Noise in Switched-Capacitor Circuits 17 March 2014 Trevor Caldwell [email protected]ECE1371 Advanced Analog Circuits 2 What you will learn • How to analyze noise in switched-capacitor circuits • Significance of switch noise vs. OTA noise Power efficient solution Impact of OTA architecture • Design example for ΔΣ ΔΣ ΔΣ ΔΣ modulator
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Noise in Switched-Capacitor Circuitsjohns/ece1371/slides/ECE...Noise in Switched-Capacitor Circuits 17 March 2014 Trevor Caldwell [email protected] ECE1371 Advanced Analog
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• How to analyze noise in switched-capacitor circuits
• Significance of switch noise vs. OTA noise
Power efficient solution
Impact of OTA architecture
• Design example for ∆Σ∆Σ∆Σ∆Σ modulator
3
Review
• Previous analysis of kT/C noise (ignoring OTA/opamp noise)
Phase 1: kT/C1 noise (on each side)
Phase 2: kT/C1 added to previous noise (on each side)
Total Noise (input referred): 2kT/C1
Differentially: 4kT/C1
4
Review
• SNR (differential)
Total noise power: 4kT/C1
Signal power: (2V)2/2
SNR: V2C1/2kT
• SNR (single-ended)
Total noise power: 2kT/C1 (sampling capacitor C1)
Signal power: V2/2 (signal from -V to V)
SNR: V2C1/4kT
5
Noise in an Integrator
• Two noise sources VC1 and VOUT
VC1: Represents input-referred sampled noise on input switching transistors + OTA
VOUT: Represents output-referred (non-sampled) noise from OTA
6
Thermal Noise in OTAs
• Single-Ended Example
Noise current from each transistor is
Assume
2 4n mI kT gγγγγ====2 / 3γγγγ ====
VIN+
VOUT
VB1
M2M1
M4M3
M5
VIN-In1
In5
In3 In4
In2
7
Thermal Noise in OTAs
• Single-Ended Example
Thermal noise in single-ended OTA
Assuming paths match, tail current source M5 does not contribute noise to output
PSD of noise voltage in M1 (and M2):
PSD of noise voltage in M3 (and M4):
Total input referred noise from M1 - M4
Noise factor nf depends on architecture
1
8
3 m
kT
g
3
2
1
8
3m
m
kTg
g
3,
1 1 1
16 161
3 3m
n eq f
m m m
gkT kTS n
g g g
= + == + == + == + =
8
OTA with capacitive feedback
• Analyze output noise in single-stage OTA
Use capacitive feedback in the amplification / integration phase of a switched-capacitor circuit
9
OTA with capacitive feedback
• Transfer function of closed loop OTA
where the DC Gain and 1st-pole frequency are
Load capacitance CO depends on the type of OTA – for a single-stage, it is CL+C1C2/(C1+C2), while for a two-stage, it is the compensation capacitor CC
1mo
O
g
C
ββββωωωω ====
1 2
11 /G C C
ββββ≈ = +≈ = +≈ = +≈ = +
,
( )1 /
OUT
n eq o
V GH s
V s ωωωω= == == == =
++++
10
OTA with capacitive feedback
• Integrate total noise at output
Minimum output noise for ββββ=1 is
Not a function of gm1 since bandwidth is proportional to gm1 while PSD is inversely proportional to gm1
22
,
0
2
1
( ) ( 2 )
16
3 4
4
3
OUT n eq
of
m
f
O
V S f H j f df
kTn G
g
kTn
C
ππππ
ωωωω
ββββ
∞∞∞∞
====
====
====
∫∫∫∫
4
3f
O
kTn
C
11
OTA with capacitive feedback
• Graphically#
Noise is effectively filtered by equivalent brick wall response with cut-off frequency ππππfo/2 (or ωωωωo/4 or 1/4ττττ)Total noise at VOUT is the integral of the noise within the brick wall filter (area is simply ππππfo/2 x 1/ββββ2)
12
Sampled Thermal Noise
• What happens to noise once it gets sampled?
Total noise power is the same
Noise is aliased – folded back from higher frequencies to lower frequencies
PSD of the noise increases significantly
13
Sampled Thermal Noise
• Same total area, but PSD is larger from 0 to fS/2
Low frequency PSD is increased by
2 2, 4 1
( )4 / 2 / 2 3 / 2
n eq OUTVout f
S S O S
G S V kTS f n
f f C fτ βτ βτ βτ β= = == = == = == = =
2
,n eqG S 31
2dB
S S
f
f f
ππππττττ
====
14
Sampled Thermal Noise
• 1/f3dB is the settling time of the system, while 1/2fS is the settling period for a two-phase clock
PSD is increased by at least
If N = 10 bits, PSD is increased by 7.6, or 8.8dB
• This is an inherent disadvantage of sampled-data compared to continuous-time systems
But noise is reduced by oversampling ratio after digital filtering
1/ 2( 1)2
SfNe ττττ
−−−− − +− +− +− +<<<<
3 ( 1)ln2dB
S
fN
f
ππππ> +> +> +> +
( 1)ln2N ++++
15
Noise in a SC Integrator
• Using the parasitic-insensitive SC integrator
• Two phases to consider
1) Sampling Phase
Includes noise from both φφφφ1 switches
2) Integrating Phase
Includes noise from both φφφφ2 switches and OTA
16
Noise in a SC Integrator
• Phase 1: Sampling
Noise PSD from two switches:
Time constant of R-C filter:
PSD of noise voltage across C1
1 2
8( )
1 (2 )ON
C
kTRS f
fπ τπ τπ τπ τ====
++++
12 ONR Cττττ ====( ) 8Ron ONS f kTR====
17
Noise in a SC Integrator
• Phase 1: Sampling
Integrated across entire spectrum, total noise power in C1 is
Independent of RON (PSD is proportional to RON, bandwidth is inversely proportional to RON)
After sampling, charge is trapped in C1
2
1, 1
1
8
4ON
C sw
kTR kTV
Cττττ= == == == =
18
Noise in a SC Integrator
• Phase 2: Integrating
• Two noise sources: switches and OTA
Noise PSD from two switches:
Noise PSD from OTA:
Noise power across C1 charges to
( ) 8Ron ONS f kTR====
,
1
16( )
3vn eq f
m
kTS f n
g====
2 2
,2 Ron n eqV V++++
19
Noise in a SC Integrator
• What is the time-constant?
Analysis shows that
For large RL, assume that
Resulting time constant
2
1
1/
1L
IN
m L
sC RZ
g R
++++====
++++
1
1IN
m
Zg
≈≈≈≈
1 1(2 1/ )ON mR g Cττττ = += += += +
20
Noise in a SC Integrator
• Total noise power with both switches and OTA on integrating phase
Introduced extra parameter
,2
1,
1 1 1
1
( )
4
16
3 4(2 1/ )
4
3 (1 )
vn eq
C op
f
m ON m
f
S fV
nkT
g R g C
nkT
C x
ττττ====
====++++
====++++
2
1, 2
1 1
1
( )
4
8
4(2 1/ )
(1 )
RonC sw
ON
ON m
S fV
kTR
R g C
kT x
C x
ττττ====
====++++
====++++
12 ON mx R g====
21
Noise in a SC Integrator
• Total noise power on C1 from both phases
Lowest possible noise achieved if
In this case,
What was assumed to be the total noise was actually the least possible noise!
2 2 2 2
1 1, 1, 1 1, 2
1 1 1
1
4
3 (1 ) (1 )
4 / 3 1 2
1
C C op C sw C sw
f
f
V V V V
nkT kT x kT
C x C x C
n xkT
C x
= + += + += + += + +
= + += + += + += + ++ ++ ++ ++ +
+ ++ ++ ++ + ==== ++++
x → ∞→ ∞→ ∞→ ∞
2
1
1
2C
kTV
C====
22
Noise Contributions
• Percentage noise contribution from switches and OTA (assume nf=1.5)
0 2 4 6 8 100
20
40
60
80
100
x=2RONgm1
Noise Fraction (%)
Switch
OTA
23
Noise Contributions
• When gm1 >> 1/RON (x >> 1)#
Switch dominates both bandwidth and noise
Total noise power is minimized
• When gm1 << 1/RON (x << 1)#
OTA dominates both bandwidth and noise
Power-efficient solution
Minimize gm1 (and power) for a given settling time and noise
Minimized for x=0
1 2
1
41 2
3m f
C
kTg n x
Vττττ
= + += + += + += + +
24
Maximum Noise
• How much larger can the noise get?
Depends on nf# (table excludes cascode noise)
Architecture Relative VEFF’s nf
Maximum
Noise (x=0)+dB
Telescopic/
Diff.PairVEFF,1=VEFF,n/2 1.5 3.kT/C1 1.76
Telescopic/
Diff.PairVEFF,1=VEFF,n 2 3.67.kT/C1 2.63
Folded
CascodeVEFF,1=VEFF,n/2 2.5 4.33.kT/C1 3.36
Folded
CascodeVEFF,1=VEFF,n 4 6.33.kT/C1 5.01
25
Separate Input Capacitors
• Using separate input caps increases noise
Each additional input capacitor adds to the total noise
Separate caps help reduce signal dependent disturbances in the DAC reference voltages
C1
VIVO
C2
1 2
12
C1a
VDAC
2
1
2 11
1 1
4 / 3 1 21 ...
1af
C
Cn xkTV
C x C
+ ++ ++ ++ + = + += + += + += + + ++++
26
Differential vs. Single-Ended
• All previous calculations assumed single-ended operation
For same settling time, gm1,2 is the same, resulting in the same total power [0dB]
Differential input signal is twice as large [gain 6dB]
Differential operation has twice as many caps and therefore twice as much capacitor noise (assume same size per side – C1 and C2) [lose ~1.2dB for nf=1.5, x=0# less for larger nf]
• Net Improvement: ~4.8dB
27
Differential vs. Single-Ended
• Single-Ended Noise
• Differential Noise
• Relative Noise (for nf=1.5, x=0)
2
1,
1
4 / 3 1 2
1f
C se
n xkTV
C x
+ ++ ++ ++ + ==== ++++
2 2 2 2
1, 1, 1, 1 1, 2
1 1 1
1
4 2 2
3 (1 ) (1 )
4 / 3 2 4
1
C diff C op C sw C sw
f
f
V V V V
nkT kT x kT
C x C x C
n xkT
C x
= + += + += + += + +
= + += + += + += + ++ ++ ++ ++ +
+ ++ ++ ++ + ==== ++++
2
1,
2
1,
4 / 3 2 4 4
4 / 3 1 2 3
C diff f
fC se
V n x
n xV
+ ++ ++ ++ += == == == =
+ ++ ++ ++ +
28
Noise in an Integrator
• What is the total output-referred noise in an integrator?
Assume an integrator transfer function
where and1
Aµµµµ ====1
2
Ck
C====
�(�) =����
� + � + � − � + ���≈
����
� − ���
29
Noise in an Integrator
• Total output-referred noise PSD
where
and
Since all noise sources are sampled, white PSDs
To find output-referred noise for a given OSR in a ∆Σ∆Σ∆Σ∆Σmodulator:
2
1( ) ( ) ( ) ( )INT C OUTS f S f H z S f= += += += +
2 4
3OUT f
O
kTV n
Cββββ====
2
1
1
4 / 3 1 2
1f
C
n xkTV
C x
+ ++ ++ ++ + ==== ++++
2
/ 2x
x
S
VS
f====
/(2 )
2
0
( )Sf OSR
INT INTV S f df
⋅⋅⋅⋅
==== ∫∫∫∫
30
Noise in a ∆Σ∆Σ∆Σ∆Σ Modulator
• How do we find the total input-referred noise in a ∆Σ∆Σ∆Σ∆Σ modulator?
1) Find all thermal noise sources
2) Find PSDs of the thermal noise sources
3) Find transfer functions from each noise source to the output
4) Using the transfer functions, integrate all PSDs from DC to the signal band edge fS/2�OSR
5) Sum the noise powers to determine the total output thermal noise
6) Input noise = output noise (assuming STF is ~1 in the signal band)
31
Noise in a ∆Σ∆Σ∆Σ∆Σ Modulator
• Example
fS = 100MHz, T = 10ns, OSR = 32
SNR = 80dB (13-bit resolution)
Input Signal Power = 0.25V2 (-6dB from 1V2)
Noise Budget: 75% thermal noise
Total input referred thermal noise: This image cannot currently be displayed.
This image cannot currently be displayed.
32
Noise in a ∆Σ∆Σ∆Σ∆Σ Modulator
1) Find all thermal noise sources
2
1
1
4 / 3 1 2
1fA A
ni
A A
n xkTV
C x
+ ++ ++ ++ +==== ++++
2
2
1
4 / 3 1 2
1fB B
ni
B B
n xkTV
C x
+ ++ ++ ++ +==== ++++
2
2
4
3no fB
B OB
kTV n
Cββββ====
2
1
4
3no fA
A OA
kTV n
Cββββ====
2 323
1 1 1 1
2 21 (1 2 1)ff
n
f f f f
CCkT kTV
C C C C
= + + = + += + + = + += + + = + += + + = + +
33
Noise in a ∆Σ∆Σ∆Σ∆Σ Modulator
2) Find PSDs of the thermal noise sources
For each of the mean square voltage sources,
3) Find transfer functions from each noise source to the output
Assume ideal integrators
2
/ 2x
x
S
VS
f====
1
1( ) ( )
1A B
zH z H z
z
−−−−
−−−−= == == == =−−−−
( ) 1STF z ====
1 2
2
1( ) (1 )
1 2 ( ) ( )NTF z z
H z H z
−−−−= − == − == − == − =+ ++ ++ ++ +
34
Noise in a ∆Σ∆Σ∆Σ∆Σ Modulator
3) Find transfer functions from each noise source to the output
From input of HA(z) to output#
From output of HA(z) to output#
(((( ))))2
1
21 2
2
( ) 2 ( ) ( ) ( )
2 ( ) ( )2
1 2 ( ) ( )
iNTF z H z H z NTF z
H z H zz z
H z H z
− −− −− −− −
= += += += +
++++= = −= = −= = −= = −
+ ++ ++ ++ +
(((( ))))1
1 1
2
( ) 2 ( ) ( )
2 ( )(1 )(2 )
1 2 ( ) ( )
oNTF z H z NTF z
H zz z
H z H z
− −− −− −− −
= += += += +
++++= = − −= = − −= = − −= = − −
+ ++ ++ ++ +
35
Noise in a ∆Σ∆Σ∆Σ∆Σ Modulator
3) Find transfer functions from each noise source to the output
From input of HB(z) to output#
From output of HB(z) to output (equal to transfer function at input of summer to output)#
2
1 1
2
( ) ( ) ( )
( )(1 )
1 2 ( ) ( )
iNTF z H z NTF z
H zz z
H z H z
− −− −− −− −
====
= = −= = −= = −= = −+ ++ ++ ++ +
1 2
2 ( ) ( ) (1 )oNTF z NTF z z−−−−= = −= = −= = −= = −
36
Noise in a ∆Σ∆Σ∆Σ∆Σ Modulator
3) Find transfer functions from each noise source to the output
Most significant is NTFi1
10-3
10-2
10-1
-60
-40
-20
0
20
Normalized Frequency
Magnitude (dB)
Signal Band
|NTFi1|
|NTFo1|
|NTFi2|
|NTFo2|
37
Noise in a ∆Σ∆Σ∆Σ∆Σ Modulator
4) Using the transfer functions, integrate all PSDs from DC to the signal band edge fS/2�OSR
Use MATLAB/Maple to solve the integrals#
/(2 )222 1
1 1
0
2
1
( )/ 2
5 2sin
/ 2 2
Sf OSR
nii i
S
S Sni
S
VN NTF f df
f
f fV
f OSR OSR
ππππππππ
⋅⋅⋅⋅
====
= −= −= −= − ⋅⋅⋅⋅
∫∫∫∫
/(2 )222 1
1 1
0
2
1
( )/ 2
7 2 9sin cos sin
/ 2
Sf OSR
noo o
S
no S S S
S
VN NTF f df
f
V f f f
f OSR OSR OSR OSR
π π ππ π ππ π ππ π ππ ππ ππ ππ π
⋅⋅⋅⋅
====
= + −= + −= + −= + −
∫∫∫∫
38
Noise in a ∆Σ∆Σ∆Σ∆Σ Modulator
4) Using the transfer functions, integrate all PSDs from DC to the signal band edge fS/2�OSR
(Some simplifications can be made for large OSR)
22 22 sin
/ 2S Sni
i
S
f fVN
f OSR OSR
ππππππππ
= −= −= −= −
2 22 2 3
2
3sin cos
/ 2no n S S
o
S
V V f fN
f OSR OSR OSR
π ππ ππ ππ πππππ
++++ = += += += +
4sinSf
OSR
ππππππππ
−−−−
39
Noise in a ∆Σ∆Σ∆Σ∆Σ Modulator
5) Sum the noise powers to determine the total output thermal noise
Assume xA = xB = 0.1 and nfA = nfB = 1.5
With an OSR of 32, first term is most significant (assume ββββA = ββββB = 1/3)