Lecture 27: Non-Linear Power...

EECS 142

Lecture 27: Non-Linear Power Amplifiers

Prof. Ali M. Niknejad

University of California, Berkeley

Copyright c© 2005 by Ali M. Niknejad

A. M. Niknejad University of California, Berkeley EECS 142 Lecture 27 p. 1/32 – p. 1/32

Efficiency of Class A

Consider the power dissipated by the transistor

pt(t) = i(t) × v(t) = (IQ + io cos ω0t) × (VCC − vo cos ω0t)

pt(t) = IQVCC − iovo cos2 ω0t + (ioVCC − IQvo) cos ω0t

The average power over a cycle is given by

pt(t) = IQVCC −iovo

2

The above result is obvious from conservation ofenergy. It simply states that the difference between DCpower and the power to the load must be the power lostto heat.


Efficiency Cont

Class A amplifiers have the undesirable property thatthe power dissipation is maximum for the absence of aninput signal, a common condition. In fact, the efficiencyis a linear function of the output power.

The minimum power dissipation occurs for themaximum output power, where half the power isdelivered to the load and the other half is converted toheat

pt(t)∣

∣

∣

min=

IQVCC

2


Average Efficiency

Since the efficiency drops like V2, the average efficiencyfor a signal distribution such as a Gaussian profile isparticularly low

ηav =

∫

∞

−∞

η(v)p(v)dv0.2

0.4

0.6

0.8

1

p(vo)

vo

A typical situation with Class A is that the peakefficiency is about 30%, but the average efficiency is aslow as 6%.


Multi-Stage PAs

vin

vo

VCC

RL

C∞

RFC

Driver Amp

Often 2-3 stages of power amplification are integratedinto a PA to achieve ∼ 30 dB of gain.

In a single-ended design, more gain is dangerous, dueto parasitic feedback paths which limit stability.

In a high power design the driver amplifier must delivera considerable amount of power to the load.


Two Stage Example

vin

vo

R0

VDD

L1

C1

L2

C2

RB1 RB2

LD

VB1 VB2

If the gain of the output stage is not too great, then thepower of the driver is not negligible and the overallefficiency must include a careful driver design.

Here, an L matching network is used to convert theinput impedance of the output stage (usually low) to ahigher value close to the Ropt of the driver stage.


Class B PA

+vin

VCCvbias

The above circuit utilizes two transistors. Each deviceonly delivers a half sinusoid pulse and the full sinusoidis recovered by phase inversion through thetransformer.

The base and collector bias voltages come from thetransformer center tap. The base (or gate) is biased atthe edge of conduction (threshold).


Class B Operation

p(t)

ic(t) vc(t)

VCC

0V

Since the voltage at the load is ideally a perfectsinusoid, the voltage on the collectors is likewisesinusoidal.

The power dissipated by each transistor is thus theproduct of a sine and a half sine as shown above.


Efficiency of Class B

The average current drawn by each transistor is givenby

IQ =1

T

∫ T

0

ic(t)dt =Ip

T

∫ T/2

0

sin ωtdt =Ip

2π

∫ π

0

sin θdθ =Ip

π

Where Ip is the peak voltage drawn from the supply.

The peak current drawn from the supply is just the loadcurrent swing reflected to the collector, Ip = io × n.

η =1

2

(

Ip

2IQ

) (

vo

VCC

)

Note that the total DC current draw is twice IQ sinceboth devices draw current from the supply.


Efficiency (cont)

Since the collector voltage swing can be as large asVCC (similar to an inductively loaded Class A), theefficiency is bounded by

η ≤1

2

(

Ip

2IQ

)

=1

4

(

Ip

IQ

)

η ≤π

4≈ 78%

This is a big improvement over the peak efficiency ofClass A.

Note that the average current naturally scales withoutput power, and so efficiency drops more gracefullyas we back-off from peak power.

A. M. Niknejad University of California, Berkeley EECS 142 Lecture 27 p. 10/32 – p

Class B Efficiency versus Voltage

η(V )

VCC

50%class B

class A

The efficiency drops linearly as we back-off from thepeak output voltage

η(v) =π

4

(

vc

VCC

)

where vc is the collector voltage swing, which is just ntimes smaller than the load voltage, vc = vo/n.


Transformerless Class B

vin

vo

VCC

RL

C∞

RFC

vbias

A tuned Class B amplifier works with a single devicesby sending half sinusoid current pulses to the load. Thedevice is biased at the edge of conduction.

The load voltage is sinusoidal because a high Q RLCtank shunts harmonics to ground.


Class B Tank

In a single transistor version, the “minus” pulse is in factdelivered by the RLC tank. The Q factor of the tankneeds to be large enough to do this. This is analogousto pushing someone on a swing. You only need to pushin one direction, and the reactive energy stored willswing the person back in the reverse direction.

The average current drawn from the supply is the sameas before, IQ = Ip/π. The harmonic current delivered tothe load is given by Fourier analysis of the half pulse

Iω1=

2

2πIp

∫ π

0

sin θsinθdθ =1

πIp

∫ π

0

1 − cos 2θ

2dθ

=1

π

π

2Ip =

Ip

2A. M. Niknejad University of California, Berkeley EECS 142 Lecture 27 p. 13/32 – p

Class B Waveforms

ic(t) vc(t)

VCC

0V

We see that the transistor is cutoff when the collectorvoltage swings above VCC . Thus, the power dissipatedduring this first half cycle is zero.

During the second cycle, the peak current occurs whenthe collector voltage reaches zero.


Class B Power Dissipation

The efficiency is therefore the same

η =1

2

Iω1

IQ

vc

VCC=≤

1

2

π

2=

π

4

The DC power drawn from the supply is proportional tothe output voltage

Pdc = IQVCC =VCCIp

π

Ip =vc

Ropt=

nvo

Ropt

The power loss in the transistor is given by

pt(t) =1

2π

∫ π

0

Ip sin θ(VCC − vc sin θ)dθ


Power Loss (cont)

Integrating the above expression we have

pt(t) =VCCIp

2π(− cos θ|π

0−

vc

Ip2π

=1

2π

(

2VCCIp −vcIp

2π

)

=Ip

πVCC −

vcIp

4

= IQ · VCC −vcIω1

2

= Pdc − PL


Conduction Angle

Often amplifiers are characterized by their conductionangle, or the amount of time the collector current flowsduring a cycle.

Class A amplifiers have 360◦ conduction angle, sincethe DC current is always flowing through the device.

Class B amplifiers, though, have 180◦ conduction angle,since they conduct half sinusoidal pulses.

In practice most Class B amplifiers are implemented asClass AB amplifiers, as a trickle current is allowed toflow through the main device to avoid cutting off thedevice during the amplifier operation.


Improving The Efficiency

ic(t) vc(t) ic(t) vc(t)

The most optimal waveform is shown above, where acurrent pulse is delivered to the load during thecollector voltage minimum (ideally zero)

As the pulse is made sharper and sharper, theefficiency improves. To deliver the same power, though,the pulse must be taller and taller as it’s made morenarrow. In fact, in the limit the current spike approachesa delta function.


Class C

vin

vo

VCC

RL

C∞

RFC

Vbias < Vthresh

Class C amplifiers are a wide family of amplifiers withconduction angle less than 180◦. One way to achievethis is to bias a transistor below threshold and allow theinput voltage to turn on the device for a small fraction ofthe cycle.


Class C Linearity

I

Q

000

001

010

011

100

101

110

111

I2 + Q2 = A2

The Class C amplifier is very non-linear, and it is onlyappropriate for applications where the modulation isconstant envelope. For instance, FM uses a constantamplitude carrier and only modulates the frequency toconvey information. Likewise, any digital modulationscheme with a constellation in a circle is constantenvelope


Collector Modulation

RL

C∞

Vbias < Vthresh

B(t) cos (ωt + φ(t))A(t)

A0 cos (ωt + φ(t))

While the amplifier is a non-linear function of the inputamplitude, the Class C amplifier can be made to actfairly linearly to the collector voltage.

By driving the amplifier into “saturation” in each cycle,e.g. with a large enough swing to rail the supplies, thenthe output power is related to the voltage supply

Collector modulation then uses the power supply tointroduce amplitude modulation into the carrier


Class F

ic(t) vc(t)

Since it’s difficult to create extremely narrow pulses athigh frequency, we can take a different approach andattempt to square up the drain voltage.


Class F Circuit

3ω0

ω0 RL

The above circuit isolates the fundamental resonantload from the drain at the third harmonic, allowing thedrain voltage to contain enough third harmonic to createa more square like waveform.

It can be shown that just adding a third harmonic boostthe efficiency from 78% (Class B) to 88%.


Quarter Wave Class F

ω0 RL

Z0

ℓ = λ0/4

RFC

In this circuit a quarter wave transformer converts thelow impedance at the load (due to the capacitance) atharmonics of the fundamental to a high impedance forall odd harmonics. Even harmonics are unaltered asthey see a λ/2 line.

In theory then we can create a perfect square wave atthe drain of the transistor and thus achieve 100%efficiency.


Class D Switching Amplifier

S1

S2

RL

VDD

If give up linearity, then we can create some intrinsicallyefficient amplifiers using switches. An ideal switch doesnot dissipate any power since either the voltage orcurrent is zero.

By varying the switching rate, we can impart frequencymodulation onto the load.

A real switch has on-resistance and parasitic offcapacitance and conductance.


MOS Class D Inverter

RL

VDD

Switching amplifiers are realized with transistorsoperating as switches. MOS transistors makeparticularly good switches.

The input voltage is large enough to quickly move theoperating point from cut-off to triode region.


Class D Waveforms

0V

id(t)VDD

vd(t)

The MOS drain voltages switches from VSS to VDD atthe rate of the input signal.

A series LCR filter only allows the first harmonic ofvoltage to flow into the load. Since current only flowsthrough a device when it’s fully on (ideally Vds = 0), littlepower dissipation occurs in the devices.


Class D Efficiency

We can see that the efficiency has to be 100% for anideal switch. That’s because there is no where for theDC power to flow except to the load.

The collector voltage can be decomposed into a Fourierseries

vd =VDD

2(1 + s(ωt))

s(θ) = sign(sin(θ)) =4

π

(

sin θ +1

3sin 3θ +

1

5sin 5θ + · · ·

)

The load current is therefore

iL =4

π

VDD

2Rsin θ =

2VDD

πRLsin θ


Class D Efficiency (cont)

The load power is thus

PL =i2LRL

2=

2

π2

V 2

DD

RL≈ 0.2

V 2

DD

RL

The drain current are half-sinusoid pulses. The averagecurrent drawn from the supply is the average PMOScurrent

IQ =Ip

π=

2

π2

VDD

RL

PDC = IQ · VDD =2V 2

DD

π2RL= PL

As we expected, the ideal efficiency is η = 100%


Class D Losses

As previously noted, a real Class D amplifier efficiencyis lowered due to the switch on-resistance

We can make our switches bigger to minimize theresistance, but this in turn increases the parasiticcapacitance

There are two forms of loss associated with theparasitic capacitance, the capacitor charging lossesCV 2f and the parasitic substrate losses.

The power required to drive the switches also increasesproportional to Cgs since we have to burn CV 2f powerto drive the inverter or. A resonant drive can lower thedrive power by the Q of the resonator.

In practice a careful balance dictates the maximumswitch size


Class S

If the switching rate is much higher than thefundamental, then amplitude modulation can beconverted to pulse-width modulation.

A low-pass filter at the output faithfully recreates theenvelope of the signal.


Class S

RL

VDD

Amplitude

to PWM

Class S amplifiers are commonly used at lowfrequencies (audio) since the transistors can beswitched at a much higher frequency than thefundamental.

This allows efficiencies approaching 100% with goodlinearity.


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