage?

• The efficiency depends on the output power

• The drain efficiency changes with the root of the output power. Recall for class-A!

©James Buckwalter 63

h =p

4

VRF ,MAX

VDD

VRF

VRF ,MAX

=p

4

VMAX

-VMIN

VMAX

+VMIN

PRF

PRF ,max

h =p

41-

VMIN

VDD

æ

èçç

ö

ø÷÷

PRF

PRF ,max

h =1

21-

VMIN

VDD

æ

èçç

ö

ø÷÷

PRF

PRF ,MAX

Page 64: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Efficiency vs. Power

©James Buckwalter 64

Very important! Note that not only is the peak power higher the efficiency is higher in backoff.

When the average power is 3 dB lower than the peak, what is the efficiency for class-A and class-B?

Page 65: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-B Loadline

©James Buckwalter 65

How do we match to this loadline?

Page 66: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-B Loadline

©James Buckwalter 66

• The piecewise loadline shown in the previous slides seems to suggest a loadline.

• However, we should consider the loadlinebased on the fundamental. Therefore, the voltage swing at the fundamental is (Vmax –Vmin )/2. The current swing is Imax/2.

• The loadline resistance is the same as class-A!

RL =Vmax -Vmin

Imax

Page 67: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Maximum Output Power for Class-B

• Again, this is the same as class-A output power

• So what is the penalty?

©James Buckwalter 67

PRF

=1

2v

RF ,MAXiRF ,MAX

PRF

=1

8V

MAX-V

MIN( ) IMAX

Page 68: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Gain for Class-B

• Compared to the gain for Class-A amplifier the power gain of class-B is ¼ (or 6 dB) less.

• What is the implication of this? Lower PAE!

©James Buckwalter 68

P

RF=

ipk

2 R

8=

gmv

in( )2

8R

Page 69: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Gain

©James Buckwalter 69

Page 70: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Features of Class-B

©James Buckwalter 70

Page 71: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

CLASS C

©James Buckwalter 71

Page 72: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-C Amplifier

©James Buckwalter 72

• Like class-B, IQ= 0

• Unlike class-B, conduction

angle < 180 deg

• Gain is low since device is

turned on for only a short

period

Iout

VoutVoVmin

Imax

Vmax

time

time

Vo

Vce

Idc

Ic

Vrf

match

match

VoRL

Harmonics are

shorted

Page 73: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-C Amplifier

©James Buckwalter 73

• IDC < IRF / p

• PRF < 1/4 VRF IRF

• PDC = VDD IDC

Iout

VOUTVDDVMIN

Imax

VMAX

time

time

VDD

Vds

IDC

Id

VRF

match

match

VoRL

Harmonics are

shorted

Page 74: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Load Impedance Is A Resonant Network

©James Buckwalter 74

RLmatch

Vo

fo

Z=RL at fo

Z=0 at 2fo, 3fo, 4fo,…

Short at all harmonics here

Z

10 WAt fo -j2W j2W

10 WAt 2fo -j1W j4W

Representative Z values

fo

2fo

Page 75: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-C Waveform

©James Buckwalter 75

timeIdc

Ic

I t( ) =iPK cosf - IS when iPKcosf > IS

0 otherwise

ìíï

îï

cosf =IS

iPK

®IS is the (negative) offset

iPK is the amplitude

IDC =1

2pI q( )

0

f

ò dq =1

2piPK cosq - IDD( )

0

f

ò dq

IDC =1

piPK sinf - IDDf( )

IFUND =1

pI q( )

0

2p

ò cosqdq =2

pI q( )

0

f

ò cosqdq

IFUND =2

piPK cos2 q - IS cosq( )dq

0

f

ò

IFUND =2

p

1

2iPKf +

1

4iPK sin2f - IS sinf

æ

èç

ö

ø÷

IFUND =2

piPK

1

2f +

1

4sin2f - cosf sinf

æ

èç

ö

ø÷

IFUND =iPK

pf -

1

2sin2f

æ

èç

ö

ø÷

Page 76: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-C Conduction

• Note the angle φ defined for the waveform analysis is related to the conduction angle of the transistor by a factor of two, i.e. 2φ=Φ

• Φ is the conduction angle during which the current conducts through the transistor. What happens for Φ of pi or 2pi?

©James Buckwalter 76

IFUND =iPK

p

F

2-

1

2sinF

æ

èç

ö

ø÷IDC =

iPK

psin

F

2-

F

2cos

F

2

æ

èç

ö

ø÷

Page 77: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-C Efficiency

• As conduction angle approaches 0, the efficiency approaches 100%.

• What is the penalty?

©James Buckwalter

h =P

RF ,MAX

PDC

=v

PKiPK

2VDD

IDC

=F- sin F( )( )

2 2sinF

2

æ

èç

ö

ø÷-Fcos

F

2

æ

èç

ö

ø÷

æ

èç

ö

ø÷

77

Page 78: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class C Waveform Analysis

• Calculate efficiency in presence of

©James Buckwalter 78

h =1

2

vPKiPK

VDC IDC

=1

2

vpk

VDC

ipk

pipk

p

F - sin F( )

2sinF

2

æ

èç

ö

ø÷-Fcos

F

2

æ

èç

ö

ø÷

=Vpk

VRF ,MAX

VRF ,MAX

VDC

f F( )

h =VFUND

VFUND,MAX

VFUND,MAX

VDC

f F( )

h = f F( ) 1-VMIN

VDD

æ

èç

ö

ø÷

POUT

POUT ,MAX

f 0( ) =1 f p( ) =p

4f 2p( ) =

1

2

f F( ) =1

2

F- sin F( )

2sinF

2

æ

èç

ö

ø÷-Fcos

F

2

æ

èç

ö

ø÷

Page 79: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Power Amplifier Comparison

©James Buckwalter 79

Page 80: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-C Amplifier Efficiency

• Class C has very good efficiency because whenever the

device has current, Vds is particularly low

©James Buckwalter 80

time

time

Vo

Vds

Idc

ID

Vrf

Page 81: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Class-C Waveform Analysis

• How about the loadlineresistance?

©James Buckwalter 81

RL =VFUND

IFUND

VFUND =VMAX -VMIN

2

IFUND = IS

F

2-

1

2sinF

æ

èç

ö

ø÷

IMAX = IS - IO = IS 1- cosF

2

æ

èç

ö

ø÷

IFUND =IS

p

F

2-

1

2sinF

æ

èç

ö

ø÷ =

IMAX

2p

F - sinF

1- cosF

2

RL = pVMAX -VMIN

IMAX

2 1- cosF

2

æ

èç

ö

ø÷

F - sinF

0

2

4

6

8

10

12

14

16

18

20

0 15 30 45 60 75 90

Theta (degrees)

RL

/ R

L C

lass A

Iout

VoutVoVmin

Imax

Vmax

Page 82: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Gain and Conduction Angle

©James Buckwalter 82

Pin (dBm)

Gain (dB)

6dB

Class A

Class B ideal

Class B real

Class AB "ideal"

Class AB "real"

Class C "real"

Page 83: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Power Amplifier Comparison

©James Buckwalter 83

• Maximum voltage swing is 2Vo – Vmin

• 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.

Page 84: POWER AMPLIFIERS - UCSB€¦ · h= P R F P D C = P R F R F , MAX P R F , MAX P D C h= P R F P R F , MAX 1 2 V MAX - V (MI N) I D D V MAX + (I N) I D h= P R F P R F , MAX 1 2 2V D

Other Classes of Amplifiers

• PA research is focused around getting high-power 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 84


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