Lect15 Distofb Ann

8/15/2019 Lect15 Distofb Ann

1/31

EECS 142/242M

Effect of Feedback on Distortion

Prof. Ali M. Niknejad

U.C. BerkeleyCopyright c 2013 by Ali M. Niknejad

October 29, 2013Niknejad Feedback and Distortion


2/31

Effect of Feedback on Disto

+ a

f

si sϵ so

We usually implement the feedback with a passive network

Assume that the only distortion is in the forward path a

s o = a1

s + a2

s 2

+ a3

s 3

+· · ·

s = s i − fs o s o = a1(s i − fs o ) + a2(s i − fs o )2 + a3(s i − fs o )3 + · · ·

Niknejad Feedback and Distortion


3/31

Feedback and Disto (cont)

We’d like to ultimately derive an equation as follows

s o = b 1s i + b 2s 2i + b 3s

3i + · · ·

Substitute this solution into the equation to obtain

b 1

s i + b 2

s 2

i + b

3s 3

i +· · ·

= a1

(s i −

fb 1

s i −

fb 2

s 2

i −fb

3s 3

i +· · ·

)

+ a2(s i − fb 1s i − fb 2s 2i − fb 3s 3i + · · · )2+ a3(s i − fb 1s i − fb 2s 2i − fb 3s 3i + · · · )3

Solve for the first order terms

b 1s i = a1(s i − fb 1s i )

b 1 = a1

1 + a1f =

a1

1 + T



4/31

Feedback and Disto (square)

The above equation is the same as linear analysis (loop gain

T = a1f )

Now let’s equate second order terms

b 2s 2i = −a1fb 2s 2i + a2(s i − fb 1s i )2

b 2(a + a1f ) = a2

1− fa11 + T

2b 2(1 + T )

3 = a2(1 + T − T )2 = a2

b 2 =

a2

(1 + T )3

Same equation holds at high frequency if we replace withT ( j ω)



5/31

Feedback and Disto (cube)

Equating third-order terms

b 3s 3i = a1(−fb 3s 3i ) + a2(−fb 22s 3i ) + a3(s i − fb 1s i )3 + · · ·

b 3(1 + a1f ) = −2a2b 2f 1

1 + T +

a3

(1 + T )3

b 3(1 + T ) = −2a2f

1 + T

a2

(1 + T )3 +

a3

(1 + T )3

b 3 = a

3(1 + a

1f )−

2a22

f

(1 + a1f )5



6/31

Second Order Interaction

The term 2a22f is the second order interaction

Second order disto in fwd path is fed back and combined with

the input linear terms to generate third order distoCan get a third order null if

a3(1 + a1f ) = 2a22f



7/31

HD 2 in Feedback Amp

HD 2 = 1

2

b 2

b 21s om

= 1

2

a2

(1 + T )3

(1 + T )2

a21s om

= 1

2

a2

a21

s om

1 + T

Without feedback HD 2 = 1

2

a2a21 s om

For a given output signal, the negative feedback reduces thesecond order distortion by 11+T



8/31


9/31

HD 3 in Feedback Amp

HD 3 = 1

4

b 3

b 31s 2om

= 14

a3(1 + T )− 2a22f (1 + T )5

(1 + T )3a31

s 2om

= 1

4

a3

a31s 2om

disto with no fb1

(1 + T )

1− 2a

22f

a3(1 + T )



10/31

F db k I A i


11/31

Feedback versus Input Attenuation

af si soso

Notice that the distortion is improved for a given outputsignal level. Otherwise we can see that simply decreasing the

input signal level improves the distortion.Say s o 1 = fs i with f


12/31

BJT With Emitter Degeneration

vi

V Q

RE

I C The total input signal applied

to the base of the amplifier is

v i + V Q = V BE + I E R E

The V BE and I E terms can be split into DC and AC currents(assume α ≈ 1)

v i + V Q = V BE ,Q + v be + (I Q + i c )R E

Subtracting bias terms we have a separate AC and DCequation

V Q = V BE ,Q + I Q R E

v i = v be + i C R E



13/31

BJT ith E itt D (II)


14/31

BJT with Emitter Degen (II)

Now we know that

i c = a1v be + a2v 2be + a3v

3be + · · ·

where the coefficients a1,2,3,··· come from expanding the

exponential into a Taylor series

a1 = g m a2 = 1

2

I Q

V 2t · · ·

With feedback we have

i c = b 1v i + b 2v 2i + b 3v

3i + · · ·


Emitter Degeneration (cont)


15/31

Emitter Degeneration (cont)

The loop gain T = a1f = g mR E

b 1 = g m

1 + g mR E

b 2 =12

q kT

2

I Q

(1 + g mR E )3

b 3 = 1

4 · 6

q kT

3I Q

(1 + g mR E )4

1−2

12

q kT

2I Q

2R E

16

q kT

3

I Q (1 + g mR E )

For large loop gain g mR E → ∞

b 3 = −1

12

q kT

3I Q

(1 + g mR E )4


Harmonic Distortion with Feedback


16/31

Harmonic Distortion with Feedback

Using our previously derived formulas we have

HD 2 = 1

2

b 2

b 21s om

= 14

î c

I Q 11 + g mR E

HD 3 = 1

4

b 3

b 31s 2om

= 124

î c I Q

2 1− 3g mR E 1+g mR E 1 + g mR E


Harmonic Distortion Null


17/31

Harmonic Distortion Null

We can adjust the feedback to obtain a null in HD 3

HD 3 = 0 can be achieved with

3g mR E

1 + g mR E = 1

or

R E = 1

2g m


HD3 Null Example


18/31

HD 3 Null Example

0 20 40 60 80 100

-70

-65

-60

-55

-50

-45

-40

RE = 1

2gmRE

HD3

Example: For I Q = 1mA, R E = 13Ω


BJT with Finite Source Resistance


19/31

BJT with Finite Source Resistance

vi

V Q

RE

I C

RS

v i + V Q − I B R B = V BE + I E R E Assuming that α ≈ 1, β = β 0 (constant). Let R B = R S + r b represent the total resistance at the base.

v i + V Q = V BE + I C

R E + R B

β 0

The formula is the same as the case of a BJT with emitterdegeneration with R E = R E + R B /β 0


Emitter Follower


20/31

Emitter Follower

vi

V Q

I C

vo

RL

The same equations asbefore with R E = R L


Common Base


21/31

Common Base

vi

RE I

C

RL

−V Q

V CC

Same equation as CE with R E

feedback

v i − V Q + I C R E = −V BE



22/31

Calculation Tools: Multi-Tone Excitation


N Tones in One Shot


23/31

N Tones in One Shot

Consider the effect of an m’th order non-linearity on an input

of N tones

y m =

N n=1

An cos ωnt

m

y m = N n=1

An2e ωnt + e −ωnt m

y m =

N

n=−N An

2 e ωnt

m

where we assumed that A0 ≡ 0 and ω−k = −ωk .


Product of sums...


24/31

The product of sums can be written as lots of sums...

=

()×

()×

() · · · ×

()

m−times

=N

k 1=−N

N k 2=−N

· · · N

k m=−N

Ak 1 Ak 2 · · ·Ak m2m

× e j (ωk 1 +ωk 2 +···+ωk m )t

Notice that we generate frequency componentωk 1 + ωk 2 +

· · ·+ ωk m , sums and differences between m

non-distinct frequencies.

There are a total of (2N )m terms.



25/31

Frequency Mix Vector


26/31

q y

Let the vector k = (k −N , · · · , k −1, k 1, · · · , k N ) be a 2N -vectorwhere element k j denotes the number of times a particularfrequency appears in a given term.

As an example, consider the frequency terms

ω2 + ω1 + ω2ω1 + ω2 + ω2ω2 + ω2 + ω1

k = (0, 0, 1, 2)


Properties of k


27/31

p

First it’s clear that the sum of the k j must equal m

N j =−N

k j = k −N + · · ·+ k −1 + k 1 + · · ·+ k N = m

For a fixed vector k 0, how many different sum vectors arethere?

We can sum m frequencies m! ways. But the order of the sumis irrelevant. Since each k j coefficient can be ordered k j ! ways,the number of ways to form a given frequency product is

given by

(m; k ) = m!

(k −N )! · · · (k −1)!(k 1)! · · · (k N )!


Extraction of Real Signal


28/31

g

Since our signal is real, each term has a complex conjugate

present. Hence there is another vector k

0 given by

k 0 = (k N , · · · , k 1, k −1, · · · , k −N )

Notice that the components are in reverse order since

ω− j = −ω j . If we take the sum of these two terms we have2

e j (ωk 1 +ωk 2 +···+ωk m )t

= 2cos(ωk 1 + ωk 2 + · · ·+ ωk m)t

The amplitude of a frequency product is thus given by

2× (m; k )2m

= (m; k )

2m−1


Example: IM 3 Again


29/31

Using this new tool, let’s derive an equation for the IM 3

product more directly.Since we have two tones, N = 2. IM 3 is generated by a m = 3non-linear term.

A particular IM 3 product, such as (2ω1 − ω2), is generated bythe frequency mix vector k = (1, 0, 2, 0).

(m; k ) = 3!

1! · 2! = 3 2m−1 = 22 = 4

So the amplitude of the IM 3 product is 3/4a3s 3i . Relative to

the fundamental

IM 3 = 3

4

a3s 3i

a1s i =

3

4

a3

a1s 2i


Harder Example: Pentic Non-Linearity


30/31

Let’s calculate the gain expansion/compression due to the 5thorder non-linearity. For a one tone, we have N = 1 and m = 5.

A pentic term generates fundamental as follows

ω1 + ω1 + ω1

−ω1

−ω1 = ω1

In terms of the k vector, this is captured by k = (2, 3). Theamplitude of this term is given by

(m; k ) = 5!

2! · 3! =

5 · 42

= 10 2m−1 = 24 = 16

So the fundamental amplitude generated is a5 1016 S

5i .


Apparent Gain Due to Pentic


31/31

The apparent gain of the system, including the 3rd and 5th, is

thus given by

AppGain = a1 + 3

4a3S

2i +

10

16a5S

4i

At what signal level is the 5th order term as large as the 3rdorder term?

3

4a3S

2i =

10

16a5S

4i S i =

1.2

a3

a5

For a bipolar amplifier, we found that a3 = 1/(3!V 3t ) and

a5 = 1/(5!V 5t ). Solving for S i , we have

S i = V t √

1.2× 5× 4 ≈ 127mV


Documents

Lect15 Distofb Ann