OVERDRIVEN AMPLIFIERS
©James Buckwalter 1
Overdriven Amplifiers
• For very large input signals, the output waveform is driven into the "saturation" region (bipolar) or "linear" region (FET) - and becomes limited by the on-resistance of the device.
• Waveform behavior is determined by harmonic terminations. Theory is not simple.
• The amplifier goes into compression (gain drops but not precipitously) and can still get good efficiency.
©James Buckwalter 2
Classical Device Model
• Underlying assumption: simple model of transistor
• Transistor acts like current source, with Iout a linear replica of vin, except for limitations of cutoff when vin<vth
=> For sinewave input, output current is a sinewave, possibly with clipping
©James Buckwalter 3
vin iout
Iout
Vout
Imax
Overdriven Device Model
• Transistor acts like current source, with Iout a linear replica of vin, except for limitations of cutoff when vin<vth
• When Vout gets low enough, transistor acts like voltage source
©James Buckwalter 4
vin vout
Iout
Vout
Imax
Overdriven Class B AmplifierIout
VoutVoVmin
Imax
Vmax
match
match
Vo
RL
time
time
Vo
Vds
Iave
Id
Irf
Harmonics
are shorted
Vdc fixed
Must be sinusoidal
Vds(t) is fixed!!
Ids(t) must change
Overdriven Class B amplifiers can have strange waveforms
Overdriven Class F amplifiers can have strange waveforms
• If output voltage “tries” to go below zero the voltage waveform becomes progressively more like a square wave
• The current is mostly zero when the voltage is nonzero. The load line is traversed only during transitions
• Overdriven Class F amplifiers approach switching mode operation
©James Buckwalter 7
time
Vo
Vce
Vrf
Iout
VoutVoVmin
Imax
Vmax
Comparison of Overdriven Classes
©James Buckwalter 8
Waveform Engineering
Waveform Engineering Spreadsheet
Model transistor as current source with constant gm, together with
saturation
Input is a sinewave of voltage with specified bias point (can set
Class A, Class AB, etc)
Specify output voltage in terms of fundamental and harmonics of voltage
Spreadsheet calculates actual current, taking into account saturation:
Iout= Ioutnom / (1+exp(Vth-Vin/Vsat)) this provides
smooth clipping
Spreadsheet computes what impedances Z1, Z2, Z3 would have to be to
create the voltage waveform assumed
For this to be a valid amplifier, you should
1) Check that the impedances have positive real part.
2) Check that the voltage waveform is positive only (otherwise adjust
voltage dc bias)
Iout
VinVt1
Vsin
Irf
Waveform Specification
Vsin is centered around zero
You specify Vt to control conduction angle
Vt2
Power Amplifier WaveformsInspired by Steve Cripps Rev 1: Still being checked out, user beware!!!
Input Parameters Magnitude Angle
Imax 1 A Vdc 1.02 V
Vth -1 V Vfund 1 0 V
Vin 1 V V2nd 0 0 V
Vknee 0 V V3rd 0 0 V
Vsat 0.00001 V
Summary of Calculated Results Pdc 1.02 W
Pout 0.499997685 W Vdc 1.02 V Pdiss 0.520001157 W
Efficiency 49.01938092 % Idc 1 A Inefficiency 50.98050562
Zfund Re 1 ohm Z2nd Re -1.019770476 ohm Z3rd Re -1.019484315 ohm
Zfund Im -9.42328E-06 ohm Z2nd Im -0.020825224 ohm Z3rd Im -0.031804559 ohm
0
0.5
1
1.5
2
2.5
0 45 90 135 180 225 270 315 360 405 450 495 540 585 630 675 720 765
V,
I
angle (degrees)
Waveforms of Transistor Voltage(blue) and Current (black)
Fundamental Voltage (blue) & Current 2nd Harmonic Voltage (blue) &
Power Amplifier WaveformsInspired by Steve Cripps Rev 1: Still being checked out, user beware!!!
Input Parameters Magnitude Angle
Imax 1 A Vdc 1.02 V
Vth 0 V Vfund 1 0 V
Vin 1 V V2nd 0 0 V
Vknee 0 V V3rd 0 0 V
Vsat 0.00001 V
Summary of Calculated Results Pdc 0.323851033 W
Pout 0.249998843 W Vdc 1.02 V Pdiss 0.073851612 W
Efficiency 77.19562923 % Idc 0.317501013 A Inefficiency 22.80419208
Zfund Re 2 ohm Z2nd Re 2.20819E-05 ohm Z3rd Re -22.08967468 ohm
Zfund Im -1.03328E-05 ohm Z2nd Im 2.22897E-07 ohm Z3rd Im -4.248460907 ohm
0
0.5
1
1.5
2
2.5
0 45 90 135 180 225 270 315 360 405 450 495 540 585 630 675 720 765
V,
I
angle (degrees)
Waveforms of Transistor Voltage(blue) and Current (black)
Power Amplifier WaveformsInspired by Steve Cripps Rev 1: Still being checked out, user beware!!!
Input Parameters Magnitude Angle
Imax 1 A Vdc 0.99 V
Vth 0 V Vfund 1 0 V
Vin 1 V V2nd 0 0 V
Vknee 0 V V3rd 0 0 V
Vsat 0.00001 V
Summary of Calculated Results Pdc #NUM! W
Pout #NUM! W Vdc 0.99 V Pdiss #NUM! W
Efficiency #NUM! % Idc #NUM! A Inefficiency #NUM!
Zfund Re #NUM! ohm Z2nd Re #NUM! ohm Z3rd Re #NUM! ohm
Zfund Im #NUM! ohm Z2nd Im #NUM! ohm Z3rd Im #NUM! ohm
-0.5
0
0.5
1
1.5
2
2.5
0 45 90 135 180 225 270 315 360 405 450 495 540 585 630 675 720 765
V,
I
angle (degrees)
Waveforms of Transistor Voltage(blue) and Current (black)
Power Amplifier WaveformsInspired by Steve Cripps Rev 1: Still being checked out, user beware!!!
Input Parameters Magnitude Angle
Imax 1 A Vdc 1.2 V
Vth 0 V Vfund 1 0 V
Vin 1 V V2nd 0 0 V
Vknee 0 V V3rd 0.2 0 V
Vsat 0.00001 V
Summary of Calculated Results Pdc 0.364334564 W
Pout 0.236109996 W Vdc 1.2 V Pdiss 0.131001798 W
Efficiency 64.80581827 % Idc 0.303612137 A Inefficiency 35.95645605
Zfund Re 2.117647199 ohm Z2nd Re 2.98575E-05 ohm Z3rd Re -7.199979614 ohm
Zfund Im -1.32273E-05 ohm Z2nd Im 1.00874E-07 ohm Z3rd Im -0.000223847 ohm
0
0.5
1
1.5
2
2.5
3
0 45 90 135 180 225 270 315 360 405 450 495 540 585 630 675 720 765
V,
I
angle (degrees)
Waveforms of Transistor Voltage(blue) and Current (black)
Power Amplifier WaveformsInspired by Steve Cripps Rev 1: Still being checked out, user beware!!!
Input Parameters Magnitude Angle
Imax 1 A Vdc 0.9 V
Vth 0 V Vfund 1 0 V
Vin 1 V V2nd 0 0 V
Vknee 0 V V3rd -0.2 0 V
Vsat 0.00001 V
Summary of Calculated Results Pdc 0.285750911 W
Pout 0.249998846 W Vdc 0.9 V Pdiss 0.035751486 W
Efficiency 87.48838106 % Idc 0.317501013 A Inefficiency 12.51141644
Zfund Re 2.000000029 ohm Z2nd Re 1.94839E-05 ohm Z3rd Re 167628.8141 ohm
Zfund Im -9.21318E-06 ohm Z2nd Im 3.57976E-07 ohm Z3rd Im -950463.012 ohm
00.20.40.60.8
11.21.41.61.8
2
0 45 90 135 180 225 270 315 360 405 450 495 540 585 630 675 720 765
V,
I
angle (degrees)
Waveforms of Transistor Voltage(blue) and Current (black)
Power Amplifier WaveformsInspired by Steve Cripps Rev 1: Still being checked out, user beware!!!
Input Parameters Magnitude Angle
Imax 1 A Vdc 0.78 V
Vth 0 V Vfund 1 0 V
Vin 1 V V2nd 0.3 -45 V
Vknee 0 V V3rd 0 0 V
Vsat 0.00001 V
Summary of Calculated Results Pdc 0.247650569 W
Pout 0.249998843 W Vdc 0.779999306 V Pdiss 0.029726209 W
Efficiency 100.9482204 % Idc 0.317501013 A Inefficiency 12.00328721
Zfund Re 2 ohm Z2nd Re -1.402955043 ohm Z3rd Re -10.43286793 ohm
Zfund Im -5.32327E-06 ohm Z2nd Im 5.21996E-06 ohm Z3rd Im -2.192618628 ohm
0
0.5
1
1.5
2
2.5
0 45 90 135 180 225 270 315 360 405 450 495 540 585 630 675 720 765
V,
I
angle (degrees)
Waveforms of Transistor Voltage(blue) and Current (black)
Power Amplifier WaveformsInspired by Steve Cripps Rev 1: Still being checked out, user beware!!!
Input Parameters Magnitude Angle
Imax 1 A Vdc 0.75 V
Vth 0 V Vfund 1 42 V
Vin 1 V V2nd 0.3 0 V
Vknee 0 V V3rd 0 0 V
Vsat 0.00001 V
Summary of Calculated Results Pdc 0.238125264 W
Pout 0.185785124 W Vdc 0.749998439 V Pdiss 0.052340198 W
Efficiency 78.01991319 % Idc 0.317501013 A Inefficiency 21.98011131
Zfund Re 1.486287875 ohm Z2nd Re 1.94481E-06 ohm Z3rd Re -1.786400741 ohm
Zfund Im 1.338262352 ohm Z2nd Im -1.402965471 ohm Z3rd Im -0.364870102 ohm
0
0.5
1
1.5
2
2.5
0 45 90 135 180 225 270 315 360 405 450 495 540 585 630 675 720 765
V,
I
angle (degrees)
Waveforms of Transistor Voltage(blue) and Current (black)
RL
Input
matching
network
VDD
Ls
Cds
Class J Amplifier
New designation introduced by
Steve Cripps
Amplifier design is very
straightforward corresponds to
what many designers do without
knowing it!
Class L - Lazy man's amplifier?
The harmonic matching is
provided by the device output
capacitance only =>
external matching is only done for
the fundamental
For many traditional transistors,
Cds provides a short to all
harmonics => class AB, B, etc.
For some modern transistors,
Cds is low (good!). Then should
change the fundamental match to
optimize efficiency!
RL
Input
matching
network
VDD
Ls
Cds
Class J Amplifier
If Cds is not very large, 2nd
harmonic is not shorted.
Use 2nd harmonic to achieve
voltage waveform with flat bottom
Higher efficiency
Best efficiency but requires Z2f with
negative real part!!
RL
Input
matching
network
VDD
Ls
Cds
Class J Amplifier
If Cds is not very large, 2nd
harmonic is not shorted.
Use 2nd harmonic to achieve
voltage waveform with flat bottom
Higher efficiency
Best efficiency but requires Z2f with
negative real part!!
Good efficiency and realizable. Use Zf inductive.
Class J Amplifier Characteristics
Fundamental impedance: RL + j X1, with X1~RL
2nd Harmonic impedance: j X2, with X2~ RL
3rd Harmonic impedance: j X3 ~ 2/3 RL
Ideal Efficiency ~ similar to Class B
peaks at ~ 78-80 %
Formal Class J Characteristics
For I(q) = cos q (-p/2<q<p/2, 0 otherwise)
Vtotal (q) = 1 - cos q - sin q + cos q sin q
Vtotal (q) = (1-cos q)*(1-sin q)
Vfund (q) ~ cos(q-p/4)~ cosq cosp/4 + sinq sin p/4~ cos q + sin q
V2fo(q) ~ sin 2q
~ cos q sin q
V(dc) = 1
Note that <Vtotal(q) > =1 (just like for Class B)½*Re {fundamental[Vtotal] * fundamental[I]} = ¼ (just like for Class B)
Ideal Efficiency ~ similar to Class B
peaks at ~ 78-80 %
Continuous ClassesMathematical formulations are emerging which show the
characteristics of some of the high efficiency regions
These are leading to new insights for broadband design
For I= cos q (-p/2<q<p/2, 0 otherwise)
V= 1- cos q for class B
V= (1-cos q)*(1-sin q) for Class J
V= (1-cos q)*(1- a sin q) for more general class
with same efficiency
Steve Cripps
Continuous Class FFor I= cos q (-p/2<q<p/2, 0 otherwise)
Broadband Continuous Class F PA Design
Are There Other Matching Configurations
That Yield High Efficiency ???
YES !!
Output Waveforms to Optimize Efficiency (1)
timeIave
IC time
Vo
VceV(t) is square wave
has fundamental + odd harmonics
I(t) is rectified sine wave
has fundamental + even harmonics
Power is only at fundamental !
V is minimum when I>0, h is max
time
Iave
IC
time
Vo
Vce“Dual” solution
Power only at fundamental
V is minimum when I>0
Output Waveforms to Optimize Efficiency (2)
There are plenty of other waveforms that can achieve efficiency
= “100%”
Don’t need square wave for V(t) or I(t).
Need to satisfy V=Z*I, where Z has non-negative real part at all
harmonics in order to be realizable.
Class E
X1=0
Harmonic Load TuningSimulated Efficiency vs Harmonic Load Reactance
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
X1=RL*0.7
Harmonic Load Tuning
Simulated Efficiency vs Harmonic Load Reactance
Class E
Basic Power Amplifier Design Process
1) Decide on Vdd, and identify power transistor with sufficient
power handling capability and breakdown voltage
2) Using dc characteristics, decide on resistive load line. Verify that
sufficient Pout can be obtained
3) Determine input impedance and match transistor input - using
bias condition of "average dc current corresponding to average
output power"
4) Determine load susceptance and match output to obtain RL and
BL
5) Provide output match at harmonic frequencies
6) Set up bias network
7) Optimize using simulator
Steps 2, 3, 4, 5, and 7 can be carried out experimentally with load
pull system