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Objectives Estimate expected amount of output noise

Determine minimum usable signal level

Gain insight into possible improvements

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RMS value over averaging interval T:

Xn2

is themean square value of noise x represents current ior voltagee

We will deal with multiple sources of noise

Different noise sources are added as vectors

Xn =1

T

T

0

x2n(t)dt

Xn =

X

2

n1+X

2

n2

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Frequency Domain

Noise RMS: computed in the frequencydomain

Thenoise power density is normally specified

From the power density the rms value is found

replaceXwithIfor current, Vfor voltage

x2n

(f) =dX2

n

df

Xn =

fH

fL

x2n(f)df

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White &1/fNoise

White: uniform spectral density

white-noise power is proportional to BW

1/f noise: varies with reciprocal off

1/f - noise power is proportional to # ofdecades or octaves

Xn = xnwfH fL

xn = K/f Xn = K

ln (f

H

/fL)

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Normally noise is described by specifyingenw

andfce(voltage) or inw andfci(current)

Source of noise appear in different places, andequivalent input noise sources are found

Input sources are filtered by amplifier toproduce output noise

To simplify calculations, actual magnituderesponse is replaced by aBrick-wall equivalent

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fo

(log)

f

gain

1/21/2

1.57fo

Brick-wall

equivalent

For single-pole system

Noise-equivalent bandwidth: 1.57f0 for above case

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Example 7.3 Piece-wise noise integration

20

eni

nV/Hz1/2

10

102

103104

105106

107

10

102

103104

105106

10710

102

103104

105106

107

10

|A|V/V

20

eno

nV/Hz1/2

10

102

103104

105106

107|A|

200

2

1

200

1 Hz to 1kHz:

En = enw

fceln(fH/fL) + fH fLwith

enw = 20nV /Hzfce = 100Hz fL = 1Hz

andfH = 1kH z

The result is Eno1 = 0.822V.

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20

eno

nV/Hz1/2

10

102

103104

105106

107|A|

200

From 1kHz to 10kHz,eno increases with f at a rate of

1dec/dec.So let

no(f) =

20nV/Hz

(f/103) = 21011f

nd

no2 = 21011104

103f2df = 11.5V

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Sources of noise

R

e

2

R=4kTR

i2R=4kT/R

R

(noiseless)

(noiseless)

!

+

VO

Vn

Vp

inn

inp

e

noiseless

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Example 7.7

!

+

R1

R2

R3

Eno

!

+

VO

Vn

Vp

inn

inp

en

noiseless

R1

R2

R3

iR1

iR2

iR3

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Use superposition to obtain:

e2

ni = e2

n + i2

npR2

3+ i2R3R

2

3+i2

nn + i2

R1+ i2R2

(R1 R2)

2

Using i2R =4kTR

,

e2

ni = e2

n + i2

npR2

3+ 4kTR3 +

i2

nn +4kTR1

+ 4kTR2

(R1 R2)

2

= e2n + i2

npR2

3+ 4kTR3 + i

2

nn (R1 R2)2 +

4kT

R1 R2(R1 R2)

2

=e2

n +i2

npR

2

3 + 4kTR3

+i2

nn (R1

R2

)

2

+ 4kT

(R1

R2

)= e2n + i

2

npR2

3+ i2nn (R1 R2)

2 + 4kT (R3 + R1||R2)

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For inp = inn = in,

e2

ni = e2

n + i2

nR2

s2 + 4kTRs

where Rs = R3 + R1||R2, R2

s2 = R2

3+ (R1||R2)

2.

Setting R3 = 0 reduces noise.

en dominates for low values ofRs: it is called the short-circuitnoise.

For Rs , eni i2

nR

2

s2: in is called the open-circuit noise.

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Noise is amplified by:

Av =1 + R2

R11 + (f/fA)

2=

A0

1 + (f/fA)

2

The total output rms noise is

Eno = A0

E21

+ E22

+ E23

+ E24

where

E21

= e2nw

fceln

fAfL

+ 1.57fA fL

E22

= R23i2npw

fcipln

fAfL

+ 1.57fA fL

E23

= (R1||R2)2i2nnw

fcinln

fAfL

+ 1.57fA fL

andE2

4= 4kT(R3 + R1||R2)(1.57fA fL)

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Book says: fLis 1/Tobs where Tobs is the averaging time used tomeasure the output.

TI Application note says: take fH/fL = NEB.For low-noise designs, use op amps with low enw and low corner

frequencies fci and fcn.The total rms input noise can be obtained by dividing by the

signal dc gain As0

Eni =Eno|As0|

and the signal-to-noise ratio from

SNR = 20log10Vi(rms)Eni