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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
Copyright Mark Rodwell, 2016
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!
Copyright Mark Rodwell, 2016
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
Copyright Mark Rodwell, 2016
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
Copyright Mark Rodwell, 2016
Class-B Loadline
©James Buckwalter
What does this say
about the possible
output power?
DDV
Copyright Mark Rodwell, 2016
Harmonic Matching for Class-B
©James Buckwalter
• At fundamental, generate loadline match
• At higher harmonics, ZL = 0
DDV
Copyright Mark Rodwell, 2016
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
Copyright Mark Rodwell, 2016
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
Copyright Mark Rodwell, 2016
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
Copyright Mark Rodwell, 2016
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
Copyright Mark Rodwell, 2016
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
Copyright Mark Rodwell, 2016
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
Copyright Mark Rodwell, 2016
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?
Copyright Mark Rodwell, 2016
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
Copyright Mark Rodwell, 2016
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
Copyright Mark Rodwell, 2016
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
Copyright Mark Rodwell, 2016
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
Copyright Mark Rodwell, 2016
Class B: how do we design the input match ?
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
Copyright Mark Rodwell, 2016
Class B: how do we design the input match ?
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
Copyright Mark Rodwell, 2016
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 ·
Copyright Mark Rodwell, 2016
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.
Copyright Mark Rodwell, 2016
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
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