Noise
David Johns and Ken Martin
University of Toronto
(johns@eecg.toronto.edu)
(martin@eecg.toronto.edu)
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Interference Noise
• Unwanted interaction between circuit and outside
world
• May or may not be random
• Examples: power supply noise, capacitive coupling
Improvement by ...
• Reduced by careful wiring or layout
• These notes do not deal with interference noise.
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Inherent Noise
• Random noise — can be reduced but NEVER
eliminated
• Examples: thermal, shot, and flicker
Improvement by ...
• Not strongly affected by wiring or layout
• Reduced by proper circuit DESIGN.
• These notes discuss noise analysis and inherent
noise sources.
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Time-Domain Analysis
0.4
0.3
0.2
v n(t)
(volts)
0.1
0
-0.1
-0.2
-0.3
0
0.2
0.4
0.6
0.8
time
1
(seconds)
• Assume all noise signals have zero mean
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RMS Value
T
1 2
V n ( rms ) ≡ --- ∫ v n(t)dt
T
1⁄2
(1)
0
• T — suitable averaging time interval
• Indicates normalized noise power.
• If v n(t) applied to 1Ω resistor, average power
dissipated, P diss ,
2
V n ( rms )
2
P diss = ------------------ = V n ( rms )
1Ω
(2)
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SNR
signal power
SNR ≡ 10 log -----------------------------noise power
(3)
2
• If signal node has normalized signal power of V x ( rms ) ,
2
and noise power of V n ( rms ) ,
2
V x ( rms )
V x ( rms )
SNR = 10 log ------------------ = 20 log -----------------2
V n ( rms )
V n ( rms )
(4)
• When mean-squared values of noise and signal are
same, SNR = 0dB .
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Units of dBm
• Often useful to know signal’s power in dB on
absolute scale.
• With dBm, all power levels referenced 1mW.
• 1mW signal corresponds to 0 dBm
• 1 µW signal corresponds to -30dBm
What if only voltage measured (not power)?
• If voltage measured — reference level to equiv
power dissipated if voltage applied to 50 Ω resistor
• Also, can reference it to 75 Ω resistor
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dBm Example
• Find rms voltage of 0 dBm signal ( 50Ω reference)
• What is level in dBm of a 2 volt rms signal?
• 0 dBm signal (50 Ω reference) implies
( 50Ω ) × 1mW = 0.2236
V ( rms ) =
(5)
• Thus, a 2 volt (rms) signal corresponds to
2.0
20 × log  ---------------- = 19 dBm
 0.2236
(6)
2
• Would dissipate 2 ⁄ 50 = 80 mW across a 50Ω resistor
• 80 mW corresponds to 10 log ( 80 ) = 19 dBm
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Noise Summation
+
v n1(t)
i no(t)
v no(t)
v n2(t)
i n2(t)
i n1(t)
-
Current
Voltage
(7)
v no(t) = v n1(t) + v n2(t)
T
2
1
= --- ∫ [ v n1(t) + v n2(t) ] dt
T
2
V no ( rms )
(8)
0
2
V no ( rms )
2
V n1 ( rms )
=
+
2
V n2 ( rms )
T
2
+ --- ∫ v n1(t)v n2(t)dt
T
(9)
0
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Correlation
• Last term relates correlation between two signals
• Define correlation coefficient, C ,
T
1
--- ∫ v n1(t)v n2(t)dt
T
0
C ≡ -----------------------------------------V n1 ( rms ) V n2 ( rms )
2
2
2
(10)
V no ( rms ) = V n1 ( rms ) + V n2 ( rms ) + 2CV n1 ( rms ) V n2 ( rms )
(11)
• Correlation always satisfies – 1 ≤ C ≤ 1
• C = +1 — fully-correlated in-phase (0 degrees)
• C = -1 — fully-correlated out-of-phase (180 degrees)
• C = 0 — uncorrelated (90 degrees)
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Uncorrelated Signals
• In case of two uncorrelated signals, meansquared value of sum given by
2
2
2
(12)
V no ( rms ) = V n1 ( rms ) + V n2 ( rms )
• Two rms values add as though they were vectors at
right angles
When fully correlated
2
V no ( rms ) = ( V n1 ( rms ) ± V n2 ( rms ) )
2
(13)
• sign is determined by whether signals are in or out of
phase
• Here, rms values add linearly (aligned vectors)
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Noise Summation Example
• V n1 ( rms ) = 10µV , V n2 ( rms ) = 5µV , then
2
2
2
V no ( rms ) = ( 10 + 5 ) = 125
(14)
which results in V no ( rms ) = 11.2µV .
• Note that eliminating V n2 ( rms ) noise source same as
reducing V n1 ( rms ) = 8.7 µV (i.e. reducing by 13%)!
Important Moral
• To reduce overall noise, concentrate on large
noise signals.
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Frequency-Domain Analysis
• With deterministic signals, frequency-domain
techniques are useful.
• Same true for dealing with random signals like noise.
• This section presents frequency-domain techniques
for dealing with noise (or random) signals.
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Spectral Density
2
V n(f)
V n(f)
1000
2 100
( µV ) 10
-------------Hz
1.0
0.1 1.0
31.6
µV 10
----------Hz 3.16
1.0
log ( f )
10 100 1000
Spectral Density
( Hz )
0.1 1.0
log ( f )
( Hz )
10 100
Root-Spectral Density
• Periodic waveforms have their power at distinct
frequencies.
• Random signals have their power spread out over
the frequency spectrum.
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Spectral Density
2
Spectral Density V n(f)
• Average normalized power over a 1 hertz bandwidth
• Units are volts-squared/hertz
Root-Spectral Density V n(f)
• Square root of vertical axis (freq axis unchanged)
• Units are volts/root-hertz (i.e. V ⁄ Hz ).
Total Power
2
V n ( rms )
∞
2
(15)
= ∫ V n(f)df
0
• Above is a one-sided definition (i.e. all power at
positive frequencies)
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Root-Spectral Density Example
31.6
10
V n(f)
3.16
µV
1.0
 -----------
 Hz
0.1 1.0
10 100
• Around 100 Hz, V n(f) =
log ( f )
( Hz )
10 µV ⁄ Hz
• If measurement used RBW = 30 Hz , measured rms
10 × 30 =
300 µV
(16)
• If measurement used RBW = 0.1 Hz, measured rms
10 × 0.1 = 1 µV
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White Noise
V n(f)
10
3.2
1.0
µV 
 --------- Hz
µV
V n(f) = V nw = 3.2 ----------Hz
0.1
1.0
log f
( Hz )
100 1000
10
• Noise signal is “white” if a constant spectral density
(18)
V n(f) = V nw
where V nw is a constant value
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1/f Noise
– 10 dB/decade
V n(f)
µV  10
 --------- Hz
3.2
1.0
1 ⁄ f noise corner
0.1
1.0
1/f noise
dominates
10
100 1000
2
V n(f)
–6 2
( 3.2 × 10 )
–6 2
≈ -------------------------------- + ( 1 × 10 )
f
log f
white noise
dominates
2
2
V n(f) = k v ⁄ f (k v is a constant)
(19)
• In terms of root-spectral density
V n(f) = k v ⁄ f
Falls off at -10 db/decade due to
(20)
f
• Also called flicker or pink noise.
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Filtered Noise
2
V ni(f)
2
2 2
V no(f) = A(j2πf) V ni(f)
A(s)
V no(f) = A(j2πf) V ni(f)
• Output only a function of magnitude of transferfunction and not its phase
• Can always apply an allpass filter without affecting
noise performance.
• Total output mean-squared value is
2
V no ( rms )
∞
2 2
(21)
= ∫ A(j2πf) V ni(f)df
0
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Noise Example
V ni(f)
R = 1kΩ
nV  20
 --------- Hz
3
4
1.0 10 100 10 10
A(j2πf)
( dB )
0
– 20
V ni(f)
log f
nV 
 --------- Hz
3
1.0 10 100 10
3
10
4
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C = 0.159µF
V no(f)
1
A(s) = -------------------s
1 + ----------2πf o
f o = 10
1
f o = --------------2πRC
V no(f)
log f
20
2
1.0 10 100 10
3
10
4
log f
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Noise Example
• From dc to 100 kHz of input signal
2
10
5
2
7
2
V ni ( rms ) = ∫ 20 df = 4 × 10 ( nV ) = ( 6.3 µV rms )
2
(22)
0
• Note: for this simple case,
2
V ni ( rms ) = 20 nV ⁄ Hz × 100kHz = ( 6.3 µV rms )
2
(23)
• For the filtered signal, V no(f) ,
–9
20 × 10
V no(f) = -------------------------f 2
1 +  ----
f 
(24)
o
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Noise Example
• Between dc and 100kHz
2
V no ( rms )
10
5
10
2
5
2
20
f
= ∫ ---------------------- df = 20 f o atan  ----
f 
f 2
o
0 1 +  ----
0
f 
o
5
2
= 6.24 × 10 ( nV ) = ( 0.79 µV rms )
2
(25)
• Noise rms value of V no(f) is almost 1 ⁄ 10 that of V ni(f)
since high frequency noise above 1kHz was filtered.
• Don’t design for larger bandwidths than required
otherwise noise performance suffers.
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Sum of Filtered Noise
V n1(f)
A 1(s)
V n2(f)
uncorrelated
noise sources
A 2(s)
V n3(f)
A 3(s)
V no(f)
• If filter inputs are uncorrelated, filter outputs are also
uncorrelated
• Can show
2
V no(f) =
∑
2 2
(26)
A i(j2πf) V ni(f)
i = 1, 2 , 3
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Noise Bandwidth
• Equal to the frequency span of a brickwall filter
having the same output noise rms value when white
noise is applied to each
Example
A(j2πf)
A brick(j2πf)
– 20 dB/decade
0
( dB )
0
( dB )
– 20
– 20
fo
--------100
fo
-----10
log f
fo
10f o
log f
fo
--------100
fo
-----10
fo
π
f x = --- f o
2
π
2
• Noise bandwidth of a 1’st-order filter is --- f o
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Noise Bandwidth
• Advantage — total output noise is easily calculated
for white noise input.
• If spectral density is V nw volts/root-Hz and noise
bandwidth is f x , then
2
2
(27)
V no ( rms ) = V nw f x
Example
• A white noise input of 100 nV/ Hz applied to a 1’st
order filter with 3 dB frequency of 1 MHz
V no ( rms ) = 100 × 10
–9
6
π
× --- × 10
2
V no ( rms ) = 125 µV
(28)
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1/f Noise Tangent Principle
• Method to determine the frequency region(s) that
contributes the dominant noise
• Lower a 1/f noise line until it touches the spectral
density curve
• The total noise can be approximated by the noise in
the vicinity of the 1/f line
• Works because a curve proportional to 1 ⁄ x results in
equal power over each decade of frequency
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1/f Tangent Example
V no(f)
200
1/f curve
20
2
1
10
100
10
3
10
4
10
5
10
6
10
7
Frequency (Hz)
N1
N2
N3
N4
• Consider root-spectral noise density shown above
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1/f Tangent Example
2
N1
100
2
100
2
5
2
200
= ∫ -----------df = 200 ln ( f ) 1 = 1.84 × 10 ( nV )
f
(29)
1
2
N2
10
3
2
2
10
3
5
= ∫ 20 df = 20 f 100 = 3.6 × 10 ( nV )
2
(30)
100
10
4
2 2
2
20 f
20
N 3 = ∫ ---------------- df =  --------

2
3
3
10
3 ( 10 )
10
∞
2
2
13
--- f
3
∞
10
4
10
3
8
= 1.33 × 10 ( nV )
2
10
2
(31)
4
2
2
200
200
N 4 = ∫ -------------------------- df = ∫ -------------------------- df – ∫ 200 df
2
2
f
 f 
4
0 1 +  --------
0
10 1 +  -------5-
 5
10
10
2  π
5
2
4
9
= 200 --- 10 – ( 200 ) ( 10 ) = 5.88 × 10 ( nV )
 2
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1/f Tangent Example
• The total output noise is estimated to be
2
2
2
2 1⁄2
V no ( rms ) = ( N 1 + N 2 + N 3 + N 4 )
(33)
= 77.5µV rms
• But ...
(34)
N 4 = 76.7µV rms
• Need only have found the noise in the vincinity
where the 1/f tangent touches noise curve.
• Note: if noise curve is parallel to 1/f tangent for a
wide range of frequencies, then also sum that
region.
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Noise Models for Circuit Elements
• Three main sources of noise:
Thermal Noise
• Due to thermal excitation of charge carriers.
• Appears as white spectral density
Shot Noise
• Due to dc bias current being pulses of carriers
• Dependent of dc bias current and is white.
Flicker Noise
• Due to traps in semiconductors
• Has a 1/f spectral density
• Significant in MOS transistors at low frequencies.
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Resistor Noise
• Thermal noise — white spectral density
R (noiseless)
2
V R(f) = 4kTR
R
element
2
4kT
I R(f) = --------R
R
(noiseless)
models
• k is Boltzmann’s constant = 1.38 × 10
– 23
JK
–1
• T is the temperature in degrees Kelvin
• Can also write
V R( f ) =
R
------ × 4.06 nV ⁄ Hz
1k
(35)
for 27°C
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Diodes
• Shot noise — white spectral density
kT
r d = --------qI D
(noiseless)
2
V d(f) = 2kTr d
(forward-biased)
kT
r d = --------qI D
2
I d(f) = 2qI D
(noiseless)
element
models
• q is one electronic charge = 1.6 × 10
– 19
C
• I D is the dc bias current through the diode
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Bipolar Transistors
• Shot noise of collector and base currents
• Flicker noise due to base current
• Thermal noise due to base resistance
2
1
V i (f) = 4kT  r b + ----------

2g 
2
V i (f)
m
(noiseless)
2
(active-region)
I i (f)
element
model
IC 
KI B

2
I i (f) = 2q  I B + --------- + --------------
2
f

β (f) 
• V i(f) has base resistance thermal noise plus
collector shot noise referred back
• I i(f) has base shot noise, base flicker noise plus
collector shot noise referred back
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MOSFETS
• Flicker noise at gate
• Thermal noise in channel
2
V g(f)
2
I d(f)
(active-region)
2
K
V g(f) = -------------------WLC ox f
2
2
I d(f) = 4kT  --- g m
 3
element
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MOSFET Flicker (1/f) Noise
2
K
V g(f) = -------------------WLC ox f
(36)
• K dependent on device characteristics, varies widely.
• W & L — Transistor’s width and length
• C ox — gate-capacitance/unit area
• Flicker noise is inversely proportional to the
transistor area, WL .
• 1/f noise is extremely important in MOSFET circuits
as it can dominate at low-frequencies
• Typically p-channel transistors have less noise since
holes are less likely to be trapped.
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MOSFET Thermal Noise
• Due to resistive nature of channel
2
• In triode region, noise would be I d(f) = ( 4kT ) ⁄ r ds
where r ds is the channel resistance
• In active region, channel is not homogeneous and
total noise is found by integration
2
2
I d(f) = 4kT  --- g m
 3
(37)
for the case V DS = V GS – V T
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Low-Moderate Frequency MOSFET Model
2
V i (f)
(noiseless)
2
2 1
K
V i (f) = 4kT  --- ------ + ------------------- 3 g
m WLC ox f
(active-region)
element
model
• Can lump thermal noise plus flicker noise as an input
voltage noise source at low to moderate frequencies.
• At high frequencies, gate current can be appreciable
due to capacitive coupling.
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Opamps
2
I n-(f)
2
V n(f)
element
(noiseless)
2
I n+(f)
V n(f), I n-(f), I n+(f)
— values depend on opamp
— typically, all uncorrelated.
model
• Modelled as 3 uncorrelated input-referred noise
sources.
• Current sources often ignored in MOSFET opamps
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Why 3 Noise Sources?
2
V n(f) ignored ⇒ V no = 0
Actual V 2no = V 2n
2
V no(f)
2
R
2
I n-(f) ignored ⇒ V no = V n
2
V no(f)
2
2
Actual V no = V n + ( I n- R )
2
2
I n+(f) ignored ⇒ V no = V n
2
R
2
V no(f)
2
2
Actual V no = V n + ( I n+ R )
2
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Capacitors
• Capacitors and inductors do not generate any noise
but ... they accumulate noise.
• Capacitor noise mean-squared value equals kT ⁄ C
when connected to an arbitrary resistor value.
R
R
C
1
f 0 = --------------2πRC
V R(f) =
4kTR
V no(f)
C
• Noise bandwidth equals ( π ⁄ 2 )f o
2
2
π
π
1
V no ( rms ) = V R(f)  --- f o = ( 4kTR )  ---  ---------------
 2
 2  2πRC
2
kT
V no ( rms ) = -----C
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Capacitor Noise Example
• At 300 °K , what capacitor size is needed to have
96dB dynamic range with 1 V rms signal levels.
• Noise allowed:
1V
V n ( rms ) = ------------------- = 15.8 µV rms
96 ⁄ 20
10
(39)
• Therefore
kT
C = ----------------- = 16.6pF
2
V n(rms)
(40)
• This min capacitor size determines max resistance
size to achieve a given time-constant.
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Sampled Signal Noise
• Consider basic sample-and-hold circuit
φ clk
v in
v out
C
•
• When φ clk goes low, noise as well as signal is held
on C . — an rms noise voltage of kT ⁄ C .
• Does not depend on sampling rate and is
independent from sample to sample.
• Can use “oversampling” to reduce effective noise.
• Sample, say 1000 times, and average results.
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Opamp Example
Vi
Cf
Cf
Rf
R1
I nf
I n1
I n-
Vo
Rf
V no(f)
R1
R2
V n2
Vn
I n+
R2
circuit
equivalent noise circuit
• Use superposition — noise sources uncorrelated
2
• Consider I n1, I nf and I n- causing V no1(f)
2
Rf
2
2
2
2
V no1(f) = ( I n1(f) + I nf(f) + I n-(f) ) -----------------------------1 + j2πfR f C f
(41)
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Opamp Example
2
• Consider I n+, V n2 and V n causing V no2(f)
2
Rf ⁄ R1
2
2
2
2
2
V no2(f) = ( I n+(f)R 2 + V n2(f) + V n(f) ) 1 + ------------------------------1 + j2πfC R
(42)
f f
• If R f « R 1 then gain ≅ 1 for all freq and ideal opamp
would result in infinite noise — practical opamp will
lowpass filter noise at opamp f t .
• If R f » R 1 , low freq gain ≅ R f ⁄ R 1 and f 3dB = 1 ⁄ ( 2πR f C f )
similar to noise at negative input — however, gain
falls to unity until opamp f t .
2
2
2
Total noise: V no(rms) = V no1(rms) + V no2(rms)
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Numerical Example
• Estimate total output noise rms value for a 10kHz
lowpass filter when C f = 160pF , R f = 100k , R 1 = 10k ,
and R 2 = 9.1k .
• Assume V n(f) = 20 nV ⁄ Hz , I n(f) = 0.6 pA ⁄ Hz
opamp’s f t = 5 MHz .
• Assuming room temperature,
I nf = 0.406 pA ⁄ Hz
(44)
I n1 = 1.28 pA ⁄ Hz
(45)
V n2 = 12.2 nV ⁄ Hz
(46)
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Numerical Example
2
• The low freq value of V no1(f) is found by f = 0 , in
(41).
2
2
2
2
2
V no1(0) = ( I n1(0) + I nf(0) + I n-(0) )R f
2
2
2
9 2
= ( 0.406 + 1.28 + 0.6 ) ( 1 × 10 ) ( 100k )
= ( 147 nV ⁄ Hz )
2
2
(47)
• Since (41) indicates noise is first-order lowpass
filtered,
2
2
π⁄2
V no1(rms) = ( 147 nV ⁄ Hz ) × -------------------------------------------------2π ( 100kΩ ) ( 160pF )
2
= ( 18.4 µV )
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Numerical Example
2
• For V no2(0)
2
2
2
2
2
V no2(0) = ( I n+(f)R 2 + V n2(f) + V n(f) ) ( 1 + R f ⁄ R 1 )
2
= ( 24.1 nV ⁄ Hz ) × 11
= ( 265 nV ⁄ Hz )
2
2
2
(49)
• Noise is lowpass filtered at f o until f 1 = ( R f ⁄ R 1 )f o
where the noise gain reaches unity until f t = 5MHz
• Breaking noise into two portions, we have
2
–9 2 π ⁄ 2
–9 2 π
V no2(rms) = ( 265 × 10 )  ------------------ + ( 24.1 × 10 )  --- ( f t – f 1 )
 2πR C 
 2
f f
= ( 74.6 µV )
2
(50)
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Numerical Example
• Total output noise is estimated to be
V no(rms) =
2
2
V no1(rms) + V no2(rms) = 77 µV rms
(51)
• Note: major noise source is opamp’s voltage noise.
• To reduce total output noise
— use a lower speed opamp
— choose an opamp with a lower voltage noise.
• Note: R 2 contributes to output noise with its thermal
noise AND amplifying opamp’s positive noise
current.
• If dc offset can be tolerated, it should be eliminated
in a low-noise circuit.
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CMOS Example
• Look at noise in input stage of 2-stage CMOS opamp
V dd
V n5
V bias
Q5
V n1
V n2
V in+
Q1
Q3
Q2
V n3
V inV no (output)
V n4
Q4
V ss
• Equivalent voltage noise sources used since
example addresses low-moderate frequency.
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CMOS Example
V no
V no
--------- = --------- = g m1 R o
V n1
V n2
(52)
where R o is the output impedance seen at V no .
V no
V no
--------- = --------- = g m3 R o
V n3
V n4
(53)
g m5
V no
=
-------------------2g m3
V n5
(54)
• Found by noting that V n5 modulates bias current and
drain of Q 2 tracks that of Q 1 due to symmetry — this
gain is small (compared to others) and is ignored.
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CMOS Example
2
2 2
2 2
V no(f) = 2 ( g m1 R o ) V n1(f) + 2 ( g m3 R o ) V n3(f)
(55)
• Find equiv input noise by dividing by g m1 R o
2
V neq(f)
=
2
2V n1(f)
+
 g m3
2
2V n3(f)  ---------
 g m1
2
(56)
Thermal Noise Portion
• For white noise portion, subsitute
2
2 1
V ni(f) = 4kT  ---  --------
 3  g 
(57)
 g m3 1
16
1
16
V neq(f) =  ------ kT  --------- +  ------ kT  ---------  ---------
 3  g   3  g  g 
m1
m1
m1
(58)
mi
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CMOS Example
• Assuming g m3 ⁄ g m1 is near unity, near equal
contribution of noise from the two pairs of transistors
which is inversely proportional to g m1 .
• In other words, g m1 should be made as large as
possible to minimize thermal noise contribution.
1/f Noise Portion
• We make the following substitution into (56),
g mi =
W
2µ i C ox  ----- I Di
 Li
 ( W ⁄ L ) 3 µ n
2
2
2
V ni(f) = 2V n1(f) + 2V n3(f)  --------------------------
 ( W ⁄ L ) 1 µ p
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CMOS Example
• Now, letting each of the noise sources have a
spectral density
Ki
2
V ni(f) = ----------------------W i L i C ox f
(61)
 µ n  K 3 L 1  
2
2  K1
V ni(f) = -----------  -------------- +  ------  -------------- 
C ox f  W 1 L 1  µ p  W L 2 
(62)
we have
1 3
• Recall first term is due to p-channel input transistors,
while second term is due to the n-channel loads
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CMOS Example
Some points for low 1/f noise
• For L 1 = L 3 , the noise of the n-channel loads
dominate since µ n > µ p and typically n-channel
transistors have larger 1/f noise than p-channels (i.e.
K 3 > K 1 ).
• Taking L 3 longer greatly helps due to the inverse
squared relationship — this will limit the signal
swings somewhat
• The input noise is independent of W 3 and therefore
one can make it large to maximize signal swing at
the output.
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CMOS Example
Some points for low 1/f noise
• Taking W 1 wider also helps to minimize 1/f noise
(recall it helps white noise as well).
• Taking L 1 longer increases the noise due to the
second term being dominant.
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