1

POWER AMPLIFIERS

General Block Diagram of Cellular Radio; where is the PA?

ELEN 665 (ESS)

TAMU-AMSC

PA

Dup

lexe

r

Transmitter

Receiver

Synthesizer

A/D and D/CCircuits

ReferenceOscillator

Micro-processor

Voice FrequencyInterface

DigitalSignalProcessor (DSP)

2

Power Amplifier System Level Considerations

MatchingNetwork

Matching Network

Matching Network

inv

1N 2N3NDriver

OutputStage

LR

Typical PA Block Diagram System

• Why can designers use only an amplifier? like this:

To deliver for instance a 1W of power to a 50 Ω antenna with common-sourceis not feasible.

DDV DDV DDV

LR

LRLR

outIRFC

XinV inV

inVM1 M1 M1

Common-Source Stages

3

• Power-Amplifier Metrics

– Power Efficiency PAη

inDC

outRF

ply

loadPA P

PPP

−

−==sup

η

%PA 100 Ideal =η

– Power-Added Efficiency ( Linear concept)

⎟⎠⎞

⎜⎝⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

−=−

=

GPP

PPPAE

PP

PP

PPP

PAE

PAload

input

ply

load

ply

input

ply

load

ply

inputload

111sup

supsupsup

η

– Output-Referred Compression Po-1dB ( non-linear concept)

dBindBo

dBindBo

PGPPGP

11

11

log10 −−

−−

+=Δ

inRF

outRF

PPG

−

−=where

4

Power Amplifier for GSM Handset

Typical Requirements of Power Amplifier:

• Positive supply only• 890 to 915 MHz transmit frequency (GSM 900)• 3 Watts power out of amplifier ( 1.2 W allowed but not popular)• Efficiency > 40% (Market place issue)• < 4 dB compression (Affects efficiency, limited by spurious signal accentuation)• Gain Control capability > 34 dB (Allows for variation in other parts of transmit path)• Supply voltage: 5 V (min.) to 8.5 V (during battery charge, in operation)Because of inductive loads, output device breakdown requirement may be > 17 V.

≈≈

System Control Requirements:

• Strict turn-on time template with limits on spurious outputs during turn-on• Control Range; 28 dB in 2 dB steps• Absolute tolerance; ± 2.5 dB for highest power and ± 6 dB for lowest power• Flatness control of output pulse (± 1.0 dB)

5

Power Amplifier for GSM Handset, Three Stages with Power Control

Voltage ControlledAttenuator(GaAs MESFET)

Medium Power2nd StageWith Distortion

High PowerThird Stage

Coupling Circuit

Ω50Class AFirst Stage

Filter forHarmonics

PackageParacitics

Coupling Circuit

Ω50Load

Key Questions:

• Q. Is the inter-stage band-pass filter necessary?A. Yes. The distortion requirements are severe. Desired efficiency cannot be met

without some compression, even on the second stage.

• Q. Why not include output coupling circuit inside package?A. Only low Q inductors are obtainable on-chip, which affects efficiency.

• Q. Why not a 2 or 3 Volt supply?A. Impedance level is so low (< 3 Ω) that package inductance and Q still present

problems.

6

Effect of PA Clipping on Phase Error

Ideal GSM (GPSK) SignalHas No AM (Flat Envelope)

Signal Vector Only

Carrier Feed-Through (Fixed)

Final RFVector

SignalVector(Rotating)

Clipping Cuts theFinal RF Vector to aConstant Amplitude

Crossover Phase (φ) is UnaffectedBut Signal Phase (θ) is Modified

QI

Clipping Causes NoChange in Phase

QI

φ φ θθ'θ

θ' > θ

7

How to relax power dissipation across the transistor?One solution to alleviate this situation follows:

In general, for practical PA designs it is required matching networks to increase (and distribute) power gain.

(i) Matching network as voltage amplifier, (ii) use of a transformer asa matching network.

CLASS A AMPLIFIERS

Rv

P

RiP

tsinRivtsiniIi

RF,oRF

RFRF

oRFRF,o

oRFDCD

2

22

2

=

=

−=ω+=

. MatchingNetwork

. ..DDV DDV

RFC RFC

VinVinM1 RL

XM1

1:4

50Ω

5016 Ω

(i) (ii)

8

RVv

P DDRF,oiplysup =

Thus, the power efficiency for a class A amplifier yields:

DD

RF,oA V

v2

=η

Observe that in order to keep the transistor in saturation, which becomes

TGSDSATDS VVVv −=>

DSATDDRF,oSAT

SATDDDDST VVv;

kLEELkIII

V −≤++

=222 2

where

( )SAT

tGSTGSDoxo LE

VV;VVkI,LWCk −

=θθ+

−=μ=

12

2

ESAT is the field strength in V/m at which the carrier has reached velocity saturation. Furthermore, remember that:

( )

⎪⎪⎩

⎪⎪⎨

⎧

μπ

−μ

=π

≅channelshort for

41

channel longfor 43

2

2

LE

LVV

Cgf

SAT

TGS

gs

mT

9

Similarly for the BJT

⎪⎪⎩

⎪⎪⎨

⎧

πωμωπ

μ

=πτ

≅saturationvelocity

4

injection level-low

21 2

,E

,qkT

f

B

SAT

B

FBIP,T

Typical values are

V.V,.,GHz~fV.VV,.~f

Vm.,Vm.,mvE

bemBBJT,T

tGSmNMOS,T

SBULKS

channelSAT

800503080250 , GHz 2520

210551051104 2226

=μ=ω

=−μ−

×=μ×=μ×≅ −−

Example. Assume VDD=3.V and VDSAT=0.42

segmA

eqmV

DD

RF,oA

RRgG

RgA

....

Vv

2

4203325033

2

=

−=

=×−

==η

10

Rs is the source resistance, and LOAD

RF,oeg P

vR

2

2

= Typically G ~ 10dB, then the PAE is

at least 90% of the power efficiency.

Let us now explore the output 1-dB compression point.

Recall that

( )( )21

501θ+θ+

−μ

=.VV

LWCg TGS

oxm

Then

s

.

dB

eqdBsmdBeq

A

RaaP

RPRgPRG

11092 050

3

11

12

12

−=

=

−

−−

a1 and a3 are the coefficients of the non-linear polynomial of gm.

.

11

Then the output compression point becomes:

( )⎥⎦

⎤⎢⎣

⎡θθ+

×=−load

eq

dBs P..logRPRa

3212

1501103384410

Example.- VDD = 3.3V, VDSAT = 0.5 (in a 0.25μm CMOS technology), and a transmit power of 600 mW. (27.8dBm).

Then ( ) dBm.P..log

..loadP

load 13050110338441050

60

32 =⎥

⎦

⎤⎢⎣

⎡θθ+

×=θ

=

Note that 30.1dBm is about 2.3dB higher thant the Po-1dB

( ) ( )Ω=

×−

=−

=

=×−

=−

==η

566025033

2

423325033

2222

....

PVVR

%...

VVV

Vv

load

DSATDDeq

DD

DSATDD

DD

RF,oA

12

The corresponding bias current is

mA...

..IeqR

DSATVDDVD 8304

565033=

−== −

From the expression of ID one can determine W

( )( )

( )mm.

......W

EVCVLEI

VVCLIW

SATDSATox

DSATSATD

tGSox

62410450003500150

5010410250430802

2012

62

66

22

=×⋅+×⋅×

×=

μ+

=−μ+

=

−

Its corresponding transconductance becomes:

( ) ( )

dB.PdB....RRgG

.g..

......

LWICg

dB,insRseqm

m

.DOX

m

813825056441

441501501

1025043080024600035001502

15012

250

2

23

2

651

=⇒=××==

=++

×××××

=θ+θ+μ

=

=

−

Ω

13

The efficiency for class-A power amplifier is rather modest, thus other efficient amplifier are investigated.

Observe that class C, D, E and F amplifiers are essentially constant-envelope amplifiers.

<%45

<%78

DD

RF,oA V

v2

=η

↑

↑=πη=π

=η

Efficiency

Distortion4

504

ABA

DD

RF,oB

GG,.Vv

( )φφ−φφ−φ

=ηcossin

sinC 4

22

switchesidealwith 100%IDEAL,D =η

RV.

RVPmax,% DDDD

LOADIDEAL,E

22

5770421

2100 ≅π+

==η

%IDEAL,F 100=η

CLASS A

0

4

CLASS B

0

4

CLASS C

0

4

CLASS D

0

4

CLASS E

0

4

Dv Di

RFDC iI +

φ2

CLASS F

0

4

θπ π20

Waveforms for Ideal PAs.

14

CLASS-B POWER AMPLIFIER

Class B stage using a transformer Transformer-coupled push-pullPA. For audio applications. Complementary PA.

For Broadband Operations.

• 50% duty cycle

RVdttsin

RV

Ti

RVP;

Rv

P

tsinRiv

RVimax;idttsintsini

Ti

DDo

DDaverage,D

DDmax,o

RF,o

oRF

RF,o

DDRF

RFooRF

πω=

==

ω≅

==ω×ω=

22 122

2

22

2

220

2DI

X2M

DDV

Y1M

1DI

1T

LR

o0

1Q

DDV+

o1802Q

1T

OUTPUTFILTER

R.OUTPUTFILTER

.

DDVoo

o1801Q

2Q

R

∫T/2

o

∫T/2

o

15

78504

thus

2 2

.P

Pmax

RVP

DC

max,oB

DDDC

≅π

==η

π=

• Class B scarifies linearity but gains in terms of efficiency.

• For practical implementations

10peak

Di

I ≅

16

CLASS E AMPLIFIER

t

XV1DI

t

XV1DI

dtdVX /

Voltage and current waveforms in a class E stage.

(a) (b)

Class E stage.

. .DDV

RFC

1M

X

inV

2C 1LoutV

LR1C

..X 2C 1L

1CLR

t

XV

Response of class E stage when the transistor turns off.

17

BASIC RF SWITCHING AMPLIFIER

dcV

dcI

swIswV

DCC

LR

( )θswIpkI

dcI

0 α−0 α π π2

( )θswVpkV

dcV

0 α− α0 π π2

Basic RF switching amplifier. Basic RF switch waveforms.

( )

( )πα−π

=

θπ

=

θθπ

=

pk

dc

pk

swdc

VV

dV

dvV

1

21

(1)

∫

∫

π

-π

π

α

18

The voltage waveform can be considered to be an alternating voltage with zero mean value if it is offset by Vdc. So the peak-to-peak current swing will be Vpk/RL and

Lpkpk R/VI = (2)

pkdc IIπα

= (3)

The fundamental even Fourier coefficient of current, Ip is given by

( ) ( ) θθθπ

= dcosiI sw 11 (4)

where( )

( )

( )πα

=

θθπ

=

π<θ<αα<π−=

α<θ<α−=θ

sinII

dcosI

,,

,Ii

pk

pk

pksw

2

20

1

(5)

∫π

−π

∫α

o

19

Similarly, the fundamental Fourier coefficient for the voltage waveform is

(6)( )πα

−=sin

VV

pk

21

Combining (6.1) through (6.6), the RF power, Prf can be expressed in terms of the dc supply terms, Vdc and Idc.

( )α−παα

=

αα

=

α−πα

=

2

1

1

2

2

2

sinIVP

sinII

sinVV

dcdcrf

dc

dc

(7)

So that the output efficiency, is given by,η

( )α−παα

=η22 sin (8)

20

π2/π00

50%

100%

1.0

2.0

RFPower(Rel.)

Effcy

LINPP/1

Conduction angle ( )α

LINPP/2

LINPP/3

RF output power and efficiency of basic RF switch. Fundamental and harmonic power are expressed relative to Plin the class A RF power having the same peak RF current at the same dc supply voltage.

and defining the linear power, Plin as

4pkdc

linIV

P =

the relative power is

( )α−ππα

=2

1 8sinPPlin

(10)

(9)

21dcV

pkV

π 2π0

0π−

)(θswV

TUNED RF SWITCHING( )

πα

=

πα

=

pk

dc

pk

II

sinII 21

The sinusoidal voltage gives the simple relationshipdcVV =1

( )θswIpkI

dcI

0π− 0 α π π2α−

Tuned RF switching amplifier.

Tuned switch waveforms.

DDV

dcI

swIswV

DCC

LR

“Tank”circuit

22

π2/π00

50%

100%

1.0

2.0

RFPower(Rel.)

Effcy

LINPP/1

Conduction angle ( )α

RF output power and efficiency of basic RF switch with harmonic short.

23

Thus the fundamental RF power is( )αα

=sinVIP dcdc1

Yielding this output efficiency:( )αα

=ηsin

The relative power can be determined as before, by expressing P1 (in 11), in terms of the peak current Ipk using the relationship of (3):

( )πα

=sinVIP dcpk1

So the ratio of fundamental RF power to Plin is becomes

( )⎟⎟⎠

⎞⎜⎜⎝

⎛=

πα

=4

41 dcpklindcpk

lin

VIPsinVI

PP

(11)

(12)

(13)