Power Amplifiers: Class B - ece.ucsb.edu · Power Amplifiers: Class B ... this is the same as class-A output power ... •Compared to the gain for Class-A amplifier the power gain

Copyright Mark Rodwell, 2016

Power Amplifiers:

Class B

Mark Rodwell, University of California, Santa Barbara

ece145c lecture notes

with Material copied from Prof. Buckwalter's 145c notes


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

Note that the current is

half-wave rectified by

the voltage is a sine

wave!


Fourier Components for Half-Wave

•Even harmonics result from half wave current

©James Buckwalter

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


Consider the currents.

iD

f( ) =i

pk

p+

ipk

2sin 2p ft( ) -

2ipk

p

cos 4pkft( )4k 2 -1k³1

å

!)2/(2 :at current signalpeak -Peak

)2sin(*)2/( :frequency at current Signal

:torin transiscurrent ousinstantanePeak

pkpkpp

pk

pk

IIIf

ftIf

I

torin transiscurrent peak

frequency signalat

ppI


Class-B Loadline

©James Buckwalter

What does this say

about the possible

output power?

DDV


Harmonic Matching for Class-B

©James Buckwalter

• At fundamental, generate loadline match

• At higher harmonics, ZL = 0

DDV


Power Dissipation in Class A and Class B

©James Buckwalter


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%

PRF

=1

2v

PKiPK

=1

2

vPK

2

R=

1

2R

iPK

2

æ

èçç

ö

ø÷÷

2

PDC

=VDD

IDC

=VDD

iPK

p


Class-A versus Class-B: Efficiency at Peak Output

•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


Class-B Loadline

©James Buckwalter

How do we match to this loadline?


Class-B Loadline

©James Buckwalter

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

• However, we should consider the loadline based 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


Maximum Output Power for Class-B

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

• So what is the penalty?

©James Buckwalter

PRF

=1

2v

RF ,MAXiRF ,MAX

PRF

=1

8V

MAX-V

MIN( ) IMAX


Efficiency vs output power: Class B

modified from J Buckwalter's 145c notes

41

414/)(/

efficiencyector Drain/coll

/

4/)(8/

: driven to *not*When

/)(22/)(but /)(

:)( conditionsoutput peak At

max,

ectordrain/coll

max

maxectordrain/coll

max

22

maxmaxmaxmax

DD

sat

out

out

DD

satpk

DDsatDDpkDCout

pkDDDC

satDDpkLoadppout

sat

satDDloadsatDDsatload

sat

V

V

P

P

V

V

I

IVIVVIPP

IVP

IVVIRIP

P

IVVRVVVIVVR

P


Efficiency vs output power: Class A

modified from J Buckwalter's 145c notes

2

11

2

112/)(/

efficiencyector Drain/coll

2/

4/)(8/

: driven to *not*When

/)(22/)(but /)(

:)( conditionsoutput peak At

max,

ectordrain/coll

2

max

2

max

2

ectordrain/coll

max

max

22

maxmaxmaxmax

DD

sat

out

out

DD

satpk

DDsatDDpkDCout

DDDCDDDC

satDDpkLoadppout

sat

satDDloadsatDDsatload

sat

V

V

P

P

V

V

I

IVIVVIPP

IVIVP

IVVIRIP

P

IVVRVVVIVVR

P


Efficiency vs. Power

©James Buckwalter

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?


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

P

RF=

ipk

2 R

8=

gmv

in( )2

8R


Gain

©James Buckwalter


Features of Class-B

©James Buckwalter


Class BDesign Example


Schematic

harmonics shorts ;at filter bandpassoutput :TL16 Note

network. tuningloadlineoutput Note

current smallat BJT bias mirror,Current .network bias base idealized theNote max

f

I


Design procedure

?) (howbandwidth over Stabilize ?). (howinput Match

network. tuningload real with load idealize replace ,determined With

loadline.correct for er) transform(inductor, load idealized Tune

loadline.output generate tosufficientpower input with Drive

???)mean matched"" does(what matched. *not*initially Input

A. class Similar to

,optLZ


Simulation results: Class B


Simulation results: Class B

?match input design the wedid how---But


Class B: how do we design the input match ?

? drive signal-largeunder measure wedo How

drive. signal-largeunder

fromgreatly differ willdrive signal-smallunder

off.almost isr transisto thedrive, RF zerounder :But

load. thefirst tunemust weso ,upon depends always, As

in

in

in

loadin

Z

Z

Z

ZZ



0signallarge

0signallarge

signallarge

1

signallargesignallarge

:tcoefficien reflectioninput signal large a Define

frequency at ofcomponent Fourier

frequency at ofcomponent Fourier

:admittanceinput signal large a Define

ZZ

ZZ

ZfV

fIY

in

in

in

in

in

inin



not work. does matching signal-small why is this; with varies that Note

match. Signal-Large a called is This

zero. to bring toadjusted isnetwork nginput tuni thechart,Smith on the displayed With

signallarge

signallargesignallarge

inin

inin

P


Class B: what about stabilization ?

.fat driven PA fmax with -DC fromZout Zin,negativefor Check

:do) tohard(but Practical

chaos. and dynamicsnonlinear :yMath/Theor Underlying

2GHz.at drive RF stronggiven 1.0GHzat unstable,y potentiall bemight circuit The

drive. RF nogiven 1.0GHzat example,for stable,nally unconditio bemight circuit The

amplitude. driveinput RF offunction a asgreatly change parameterscircuit The

:discussiondifficult a inherently is This

signal

Low distortion: transmitters or receivers ?

-150

-100

-50

0

50

100

-50 -25 0 25 50 75

Pou

t (d

Bm

)

Pin (dBm)

PDC

= 979 mW

f1,2

= 1.96, 1.98 GHz

OIP2 = 76.5 dBm

OIP3 = 52.4 dBm

IM3IM2

Linear

Receivers:very low IM3: jammer rejection→ very high IP3 @ low PDC.

http://www.mathworks.com/

Griffith, 2008 CSICSTransmitters:

IM3→ out-of-band power (ACPR).want low IM3 given output near Psat.

Inoue et al, SEI Technical Review, 2014 ·


Class AB or B ? Push-Pull or just Pull ?

Class AB or class B ? Pick bias point for smallest V3 term at crossover

Push-pull or just pull ? Push-pull & 2nd-harmonic short-circuit are equivalent.

Push-pull

2nd harmonicshort-circuit

V Paidi et al (UCSB)IEEE Trans MTTFeb. 2003

vs.


Linear W-band HBT class A, AB power amplifier100 GHz PA simulations10 mm HBT emitter finger (TSC 250nm InP HBT)cell within larger PA; power scales by # fingers.recall: 2-tone clipping point is 3dB below single-tone

Performance (simulated)class A: -43 dBc IM3 at hard amplifier clipping ; less below.class AB: -40dBc IM3, or better, below clipping point

-80

-70

-60

-50

-40

-30

-20

-10

0

0.2

0.4

0.6

0.8

1

-15 -10 -5 0 5 10 15

Class A: Simple vs. Emitter Degeneration

IM3

, d

Bc, (t

wo-t

one

in

put)

PA

E (o

ne

-ton

e in

pu

t)

Output power, dBm (single tone, or sum of two tones)

100GHz

250nm InP HBT

two-tone

clipping-80

-70

-60

-50

-40

-30

-20

-10

0

0.2

0.4

0.6

0.8

1

-15 -10 -5 0 5 10 15

Class AB: Simple vs. Emitter Degeneration

IM3

, d

Bc, (t

wo-t

one

in

put)

PA

E (o

ne

-ton

e in

pu

t)

Output power, dBm (single tone, or sum of two tones)

100GHz

250nm InP HBT

two-tone

clipping

Park et al, JSSC Oct. 2014


Class B and linearity

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from J Buckwalter's 145c notes


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Documents

Power Amplifiers: Class B - ece.ucsb.edu · Power Amplifiers: Class B ... this is the same as class-A output power ... •Compared to the gain for Class-A amplifier the power gain