Transcript Document

AN OVERVIEW OF SIGMA-DELTA CONVERTERS
G. S. VISWESWARAN
PROFESSOR
ELECTRICAL ENGINEERING DEPARTMENT
INDIAN INSTITUTE OF TECHNOLOGY, DELHI
NEW DELHI 110 016
Email: [email protected]
Telephone: (011) 2659 1077; (011) 2685 2525
1
DOMAIN OF CONVERTERS
Sigma Delta
Successive Approx
Subranging/Pipelined
Flash
Signal bandwidth converted
2
PCM NYQUIST RATE A/D CONVERTERS
E[n] is a sample sequence of a random process
uncorrelated with the sequence x[n].
The probability density of the error process is uniform
over the range of quantization error i.e over /2
The error is a white noise process
3
PCM NYQUIST RATE A/D CONVERTERS
The variance of the noise power for a quantization
level  is given by
2
2



1
2
V
1
2
V


Se2    N  


12 12  2  1 
12  2N 
2
This gives us an SNR
 Sx2 
 Sx2 
  10log 
  4.77  6.02N (dB)
SNR 10log 
 S2 
 V2 
 e


4
PCM NYQUIST RATE A/D CONVERTERS
In a Nyquist converter, the maximum signal to noise
ratio that can be obtained for a sinusoidal input with a
peak voltage of V is given by:
SNR  6.02N  1.76dB
 Every additional bit  6dB of SNR.
eg. Digital audio with signal bandwidth = 20kHz.
If desired resolution = 18 bits
 SNR  110dB.
5
PCM NYQUIST RATE A/D CONVERTERS
What is the problem with getting 18 bits of resolution ?
1.
Nyquist rate converters essentially obtain output
by comparing the input voltage to various reference
levels. These reference levels are obtained by a
process of reference division; using resistors or
capacitors. Any mismatch in the resistors/capacitors
results in loss of accuracy.
2.
For an ‘N’ bit converter, the required matching of
elements is at least 1 part in 2N. Matching of
components to > 10 bits (or > 0.1 %) is difficult.
3.
Nyquist rate converters require a sharp cutoff
anti-aliasing filter.
6
OVERSAMPLED PCM CONVERTERS
Oversampled converters attempt to use relatively
imprecise analog components with additional digital
signal processing circuits to achieve high resolution.
This is done using

Oversampling - the sampling frequency is much
higher than the signal frequency
7
OVERSAMPLED PCM CONVERTERS
8
OVERSAMPLED PCM CONVERTERS
Noise spectrum when sampled at fS >> 2fB

Assume
quantization
noise
is
uniformly
distributed, white and uncorrelated with the signal.

Noise power folds back to –fS/2 to fS/2, 
oversampled converters have lower noise power within
the signal band.

Out of band noise can be removed by a digital
filter following the PCM converter.
9
OVERSAMPLED PCM CONVERTERS
We define Power Spectral Density of the output
random Process is given by
Pxy  Px (f) Hx (f)
2
and Pey  Pe (f) He (f)
2
For an oversampled PCM converter |Hx(f)| = |He(f)| = 1.
White noise assumption states that Pe(f) = Se2(f)/fs
which implies Pey(f) = Sey2(f)/fs. Thus the in band noise
power is given by
2
Sey

fB
fB
2  2fB 
 Pey (f) df  2  Pey (f) df  Se  f 
 S 
fB
0
10
OVERSAMPLED PCM CONVERTERS
We now see that the SNR ratio for this converter is
 S2 


 Sx2 
Sx2


x


SNR  10log 2  10log 2
 10log  2   4.77  6.02N  3.02OSR (dB)
S 
V 
 S 2f f  
ey
e
S


B




The spectrum of the (over) sampled signal can
represented as follows:
11
OVERSAMPLED PCM CONVERTERS

“16-bit resolution digital audio” Oversampled 8bit converter to be used. To get an SNR = 110dB with
fB = 20kHz, we need fS  2.64GHz.
This is still not good enough since the sampling
frequency is too high. Further improvement can be
obtained if noise shaping is used.
12
NOISE SHAPED OVERSAMPLED
PCM CONVERTERS
We see that for an A/D converter the output is given in
general by Y(z) = X(z)Hx(z) + E(z)He(z)
We have seen OS PCM converter using | Hx(z)| = | He(z)| = 1.
We can however realize another converter using | Hx(z)| = 1
but choose He(z) to shape the noise spectrum to improve the
noise performance. Noise shaping or modulation further
attenuates noise in the signal band to other frequencies.
The modulator output can be low pass filtered to attenuate
the out of band noise and finally down sampled to get
Nyquist rate samples.
13
OVERSAMPLED NOISE SHAPING
14
NOISE SHAPED OVERSAMPLED
PCM CONVERTERS

Noise is high pass filtered to get additional
resolution

Simplest z- domain high pass filter: 1 –z-1 We
want an output Y(z) that contains the sun of the input
and quantzation noise that is high pass filtered. i.e.
Y(z) = X(z) + (1-z-1)E(z)
or
= z-1X(z) + (1- z-1)E(z)
15
NOISE SHAPED OVERSAMPLED
PCM CONVERTERS
1
1  z 1
Analog
Digital
One possibility is to first integrate the analog input,
quantize it and then high pass filter it.
16
FIRST ORDER  MODULATION
The naïve system proposed has its own problems. The
first problem is that since it is an open loop system,
the integrator will saturate. It also requires matching
between analog and digital portions of the circuit.
Y(z) = z-1X(z) + (1 – z-1) E(z)
Y(z)
z 1

X(z)  E(z)
1
1
1z
1z

z 1 
z 1
Y(z) 1 

X(z)  E(z)
1 
1
 1z  1z
z 1
Y(z)  (X(z)  Y(z))
 E(z)
1
1z
17
FIRST ORDER  MODULATION
18
FIRST ORDER  MODULATION
Linearized ‘z’ domain model gives
Hx(z) = STF = z-1
He(z) = NTF = 1-z-1
Assuming that the quantization noise is uncorrelated
with the signal,
Sxy(f) = Sx(f)Hx(f) 2
Sey(f) = Se(f)He(f) 2
19
FIRST ORDER  MODULATION
If fB<< fS
1 2 2 f2
Sey (f) 
4 2
fs 12
fs
Thus we obtain the Noise Power as
Pnoise 
fB
 Sey (f)df
 fB
( 2 2 )
1
Pnoise 
36 OSR3
20
Taking OSR to be of the form 2r we can obtain the
SNR as
 Sx2 
 2 
  10log   9.03r (dB)
SNR 10log
2
 3 
S 
 
 e
21
FIRST ORDER  MODULATION
Noise power coming out of First Order Modulator for an OSR of 128.
22
FIRST ORDER  MODULATION
Before we proceed to implement the transfer function
we need to look in to certain realizatios in the sampled
data domain. As the word  implies there is an
integration involved. In the continuous domain, this
requires resistance and capacitance.
As a designer we have the Capacity to Design but not
the Resistance.
23
SWITCHED CAPACITOR CIRCUITS
DOYEN OF SAMPLED DATA DESIGNS
Sampled Signals:

1  jst
xs (t)  x(t)  (t  kTs ) 
e
Ts k  
k  
This gives a z transform

Xs (z)   x(kTs )z k
k  
24
Realizing resistors for Sampled Data Circuits
i1
i2
The average value of current i1 or i2 is given by
1 T /2
1 T /2
1
i1 
i
dt

dq

C (V1  V2 )


1
1
T 0
T 0
T
This emulates a resistance of value R = T/C = 1/fC
25
OTHER REALIZATIONS OF R
26
SWITCHED CAP INTEGRATORS
27
SWITCHED CAP INTEGRATORS
During 1
VC (nT  T / 2)  V1 (nT  T / 2)  V1 (nT)
s
During 2
Cs V1 (nT)   CF (Vo ((n  1)T)  Vo (nT)
Using z transforms, this reduces to
 (Cs / CF ) z 1
Vo (z)
1
 H(z) 

H
(
z
)
z
1
1
V1 (z)
(1  z )
28
SWITCHED CAP INTEGRATORS
If  << 1/T, and using z = exp(jT) we get H(ejT) as
H(e
jT
 Cs 1
1 1
)

CFT j
j RCF
This circuit is then an integrator with a delay using
the transformation s = (z-1)/T and is called the
Forward Euler Integrator.
29
SWITCHED CAP INTEGRATORS
This is another integrator that gives a non inverting
integration at the output and uses the transformation
s = (1-z-1)/T and is called the Backward Euler
Integrator.
30
SWITCHED CAP INTEGRATORS
The sampling capacitor Cs is now effectively Cs + CP,
thus making the realized resistance R = T/(Cs + CP),
different from the intended value --- needs
correction,
look
for
parasitic
insensitive
configuration.
31
SWITCHED CAP INTEGRATORS
32
SWITCHED CAP INTEGRATORS
At 1 Cs gets charged to Vin(nT) and
During 2
Cs (Vin(nT))  CF (Vo ((n  1)T)  Vo (nT)
Giving us
(Cs / CF ) z 1
Vo (z)
 H(z) 
V1 (z)
(1  z 1 )
33
SWITCHED CAP INTEGRATORS
This configuration gives
 (Cs / CF )
Vo (z)
 H(z) 
V1 (z)
(1  z 1 )
34
BACK TO SIGMA DELTA CONVERTERS
Implementation Imperfection in the first order sigmadelta modulator
§
Finite op-amp gain
§
Capacitance mismatch
§
Incomplete settling
35
FINITE OPAMP GAIN
36
FINITE OPAMP GAIN
37
FINITE OPAMP GAIN
Using charge conservations at the nth clock cycle, we have:
CSVI[n]- CSVd[n] = CF [Vo[n]+ Vd[n] – Vo[n-1] - Vd[n-1]]
CS
CS
 Vo [n] 
Vi[n]  Vo [n  1] 
Vd[n]  Vd[n]  Vd[n  1]
CF
CF
Using Vo[n] = Avd[n] and writing in z domain
Vo (z)

Vi n(z)
for
CS 1
z
CF


1 

1





CS  
1 
A




1 
 1  C   1 
A

F 



C
1 

1  S
1


A
CF

CS  CF ,
g
1
2
1
A
;
1
A
 
2
1
A




z 1









gz 1
1  z 1
1
38
FINITE OPAMP GAIN
Output of the modulator is now given by
 Y (z) 
STF 
H(z)
E(z)
X (z) 
1  H(z)
(1  H(z))
gz1
1  (g  )z
1
and NTF 
1   z 1
1  (g  )z 1
where NTF denotes the noise transfer function and
STF denotes the signal transfer function,
NTF ‘0’ is shifted away from DC. Neglecting the
effect of the pole in the NTF,
39
FINITE OPAMP GAIN
| NTF |2  | 1   z 1 |2z  e j ,  
2 f
fs
= 1 +2-2 cos 
2
cos  1 
2
For small 
 Noise power at the output is then
f
f
1 B 2
1 B 2
2
2
Pnoise 
(
1


)
df



df


fs  f 12
fs  f 12
B
B

2
2
1
2

4

1

(1  )2 
OSR 12
12 3 OSR3
40
FINITE OPAMP GAIN
1

1 
2 
A
(1  )  1 
2
 1 
A

2
CS
for
1
CF
2
 1 
 A 
1

 2
2
A
1  
A

2
2
1
2

4

1
Pnoise 
(1  )2 
OSR 12
12 3 OSR3
1
2 1
2 42
1


OSR 12 A2
12 3 OSR3
41
EFFECT OF FINITE BANDWIDTH
Vo

Vi

1 
1   1 
 A 
CS
CF
CS  S  CF  CS 



CF  u  CF 
42
EFFECT OF FINITE BANDWIDTH
Vo
Vi
CS
CF
C

 Vo (t)  Vi S (1  e ut )
s
CF
1
 u
Larger feedback factor  lower gain  faster setting
Settling determines maximum clock frequency
eg: CS = CF = 1pF
  = 0.5
Assume u = 100 MHz
If we want setting to 1% error, time required 
14.6ns  clock frequency = 34MHz.
43
TIME DOMAIN BEHAVIOUR
Y[n] = Y [n-1] + (X[n-1] – V[n-1])
if
Y[n]  0
Y[n] = 1.0
else
Y[n] = - 1.0
44
TIME DOMAIN BEHAVIOUR
For example, for a DC input = , the time domain output
for the first six clock cycles is given by:
Y[n]
V[n]
0
0.0
1
0.33
1
2
-0.33
-1
3
1
1
4
0.33
1
5
-0.33
-1
6
1
1
It can be seen that the average value of the output is 1/3
45
TIME DOMAIN BEHAVIOUR (Non Linear)
§ Quantization error spectrum is not white; successive
output levels may be correlated.
§ Limit cycle oscillations that lead to tones in the output
eg. DC input X[n] = x
For a limit cycle of period T;
V[n] = V[n+T]

Y[n] = Y[n+T]
Since the input is DC, the input to the integrator will
also be periodic.
46
TIME DOMAIN BEHAVIOUR (Non Linear)
Now Y[n] – Y[n-1] = X – V[n-1].
Write this equation for ‘T’ time instances and add; we
get
T
T
i1
i1
Y[T]  Y[0]   X   V[n  1]
but Y[T] = Y [0]
1 T
P N
 X   V[n  1]  
V
T i1
 T 
47
PATTERN NOISE IN  MODULATOR
It should be clear that the  MODULATOR is
expected to give out the output equal to the DC input.
Only limited no. of levels are allowed to the output ,
therefore output has to toggle from one level to
another in order to keep average output equal to the
DC input.
For eg. Input=0.5 Levels allowed are 0 and 1
Then the output will toggle between 0 and 1. If
average is taken then the value of output of SDM is
0.5.
Therefore the output is oscillating with a frequency
half of that of fs. That means in frequency domain
48
the output will have tones at fs/2 and fs.
PATTERN NOISE IN  MODULATOR
Similarly for dc level of 1/256, the output will have, one
one and 255 zeroes in 256 clocks (fs) this means the
output will oscillate at a frequency of (fs/256). Hence it
will have tones lying at multiples of this frequency. As the
dc level comes closer to zero the tonal frequency
decreases. The tones are completely harmless till they are
out of the signal bandwidth.
The thing to note over here is that these tones represent
noise as the information or signal is at 0 frequency rest of
the frequency components are noise. This effect is very
much prominent in I order modulators. Another important
fact is that the amplitudes of the tones decrease as they
come closer to the signal bandwidth. It is always better to
49
analyze them by using simulations.
PATTERN NOISE IN  MODULATOR
The question to be asked is why are this tones dangerous
in the signal bandwidth? The answer to this question lies
in the fact that all the analysis made earlier on was
based on the white noise approximation and the problem
with the tones is that they are much above the expected
noise floor. Hence the true signal to noise ratio is much
lesser than what was expected from the analysis.
50
PATTERN NOISE IN  MODULATOR
It’s generally said that the pattern noise is visible only
for slow moving inputs (not just DC). To understand
this more clearly assume the input signal is a sinusoid
with an input frequency of fm. If fm is a factor of fs
then every time a new period of the sine wave starts
the SDM will generate the same output as it generated
in the earlier period. This means the output will also be
changing with a frequency of fm. Hence the output will
have tones at the harmonics of the input sinusoidal
signal. If fm is very small then some of these harmonics
will lie in signal bandwidth and the SNR will be lesser
than expected.
51
PATTERN NOISE IN  MODULATOR
Pattern Noise Reduces Effective Bits.
The frequency domain output of the SDM shows
tones and a noise floor. Consider them this noise to
be made of two components 1. Tones 2. Random
noise. Therefore in time domain these tones will give
rise to impulses (if a large number of tones exist in
the signal bandwidth). Since there is random noise,
the impulse train will have a slightly varying
magnitude but the frequency of repetition will be
equal to the fundamental frequency. When these
impulses are of the order of 2 or 3 LSBs. This
means ENOB is lesser then was expected.
52
SECOND ORDER  MODULATOR
The 2nd order modulator has one delaying and one
non-delaying integrator. Note that the last loop with
the quantizer must have one unit of delay for
stability. The z-domain transfer function of the
second order modulator is given by:
Y(z) = z-1X(z) +(1-z-1)2 E(z)
NTF = (1-z-1)2
53
SECOND ORDER  MODULATOR
We can calculate the in band noise power of a second
order  modulator to obtain
5
2  4  1 
Pnoise 


12 5  OSR 
Giving us a noise figure of
 4 
 Sx2 
  15.05r (dB)
  10log
SNR 10log
2
S 
 5 
 e


54
SECOND ORDER  MODULATOR
55
INTEGRATOR OVERLOAD
In second order modulator with a single delaying
integrator, simulations show that the maximum
outputs of the two integrators increase as the signal
level increase. Very often, they are several times the
full scale analog input range. The following table
contains data from simulations. The output levels
indicated are the maximum levels at the output of the
two integrators.
56
INTEGRATOR OVERLOAD
Input
level (dB)
Ist
integrator
output level
2nd
integrator
level
-40
-20
-13.9
-10.45
-7.95
-6.02
-4.43
-3.09
-1.9
0.33
0.96
0.99
1.09
1.22
1.33
1.37
1.49
1.43
2.62
2.77
2.8
3.03
3.51
3.99
4.08
5.38
5.21
It is seen that the levels increase as the input value
increases. This reduces the dynamic range of the
modulation since the integrations will now saturate.
The 2nd order modulator can be modified as follows:
57
INTEGRATOR OVERLOAD
The linearized transfer function is
Y(z) = X (z) . z-2 + (1 – z-1)2E(z)
The signal levels at the output of the integrators are
now the following
58
INTEGRATOR OVERLOAD
Input
level (dB)
Ist
integrator
output level
2nd
integrator
level
-40
-20
-13.9
-10.45
-7.95
-6.02
-4.43
-3.09
-1.9
0.33
0.96
0.99
1.09
1.22
1.33
1.37
1.49
1.43
2.62
2.77
2.8
3.03
3.51
3.99
4.08
5.38
5.21
The signal levels at the first integrator output is reduced.
However the second integrator output levels are still high.
§
The SNR in the two cases remains the same.
§
The circuit specifications are now more relaxed since there
are two units of delay in the loop.
59
INTEGRATOR OVERLOAD
We need to reduce the output levels in the second
integrator. For this we need to alter the gain just before
the second integrator. Let us see the effect of altering
this gain.
60
INTEGRATOR OVERLOAD
[X(z)  Y(z)]
kz 1
1z
1
 E(z)  Y(z)
kz 1
 1  (1k)z 1 

 X(z)
 E(z)  Y(z)
1

1
 1z

1z


 Y(z) 
X(z)kz 1
1  (1  k)z
1

E(z)(1  z 1 )
1  (1  k)z 1
Clock
cycle
Output
2
3
4
5
6
7
8
1
1
-1
1
1
1
-1
61
INTEGRATOR OVERLOAD
Therefore, even though the linearized transfer
function has changed, there is no change in the actual
output. This is because we have a two level quantizer,
the output of which depends only on the polarity and
not the magnitude of the input. The quantizer
effectively acts as an AGC and makes the overall gain 1.
The second integrator gain can be adjust to reduce
the integrator output levels. Typically it is made less
than one. For a gain of ½, the integrator output levels
are the following
62
INTEGRATOR OVERLOAD
Signal
level (dB)
Ist
integrator
output level
2nd
integrator
output level
-40
-20
-13.9
-10.45
-7.95
-6.02
-4.43
-3.09
-1.9
0.83
0.96
0.99
1.09
1.22
1.33
1.37
1.49
1.43
0.655
0.69
0.7
0.75
0.87
0.99
1.02
1.34
1.3
63
CIRCUIT NOISE
The sizes of the input capacitors should be chosen both
on the basis of slow rate as well as thermal noise
considerations. Thermal noise is basically introduced by
non-zero resistance of the sampling switches.
The baseband component of this noise is approximately
proportional to (kT/C)(1/OSR) where ‘C’ is the sampling
capacitor. If the OSR = 256, C = 1pF , the noise power
will be 1.625 x 10-11 Joules. The total quantization noise
power in baseband at this OSR, with quantizer levels = 
1 is 5.9 x 10-12 Joules. Choose larger capacitance.
64
SAMPLING JITTER
Sampling Clock Jitter results in non uniform sampling,
increasing total noise power in the quantizer output.
For a sinusoidal input with amplitude A and frequency fx
dX
dt
 .2fx .A cos(2fx t)
X( t  )  X( t)  .
65
SAMPLING JITTER
If the jitter is assumed to be an uncorrelated
Gaussain random process (‘white’), with standard
deviation t, the average power of this error signal is
A2
P 
(2fx)2
2
Since this is assumed to be white, the total error power
in baseband is
2 (2fx)2
P 
8 OSR
66
IMPLEMENTATION IMPERFECTIONS
Supposing the two integrators have the
following transfer functions
g1
1  1z
1
and
g2z 1
1  2z 1

 g z 1
g1
 Y 2
 E(z)  Y(z)
(X(z)  Y(z))
1
1
1  1z

 1  2z
 STF 
& NTF 
g1g2z 1
1  (1  a2  g2  g1g2 )z 1  1 ( a2  g2 )z 2
(1  1z 1 )(1  2z 1 )
1  (1  a2  g2  g1g2 )z 1  1 ( a2  g2 )z 2
67
IMPLEMENTATION IMPERFECTIONS
Assume A1=A2 (the two opamp have the gain). Generally
we can neglect the effect of the denominator and
obtain NTF = (1 – z-1)2
2
| NTF | | 1  z
2f
| | j ,  
z e
fs
1 4
 (1  ) 4  2(1  )2 2  22
(1-)4 is the unshaped noise, 2(1-)2  2 is the 1st order
shaped noise and 2 4 is the 2nd order shaped noise.
To make sure we get second shaped, we need A  OSR.
68
IMPLEMENTATION IMPERFECTIONS
Attenuation
Maximum
Integrator
Output levels
SNR (OSR =
256) Input = 20dB
0.9
5.98, 9.9
70
0.8
3.56, 4.26
83.15
0.7
2.05, 1.86
84.73
0.6
1.29, 0.9
83.59
0.5
0.962, 0.693
86.69
0.4
0.693, 0.625
86.71
0.3
0.510, 0.589
85.93
0.2
0.34, 0.562
77.45
69
 AD Converter
SIGNAL OUTPUTS OF  MODULATOR
71
 D/A CONVERTER
The sigma Delta D/A converter has a similar topology
to the A/D converter. Here the input digital signal
first goes through an interpolation filter, where it is
upsampled and low pass filtered. After this it is fed
to the modulator. The output of the modulator is a
single bit signal, that comes at rate much higher than
the Nyquist rate. The output of the modulator is
ample and held and low pass filtered to give the analog
output.
72
 D/A CONVERTER
73
 D/A CONVERTER
The input to the modulator is a 12 bit signal that is
upsampled. The clock rate is much higher than the
Nyquist rate. The modulator is a second order modulator
and the topology is the same as the A/D converter. All
numbers are in the 2’s complement form. A one bit
quantizer in this case, would simple keep the MSB and
throw out all the other bits. The D/D converter converts
the one bit quantized output to 14 bit positive or negative
74
number as shown.
AT LAST
75