Overdriven Amplifiers - UCSB · 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

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.


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


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


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


time

Vo

Vce

Vrf

Iout

VoutVoVmin

Imax

Vmax

Comparison of Overdriven Classes


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) &



Imax 1 A Vdc 1.02 V

Vth 0 V Vfund 1 0 V

Vin 1 V V2nd 0 0 V


Vsat 0.00001 V



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)




Imax 1 A Vdc 0.99 V

Vth 0 V Vfund 1 0 V

Vin 1 V V2nd 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)




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




Zfund Re 2.117647199 ohm Z2nd Re 2.98575E-05 ohm Z3rd Re -7.199979614 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)




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




Zfund Re 2.000000029 ohm Z2nd Re 1.94839E-05 ohm Z3rd Re 167628.8141 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)




Imax 1 A Vdc 0.78 V

Vth 0 V Vfund 1 0 V

Vin 1 V V2nd 0.3 -45 V


Vsat 0.00001 V




Zfund Re 2 ohm Z2nd Re -1.402955043 ohm Z3rd Re -10.43286793 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)




Imax 1 A Vdc 0.75 V

Vth 0 V Vfund 1 42 V

Vin 1 V V2nd 0.3 0 V


Vsat 0.00001 V




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)


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

Documents

Overdriven Amplifiers - UCSB · 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