Document 7651437
Download
Report
Transcript Document 7651437
EE 616
Computer Aided Analysis of Electronic Networks
Lecture 9
Instructor: Dr. J. A. Starzyk, Professor
School of EECS
Ohio University
Athens, OH, 45701
1
Outline
2
Sensitivities
-- Network function sensitivity
-- Zero and pole sensitivity
-- Q and 0 sensitivity
Multiparameter Sensitivity
Sensitivities to Parasitics and Operational Amplifiers
Sensitivities
Normalized sensitivity of a function F w.r.t parameter
ln F h F
ln h
F h
S Fh
Two semi-normalized sensitivities are discussed when either F
or h is zero.
S hF
and
S hF
F
F
h
ln h
h
ln F 1 F
h
F h
F can be a network function, its pole or zero, Q etc., while h can be
component value, frequency s, operating temperature or humidity,
etc.
3
SENSITIVITIES - Example
Resonant circuit
1
Z
C
s
1
s
G
1
o
C 2
2
s s
s s
o2
C LC
Q
where
1
0
o C
1
1
Q
G
o GL G
LC
We have
C
L
1
1
ln o ln L ln C
2
2
1
1
ln Q ln G ln C ln L
2
2
so
o
L
1
&
2
also
o
G
S
4
S
S Co S QL S QC
0
SQG 1
1
2
SENSITIVITIES
The use of sensitivities can be demonstrated when we replace
differentials by increments. Using the above example we have
S QL
Q L
L Q
and since SQL 1 => Q 1 L
2
Assume that there is a 1% increase of
then
Q
L
Q
1
0.01 0.5%
Q
2
so we can expect Q to decrease by 0.5%.
5
2 L
L
0.01
L
Network function sensitivity
If network function is
if
N
D
ln T ln N ln D
then
so
T
ln T
S
S nN S Dh
ln h
T
n
T T exp ( j )
ln T
then
ln T ln T j
T
S
j
S h j S h
ln h
ln h
T
n
6
so
Example:
we have
KCL at node v1 :
(v1 - E)G1 + (v1 - vout)G2 + (v1 - v2)sC2 + v1sC1 = 0
KCL at node v2 :
(v2 - v1) sC2 + v2G3 = 0
or
G1 G 2 s (C1 C 2 ) AG 2 sC 2 v1 EG 1
v 0
sC 2
sC 2 G 3 2
and from here the transfer function
7
Example (cont’d)
AG 1 sC 2
v out
T
E G 1 G 2 s(C1 C 2 ) sC 2 G 3 sC 2 AG 2 sC 2
AG 1C 2 s
2
s C1C 2 sC 2 G1 G 2 G 3 AG 2 C1G 3 G 3 G1 G 2
For C1 = C2 = 1, G1 = G2 = G3 = 1, and A = 2 we have
T(S)
2s
s 2 2s 2
A
A
2s
S S S
G1 sC2 G2 C2 s 1 2
A G1 sC2
D
s 2s 2
T
A
8
N
A
D
A
Zero and pole sensitivity
Zeros and poles give good characterization of network
response for different frequencies.
The sensitivity of the zero of a polynomial is obtained
through expressing zero as function of parameter h.
Ph, s(h) s z 0
Since zero of the polynomial is not known analytically (it can
be obtained by nonlinear iterations), the problem which must
be solved is how to find derivative for evaluation of its
sensitivity without explicit knowledge of the zero or its
functioan dependence on the parameter h.
9
Zero and pole sensitivity (cont’d)
Differentiating P w.r.t. h gives
P P ds
0
h s dh sz
ds
dz
P / h
dh s z dh
P / s s z
=>
This expression is valid for simple zeros and can be used to get
h dz
S
z dh
z
h
if z = a + jb we obtain
S ah
10
h
dz
Re
a
dh
&
S bh
h
dz
Im
b
dh
Zero and pole sensitivity - example
Suppose a transfer function of the network is (compare with
the previous example)
T (s)
AG 1C2 s
s 2C1C2 sC2 G1 G2 G3 AG2 C1G3 G3 G1G2
N
2s
D 2 1 js 1 j
Find the sensitivity of a pole sp = -1+j w.r.t. A
11
Zero and pole sensitivity - example
Using the derived formula we have
ds p
dA
D D
/
A s
ss p
sC 2 G2
2 sC1C 2 C 2 G1 G2 G3 AG2 C1G3
s 1 j
For C1 = C2 = 1, G1 = G2 = G3 = 1, and A = 2 we have
ds p
dA
1 j
1 1
j
22j 2 2 2
so the zero sensitivity w.r.t. A is
S
sp
A
A ds p
2 1 j 1
2jj
s p dA 1 j 2
2
and for sp=a+jb=-1+j
A ds p
2 1
S Re
* 1
a
dA 1 2
a
A
12
, and
S Ab
A ds p 2 1
Im
* 1
b
dA 1 2
Q and 0 sensitivity
In filter design Q and o are easier to work with. For a pair of
_
complex zeros z and z
o
(s z) (s z) s (z z) s z z s s o2
Q
_
where
o
Q
_
2
w o2 z
&
_
( z z)
or for
z a jb
Q
_
o
2a
2
2
o2 a 2 b 2
using zero's sensitivity we obtain
S
Q
h
S
o
h
S
a
h
S
o
h
1
o
a S
2
2
a
h
b 2Sbh Sbh
for a 2 b 2 (high Q circuits)
13
Q and 0 sensitivity (cont’d)
Derivation
o
Sh
1 h 2a a 2bb 1 2 h a 2 h b
b
2 a
2
h o a h
b h
o 2 2h
14
ln a 2 b 2 1 h a 2 b 2
2
2
ln h
2 a b
h
1
2
o
a
2
S ha b 2 S hb
Example
In the case of transfer function from previous example
2s
T 2
s 2s 2
we have z = a+jb = -1+j
so
a b 2 and
2
o
2
2
2
1
Q
2(1)
2
Using S Aa 1 and S Ab 1 we have
1
1
S Ao 2 a 2 S Aa b 2 S Ab (1 1) 0
4
o
In this case
15
S Ao S Ab but Q was low so approximation did not hold.
SQA SAo Sah 0 1 1
Example 2
Derive the transfer function of the network shown in figure.
Find the transfer function sensitivity S Th w.r.t. the capacitors
and the amplifier
KCL at v1:
KCL at v2:
16
G1 G2 sC1 v1 G2v2 sC1vout G1E
G2v1 G2 sC2 v2 0
Example 2 (cont’d)
so
sC
sC V
v1 1 2 v 2 1 2 out
G2
G2 A
T
v out
E
AG 1
G2
G1 G 2 sC1 1 sC 2 G 2 sC1A
AG 1G 2
G1G 2 sC1G 2 sC 2 (G1 G 2 sC1 ) sC1AG 2
Using the formula for transfer function sensitivity
S
T
C1
S S
N
C1
D
C1
C1 D sC1G2 s 2 C1C2 sC1 AG2
D C1
D
S CT2 S CN2 S CD2
17
STA
C2 D sC 2 (G1 G2 sC1 )
D C2
D
sC A
A N A D
1 1
N A D A
D
Multiparameter Sensitivity
The function F generally depends on several parameters
F F h1 , h2 , ... , hm F h
The change in F due to infinitesimally small changes in parameters
is expressed by the total differential
m
dF
i 1
or
F
d hi
hi
m
F hi d hi
dF
d hi
S hiF
F
hi
i 1 hi F hi
To compare different designs we introduce multiparameter
sensitivity measures.
18
Multiparameter Sensitivity (cont’d)
The worst case multiparameter sensitivity
WCMS S Fhi
For incremental changes of parameters within their tolerance
hi
ti
hi
we have
F
S hiF t i
F
or in case all ti are equal to t
F
t * WCMS
F
19
This is a very pessimistic estimate of the function deviation from
its nominal value.
Multiparameter Sensitivity (cont’d)
In IC fabrication design parameters like resistor or capacitor values
track each other – i.e. change in their values are strongly correlated.
So, to design these circuits we use the multiparameter tracking
sensitivity
all elements of type k
MTS k
F
S
hi
i 1
Since all elements of the same kind (e.g. capacitors) have similar
values of hi / hi and for such elements (only)
F h
S Fhi
F
h
then, for all types of elements, the worst case variation with
tracking is given by
F
t k * MTS k
F
k
20
Multiparameter Sensitivity (cont’d)
However, worst case situation is very unlikely to happen in
practice. Fabricated device parameter deviations hi / hi
follow a statistical distribution. Two commonly used distributions to
model parameter deviations are uniform and normal distributions
h
prob
h
-t
0
t
For uniform distribution:
1
h
prob
h 2 * to
21
h
h
Multiparameter Sensitivity (cont’d)
For normal distribution:
prob (
h
1
)
e
h
2
1 h 2
2 h
The function deviation F / F becomes a random variable with its
own distribution. For large circuits F / F has approximately 2
normal distribution with zero mean and variance 2F S hiF h
F
h
provided that the component variations are statistically independent,
where
hi
hi
22
ti2
32
ti
9
for uniform
distributi on
for
distributi on
normal
Multiparameter Sensitivity (cont’d)
If all the tolerances are equal, and hi / hi have the same distribution
then the standard deviation can be calculated from
F h MSS
F
h
where MSS is the multiparameter statistical sensitivity
S
F 2
hi
MSS
i
Actual variation will lie in the interval F 68% of the time,
F
2 F
F
23
95%, and
2 F
F
99.7%.
Example
We have KCL1:
G1 sC2 v1 G1E sC2vout 0
KCL2:
G 3 G 4 v1 v out G 3 0
so
v1
G3
vout
G3 G4
T
24
&
G
G1 sC 2 3 sC 2 vout G1 E
G3 G4
v out
G G G 4
1 3
E
G1G 3 sC 2 G 4
Example (cont’d)
Let us assume that all elements have tolerances t = 1% and s = 1. Let’s calculate
various multiparameter sensitivities and use them to predict deviations of the
transfer function T from its nominal value.
STG1 SGN1 SGD1 1
G1
sC 2 G 4
4 4
G3
D
G1G 3 sC 2 G 4 3 3
C2
4
4
sG 4
D
3
3
G 3 G1
G
1 1
8
STG3 SGN3 SGD3
3 G1
G1 G 3 G 4 D
5 3 15
G 4 G1
G
4 4
STG 4 SGN4 SGD4
4 sC 2
G1 G 3 G 4 D
5 3
G 4 G1
G
4 4
STG 4 SGN4 SGD4
4 sC 2
G1 G 3 G 4 D
5 3
4 4 8 8
28 56
WCMS Si
2
3 3 15 15
15 15
STC2 SCN2 SCD2
25
MTS i S TG1 S TG 3 S TG 4
4
3
MSS
8
15
8
15
16 16 G4
G
4 2.03
9 9 225 225
Example (cont’d)
For the nominal values the transfer function can be evaluated as
5
1.667
3
Let us discuss the effect of 1% changes assuming
G1 = .99
C2 = 1.01
G3 = 1.01
G4 = 3.96
The actual transfer function value can be calculated as
T
T'
.99(1.01 3.96)
4.92
1.640
.99 1.01 1.01 3.96
3
so
T
0.0267
0.016 1.6%
T
1.667
26
Example (cont’d)
while the estimate for such a change using different multiparameter
sensitivities is as follows.
Worst case analysis
T
56
t *WCMS 1% * 3.73%
T
15
(too pessimistic)
Worst case analysis with tracking
T
4
t * MTS 1 t * MTS 2 1% 2 2.67%
T
3
27
(still too big)
Statistical analysis
If tolerances are distributed uniformly then the standard deviation
MSS
T
T
h
h
t
2.03 0.0117
3
and if the tolerances are distributed normally then the
standard deviation
t
T h MSS 2.03 0.0068
3
T
h
indicating that 95% of the time
will be less than 2 T 2.34 %
T
T
in the uniform case and less than 2 T 1.35 %
T
in the normal case
T
28
Since the true deviation that was 1.6% exceeded the 95% limits for
the standard deviation of the normal distribution so our case was
rather uniform than normal.
Sensitivities to Parasitics and Operational Amplifiers
Since parasitics have nominal values equal zero we cannot calculate
sensitivities to these elements in the regular way. Denote parasitics by
i . We have
F all nozero elements F hi hi all parasitics F 1
i
F
i 1
i 1
hi F hi
i F
or equivalent
hi
F
S Fhi
S Fi i
F
hi
hi
hi
semi-normalized sensitivity
are fixed for a specific technology, so the
as well as i
only way to reduce functional variation of F is to have design with
small S F and S F
i
hi
29
Sensitivities to Parasitics and Operational Amplifiers (cont’d)
F
To evaluate S i we analyze the network in the regular way, calculate
F
i
and finally substitute i = 0 at the final result.
In the case of Op Amp we may consider the inverse of its
amplification as a parasitic
where
B
1
A
we have
v k v i v j A
or
30
v i v j Bv k 0
B is parasitic. If B 0 then we obtain ideal Op Amp.
Example:
Find the sensitivity S BT for the transfer function T of the
network shown where B 1
A
from the previous example
T
31
v out
G 1G 2
E
BG1G 2 sC1G 2 sC 2 G1 G 2 sC1 sC1G 2
S BT
T 1
B T
D 1
B D
B 0
B 0
G1G2 G1G2 sC1G2 sC2 G1 G2 sC1
D2 T
G1G 2 sC1G 2 sC 2 G1 G 2 sC1
sC1G 2