mathematical models for class-d ampliﬁers€¦ · · 2010-04-21mathematical models for class-d...

Mathematical models forclass-D amplifiers

Stephen Cox

School of Mathematical Sciences, University of Nottingham, UK

3 March 2010

Stephen Cox Mathematical models for class-D amplifiers 1/36

Background


BackgroundHalcro manufacture “reference quality, super-fidelity” amplifiers

La Revue du Son et duHome Cinéma




Their amplifiers are class D

Class-D amplifiers are highly efficient, and (non-Halcroexamples) are used in mobile phones, laptops, hearing aids, …




Their amplifiers are class D

Class-D amplifiers are highly efficient, and (non-Halcroexamples) are used in mobile phones, laptops, hearing aids, …

In practical designs, there is negative feedback, but thisintroduces distortion

We will quantify this distortion through a mathematical modelfor the amplifier design(s), and seek ways to eliminate it

Bruce Candy, Tan Meng Tong (NTU, Singapore), Stephen CreaghStephen Cox Mathematical models for class-D amplifiers 5/36

The ideal class-D amplifier


The ideal Class-D amplifier


s(t)

v(t)

g(t)−+

Audio signal s(t ) compared withhigh-frequency triangular wavev (t ) to generate high-frequencypulse-width-modulated (PWM)square wave g(t )

g(t)

−1

+1

−1

+1

tn−1 n+1n

v(t) s(t)


The ideal Class-D amplifier


s(t)

v(t)

g(t)−+

Audio signal s(t ) compared withhigh-frequency triangular wavev (t ) to generate high-frequencypulse-width-modulated (PWM)square wave g(t )

g(t)

−1

+1

−1

+1

tn−1 n+1n

v(t) s(t)

The ideal class-D amplifier is distortion-free……unfortunately negative feedback is desirable in practice, andthis introduces distortion…


A first-order class-Damplifier with negative

feedback


Class-D amplifier with –ve feedback

g(t)+ +

−+dt−c ∫

0s(t)

h(t)

v(t)

In the mathematical model:we scale voltages so that the supply voltage is ±1we scale time so that high-frequency carrier wave period is 1


Negative-feedback model

g(t)+ +

−+dt−c ∫

0s(t)

h(t)

v(t)



g(t)+ +

−+dt−c ∫

0s(t)

h(t)

v(t)

carrier wave: v(t+1)=v(t)

v (t ) =

{1 − 4t 0 ≤ t < 1

2−3 + 4t 1

2 ≤ t < 1



g(t)+ +

−+dt−c ∫

0s(t)

h(t)

v(t)


v (t ) =

{1 − 4t 0 ≤ t < 1

2−3 + 4t 1

2 ≤ t < 1

integrator output

dhdt

= −c(g(t ) + s(t ))



g(t)+ +

−+dt−c ∫

0s(t)

h(t)

v(t)


v (t ) =

{1 − 4t 0 ≤ t < 1

2−3 + 4t 1

2 ≤ t < 1

integrator output

dhdt

= −c(g(t ) + s(t ))

comparator output

g(t ) =

{1 if h(t ) + v (t ) > 0−1 if h(t ) + v (t ) < 0



g(t)+ +

−+dt−c ∫

0s(t)

h(t)

v(t)


v (t ) =

{1 − 4t 0 ≤ t < 1

2−3 + 4t 1

2 ≤ t < 1

integrator output

dhdt

= −c(g(t ) + s(t ))

comparator output

g(t ) =

{1 if h(t ) + v (t ) > 0−1 if h(t ) + v (t ) < 0

Aim: to find the Fourierspectrum of the outputg(t ), in particular its low-frequency (audio) part



g(t)+ +

−+dt−c ∫

0s(t)

h(t)

v(t)

v(t)

n−1

g(t)

−h(t)−1

+1

−1

+1

tn+1n


Solving the negative-feedback model

Aim: difference equations,to allow iteration from onecarrier-wave period to thenext

g(t)

n βn

−1

+1

αn+1αn n+n+

αn

βn

n+1 n+1+

αn+1t




g(t)

n βn

−1

+1

αn+1αn n+n+

αn

βn

n+1 n+1+

αn+1t

Switching conditions are given by h(t ) + v (t ) = 0:

h(n + βn) = 3 − 4βn h(n + 1 + αn+1) = −1 + 4αn+1




g(t)

n βn

−1

+1

αn+1αn n+n+

αn

βn

n+1 n+1+

αn+1t

Switching conditions are given by h(t ) + v (t ) = 0:

h(n + βn) = 3 − 4βn h(n + 1 + αn+1) = −1 + 4αn+1

Integrate ODE for h(t ) from one down-switching to the next:

h(n + βn) = h(n + αn) + c(βn − αn)

+ c(−f (n + βn) + f (n + αn))

h(n + 1 + αn+1) = h(n + αn) − c(αn+1 − 2βn + αn)

+ c(−f (n + 1 + αn+1) + f (n + αn))

where f ′(t ) = s(t ), so f (t ) is the integrated input signalStephen Cox Mathematical models for class-D amplifiers 12/36


Eliminating h(n + βn), wefind the three equations

h(n + 1 + αn+1) = −1 + 4αn+1

g(t)

n βn

−1

+1

αn+1αn n+n+

αn

βn

n+1 n+1+

αn+1t

3 − 4βn = h(n + αn) + c(βn − αn) + c(−f (n + βn) + f (n + αn))


+ c(−f (n + 1 + αn+1) + f (n + αn))

for the three quantities

βn αn+1 h(n + 1 + αn+1)



Eliminating h(n + βn), wefind the three equations

h(n + 1 + αn+1) = −1 + 4αn+1

g(t)

n βn

−1

+1

αn+1αn n+n+

αn

βn

n+1 n+1+

αn+1t



+ c(−f (n + 1 + αn+1) + f (n + αn))

for the three quantities

βn αn+1 h(n + 1 + αn+1)

We can iterate these equations to determine the behaviour ofthe amplifier - but this doesn’t tell us the frequency spectrum!



How do we find the audio component of the output g(t ) using thedifference equations

h(n + 1 + αn+1) = −1 + 4αn+1



+ c(−f (n + 1 + αn+1) + f (n + αn))?



g(t)

n βn

−1

+1

αn+1αn n+n+

αn

βn

n+1 n+1+

αn+1t

n+3

Bβ β βββ n n+1n−1 n+2

n−1 n n+1 n+2 n+3 t

α α α ααn−1 n n+1 n+2 n+3

A

We introduce interpolating functions A(ǫt ) and B(ǫt ), where

ǫ =typical signal frequency

switching frequency≪ 1



g(t)

n βn

−1

+1

αn+1αn n+n+

αn

βn

n+1 n+1+

αn+1t

n+3

Bβ β βββ n n+1n−1 n+2

n−1 n n+1 n+2 n+3 t

α α α ααn−1 n n+1 n+2 n+3

A

We introduce interpolating functions A(ǫt ) and B(ǫt ), where

ǫ =typical signal frequency

switching frequency≪ 1

We also introduce H(ǫt ) with H(ǫn) = h(n + αn)

Only sampled values A(ǫn), B(ǫn) and H(ǫn) have real meaning


Solving the negative-feedback modelThe interpolating functions A(ǫt ), B(ǫt ) and H(ǫt ) satisfy thefunctional equations

H(ǫ(t + 1)) = −1 + 4A(ǫ(t + 1))

(4 + c)A(ǫt ) = 1 + H(ǫt )

+ cǫ−1 [F (ǫ(t + A(ǫt ))) − F (ǫ(t + 1 + A(ǫ(t + 1))))]

(4 + c)B(ǫt ) = 3 − H(ǫt )

+ cǫ−1 [F (ǫ(t + B(ǫt ))) − F (ǫ(t + A(ǫt )))]

where ǫ−1dF (ǫt )/dt = s(t ) is known audio input


Solving the negative-feedback modelThe interpolating functions A(ǫt ), B(ǫt ) and H(ǫt ) satisfy thefunctional equations

H(ǫ(t + 1)) = −1 + 4A(ǫ(t + 1))

(4 + c)A(ǫt ) = 1 + H(ǫt )

+ cǫ−1 [F (ǫ(t + A(ǫt ))) − F (ǫ(t + 1 + A(ǫ(t + 1))))]

(4 + c)B(ǫt ) = 3 − H(ǫt )

+ cǫ−1 [F (ǫ(t + B(ǫt ))) − F (ǫ(t + A(ǫt )))]

where ǫ−1dF (ǫt )/dt = s(t ) is known audio input

To solve these nonlinear equations we make perturbationexpansions in the small parameter ǫ:

A(ǫt ) =

∞∑

m=0

ǫmAm(ǫt ) B(ǫt ) =

∞∑

m=0

ǫmBm(ǫt ) H(ǫt ) =

∞∑

m=0

ǫmHm(ǫt )


Solving the negative-feedback modelWe collect terms at successive orders in ǫ, after writing Taylorseries for the various functions (about the time ǫt )

Details of calculation are messy (done by computer algebra)




Special case: s(t ) = s0 constant input signal





A = 116(1 − s0) [4 − c(1 + s0)]

B = 12 + 1

16(1 + s0) [4 − c(1 − s0)]t−1

+1g(t)

n n+n+

B

A

BA n+1





A = 116(1 − s0) [4 − c(1 + s0)]

B = 12 + 1

16(1 + s0) [4 − c(1 − s0)]t−1

+1g(t)

n n+n+

B

A

BA n+1

Time-average of the output (the only low-frequency part) is

〈g(t )〉 =

[

+1 × (1 − B + A)

]

+

[

−1 × (B − A)

]

= −s0

so there is no distortion




General case: s(t ) = general input signal

Details very messy, but can calculate A = A0 + ǫA1 + · · · , etc:

A0(ǫt ) = 116(1 − s(t )) [4 − c(1 + s(t ))]

B0(ǫt ) = 12 + 1

16(1 + s(t )) [4 − c(1 − s(t ))]

…



Once we have found the first few terms in

A = A0 + ǫA1 + · · ·

B = B0 + ǫB1 + · · ·

H = H0 + ǫH1 + · · ·

how do we find the audio part of the output g(t )?


Finding the audio output

For general input s(t ), we need to findthe audio component of the output

Notation:ψ(t ; t1, t2) = top-hat function

t

t

1

0t t1 2

ψ( )


Finding the audio output

For general input s(t ), we need to findthe audio component of the output

Notation:ψ(t ; t1, t2) = top-hat function

t

t

1

0t t1 2

ψ( )

The amplifier output

g(t ) =∞∑

n=−∞

{

ψ(t ; n + βn ,n + 1 +αn+1)− ψ(t ; n +αn,n + βn)

}

−1tn−1 n+1n

g(t)+1


Finding the audio outputThe Fourier transform of the output is

g(ω) =

∫∞

−∞

e−iωt g(t ) dt

=

∞∑

−∞

2−iω

(

e−iω(n+αn) − e−iω(n+βn))

=

∞∑

−∞

2−iω

(

e−iω(n+A(ǫn)) − e−iω(n+B(ǫn)))


Finding the audio outputThe Fourier transform of the output is

g(ω) =

∫∞

−∞

e−iωt g(t ) dt

=

∞∑

−∞

2−iω

(

e−iω(n+αn) − e−iω(n+βn))

=

∞∑

−∞

2−iω

(

e−iω(n+A(ǫn)) − e−iω(n+B(ǫn)))

=∞∑

−∞

∫∞

−∞

2−iω

e2πniτe−iωτ(

e−iωA(ǫτ) − e−iωB(ǫτ))

dτ

where the last step follows using Poisson resummation:∞∑

n=−∞

f (n) =∞∑

n=−∞

∫∞

−∞

e2πniτ f (τ) dτ


Finding the audio outputThe low-frequency, audio part of the output is given by n = 0terms:

gaudio(ω) =

∫∞

−∞

2−iω

e−iωτ(


dτ

=

∫∞

−∞

e−iωτ

∞∑

0

2(−iω)n

(n + 1)!

(

An+1(ǫτ) − Bn+1(ǫτ))

dτ

=

∫∞

−∞

e−iωτ

∞∑

0

2(−1)n

(n + 1)!

dn

dτn

(


dτ


Finding the audio outputThe low-frequency, audio part of the output is given by n = 0terms:

gaudio(ω) =

∫∞

−∞

2−iω

e−iωτ(


dτ

=

∫∞

−∞

e−iωτ

∞∑

0

2(−iω)n

(n + 1)!

(


dτ

=

∫∞

−∞

e−iωτ

∞∑

0

2(−1)n

(n + 1)!

dn

dτn

(


dτ

and hence the audio part of the output is

gaudio(t ) = 1 +

∞∑

0

2(−1)n

(n + 1)!

dn

dtn

(

An+1(ǫt ) − Bn+1(ǫt ))

∼ (1 + 2A(ǫt ) − 2B(ǫt )) −ddt

(

A2(ǫt ) − B2(ǫt ))

+ · · ·


Finding the audio outputResult for the audio output

From much calculation (!), the amplifier output gaudio(t ) is

−s(t ) +1c

s′(t )︸︷︷︸

slight delay: ≈ −s(t − c−1)




−s(t ) +1c

s′(t )︸︷︷︸


[O(ǫ2)

]+

[(148

−1c2

)

s(t )︸︷︷︸

linear distortion

+148

s3(t )︸︷︷︸

nonlinear distortion

]′′

+ · · ·

Remember: the input signal s(t ) is slowly varying




−s(t ) +1c

s′(t )︸︷︷︸


[O(ǫ2)

]+

[(148

−1c2

)

s(t )︸︷︷︸

linear distortion

+148

s3(t )︸︷︷︸

nonlinear distortion

]′′

+ · · ·

Remember: the input signal s(t ) is slowly varying

Distortion O(ǫ2) = O((

typical signal freq.switching freq.

)2 )

∼

(1kHz

500kHz

)2


Removing the distortion

How do we modify the amplifier design to

retain the negative feedback

butremove the distortion?


Removing the distortion: IOne way to remove the distortion is to modulate the carrier wave



v (t ) = −k∫ t {

w (τ) + r (τ)}

dτ

k = 4

w(t)dt−k+ ∫

r(t)v(t)



v (t ) = −k∫ t {

w (τ) + r (τ)}

dτ

k = 4

w(t)dt−k+ ∫

r(t)v(t)

Working through all the details again, with this morecomplicated carrier wave, gives gaudio(t ) to be

−s +1c

s′

+

[(116

c2s2(s′ − 4r )′)

+

(

[148

−1c2 ]s′ + r ′

)′]

+ · · ·



v (t ) = −k∫ t {

w (τ) + r (τ)}

dτ

k = 4

w(t)dt−k+ ∫

r(t)v(t)

If we choose the modulating function to be r (t ) = s′(t )/4 thenthe nonlinear distortion term is eliminated and gaudio(t ) is simply

−s +1c

s′ +

([148

−1c2

]

s′ +s′′

4

)′

+ · · ·

This has only linear distortion

This is Bruce Candy’s method for removing the nonlineardistortion: remember the Halcro dm88?


Removing the distortion: IIAnother way to remove the distortion arises from themathematical model



It involves a sample-and-hold element and a multiplier in thecircuit

dt−c ∫−++

s(t)

h(t)

g(t)0

v(t)

S/H

+

Again gaudio(t ) has only linear distortion



It involves a sample-and-hold element and a multiplier in thecircuit

dt−c ∫−++

s(t)

h(t)

g(t)0

v(t)

S/H

+

Again gaudio(t ) has only linear distortion – it works!


Pulse skipping in asecond-order class-Damplifier with negative

feedback


A second-order class-D amplifier

0

v(t)

−c

−c

dt∫1s(t) g(t)p(t)

m(t)

2 dt∫

+ +−+



0

v(t)

−c

−c

dt∫1s(t) g(t)p(t)

m(t)

2 dt∫

+ +−+

For an audio input

s(t ) = 0.7 sin(ωt ) freq. 5000Hz (ω = 10000π)

the audio output of the amplifier is

0.702 sinωt + 5.08 × 10−4 sin 3ωt mathematical model

0.701 sinωt + 5.11 × 10−4 sin 3ωt engineering simulations

for v (t ) with frequency T −1 ≡ 250kHz; thus ǫ = ωT ≈ 0.126



So agreement is excellent between themathematical model and engineering

simulations


Pulse-skipping in a class-D amplifiers(t ) = 0.7 sin(200πt ) [100Hz] v (t ) [250kHz]

Yu Jun, Tan Meng Tong (NTU, Singapore)



Pulse skipping occurs when thecalculated switching times falloutside the ranges

0 < αn <12

12 < βn < 1

g(t)

n βn

−1

+1

αn+1αn n+n+

αn

βn

n+1 n+1+

αn+1t



But the mathematical model predicts switching times with

αn ∼ 116(1 − s(ǫn)) [4 − c1(1 + s(ǫn))]

βn ∼ 12 + 1

16(1 + s(ǫn)) [4 − c1(1 − s(ǫn))]

and hence (since −1 < s(t ) < 1 and 0 < c1 < 2)

0 < αn <12

12 < βn < 1

So the mathematical model predicts no pulse skipping


Pulse-skipping in a class-D amplifier

Not such good agreement between themathematical model and engineering

simulations, then?


Pulse-skipping in a class-D amplifierIn fact, the mathematical model and the simulations do agree



The analytical solutionturns out to be unstablefor large enough signalamplitudes: |s(t )| > 0.7

This can lead to pulseskipping, as observed inengineering simulations

0.4 0.6 0.8

1 1.2 1.4 1.6 1.8

2 2.2 2.4

0 0.2 0.4 0.6 0.8 1

Λ

s0





0.4 0.6 0.8

1 1.2 1.4 1.6 1.8

2 2.2 2.4

0 0.2 0.4 0.6 0.8 1

Λ

s0

0

0.2

0.4

0.6

0.8

1

1.2

0 0.002 0.004 0.006 0.008 0.01 0.012 0.014

t

αn

βn

Λ

Need to decrease the precision in Maple to see the pulseskipping





0.4 0.6 0.8

1 1.2 1.4 1.6 1.8

2 2.2 2.4

0 0.2 0.4 0.6 0.8 1

Λ

s0

-1

-0.5

0

0.5

1

0.0038 0.00385 0.0039

g(t)

tNeed to decrease the precision in Maple to see the pulseskipping


Pulse-skipping in a class-D amplifier

In summary, for this second-order amplifier

◮ The mathematical model accurately predicts the amplifier’saudio output

◮ A stability analysis of the clean analytical solution showswhy pulse skipping is observed in practice (quantisationnoise)

◮ Next we need to find a way to eliminate the nonlinear(third-harmonic) distortion and eliminate pulse skipping!


Summary◮ Negative-feedback class-D amplifiers are all around us, in

part because they are so efficient




◮ Mathematical models for these amplifiers provide usefulinformation to engineers (on distortion, pulse skipping, etc)





◮ Mathematical models can be developed to help improveamplifier designs





◮ Mathematical models can be developed to help improveamplifier designs

Current focus:

◮ How to eliminate distortion in the second-order amplifier

◮ How to stabilise the behaviour (to eliminate pulse skipping)

◮ Better ways to analyse the amplifier? …


mathematical models for class-d ampliﬁers€¦ · · 2010-04-21mathematical models for class-d...

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