Upload
buicong
View
227
Download
0
Embed Size (px)
Citation preview
MAE140 Linear Circuits 188
Designing Circuits – Synthesis - Lego
Port = a pair of terminals to a cct One-port cct; measure I(s) and V(s) at same port
Driving-point impedance = input impedance = equiv impedance =Z(s) Two-ports
Transfer function; measure input at one port, output at another
I(s)
V(s) +
- sCR
sLsIsVsZ
++==
−11
)()()(
I1(s)
V1(s) +
- V2(s)
I2(s) +
-
Inputs Outputs
MAE140 Linear Circuits 189
Transfer functions
Transfer function; measure input at one port, output at another
I1(s)
V1(s) +
- V2(s)
I2(s) +
-
Inputs Outputs
€
Transfer function =zero - state response transform
input signal transform
(I.e., what the circuit does to your input)
MAE140 Linear Circuits 190
Example 11-2, T&R, 6th ed
€
TV(s) =
V2(s)
V1(s)
=R
1
R1+1sC
1
=s
s+1R
1C
1
Transfer function? Input impedance?
€
Z(s) =V
1(s)
I1(s)
= R1+1sC
1
MAE140 Linear Circuits 191
Cascade Connections
We want to apply a chain rule of processing
When can we do this by cascade connection of OpAmp ccts? Cascade means output of ccti is input of ccti+1
This makes the design and analysis much easier
This rule works if stage i+1 does not load stage i Voltage is not changed because of next stage
Either Output impedance of source stage is zero
Or Input impedance of load stage is infinite
Works well if Zout,source<<Zin,load
)(...)()()()( 321 sTsTsTsTsT VkVVVV ××××=
MAE140 Linear Circuits 192
Cascade Connections
Is chain rule valid? + _ R1 V1(s) R2
2
1sC
1
1sC
V2(s) V3(s)
+ + +
Mesh analysis
€
TVtotal(s)=?T
V 1(s)×T
V 2(s) =
R1C
1s
R1C
1s+1
×1
R2C
2s+1
=R
1C
1s
R1R
2C
1C
2( )s 2 + R1C
1+ R
2C
2( )s+1
I2(s) I1(s)
0)(1)(
)()()(1
22
2111
121111
=!!"
#$$%
&+++−
=−!!"
#$$%
&+
sIsC
RRsIR
sVsIRsIRsC
!"
#$%
&
!!!!
"
#
$$$$
%
&
++−
−+=!!
"
#$$%
&
−
0)(
1
1
)()( 1
1
212
1
111
2
1 sV
RRsC
R
RRsC
sIsI
( ) ( )
( ) ( ))(
1)(1)(
)(1
)(
1222111
22121
112
23
1221211
22121
1212
2
sVsCRCRCRsCCRR
sCRsIsC
sV
sVsRCRCRCsCCRR
RCCssI
++++==
++++=
No! Why?
MAE140 Linear Circuits 193
Cascade Connections – OpAmp ccts
OpAmps can be used to achieve the chain rule property for cascade connections The input to the next stage needs to be driven by the
OpAmp output Consider standard configurations
Noninverting amplifier No current drawn from V1 – no load
Inverting amplifier
Current provided by V1(s) Need to make sure that stage is
driven by OpAmp output to avoid loading V1(s)
V1(s) + -
V2(s) +
Z2 +
Z1
V1(s) + -
Z1 Z2
V2(s)
+ + I(s)
)()()(
11sZsVsI =
MAE140 Linear Circuits 194
OpAmp Ccts and transfer functions Node B: Node B:
+ _ V1(s) + -
Z1 Z2
V2(s)
+ A B C
)()(
)()()(0)(
0)()()(
)()()(
12
12
22
11
sZsZ
sVsVsTsV
sZsVsV
sZsVsV
VB
BB
−==⇒=
=−
+−
+ _ V1(s)
- +
V2(s) + A
B
C
Z2
Z1
)()()()()()(
0)()(
)()()(
121
1
122
sZsZsZsTsVsV
sZsV
sZsVsV
VB
BB
+=⇒=
=+−
MAE140 Linear Circuits 195
Example 11-4, T&R, 5th ed, p511
Find the transfer function from V1(s) to V2(s)
1 1 1
1 1 1 1 ) (
sC C sR
sC R s Z1 +
= + =
2 2 1 1 ) 1 )( 1 ( ) ( 1 2 + + - =
C sR C sR C sR s T V
R2
R1
V1(s)
+ -
1
1sC
2
1sC
V2(s)
+ +
1 1 ) ( 2 2 2
2 2 2
2 2 + =
+ =
s C R R
sC R sC
R s Z
MAE140 Linear Circuits 196
Circuits as Signal Processors Design a circuit with transfer function R1=R2=100Ω, C1=C2=1µF, C3=100µF, L=70mH, R3=1Ω
( ) ( ) ( )
2 4 8 4 2 5 2
10 10 10 2 10 42 . 1
+
- + =
+ × + × +
s
j s j s s s
s 408 408
VO(s)
R2
R1
VI(s)
+ - 1
1sC
2
1sC
V2(s)
+
+ - +
3
1sC
R3
sL
( ) ( ) ( )
Ls s LC R
C sR C sR C sR s T V
1 1 1 ) (
2 3 3
2 2 1 1 1 2 + -
× + +
- =
MAE140 Linear Circuits 197
Transfer Function Design – OpAmp Stages
First order stages α
+ -
1 Κγ Κ
α
γ
+
+ - =
s
s K s T V ) ( Series RL design
Series RC design
1
+ -
Κ γK1
α1
MAE140 Linear Circuits 198
First-order stages
α
γ
+
+ - =
s
s K s T V ) (
1
+ -
γK1
α1
K1
Parallel RL design
1
+ -
γK1
α1
Κ
Parallel RC design
MAE140 Linear Circuits 199
Design Example 11-20, T&R, 5th ed, p 542
Design two ccts to realize
Unrealistic component values – scaling needed
) 4000 )( 1000 ( 3000 ) (
+ + =
s s s s T V
1000µF + -
1Ω 1Ω
+ - 250µF
1Ω 3Ω
Stage 1 Stage 2
[ ]100010001
1011101
)( 1
3
31 +
−=
"""
#
$
%%%
&
'
+−= −
−
−
ss
ssTV [ ][ ]40003400013)(
12 +
−=+−=
−
ss
ssTV
MAE140 Linear Circuits 200
Design Example 11-19, T&R, 5th ed, p 539
Non-inverting amplifier design
Less OpAmps but more difficult design Three stage: last stage not driven
Unrealistic component values still – scaling needed
1Ω + -
1000µF
1Ω 2Ω
250µF
1Ω
Stage 1 Voltage Divider
Stage 2 OpAmp
Stage 3 Voltage Divider
) 4000 )( 1000 ( 3000 ) (
+ + =
s s s s T V
MAE140 Linear Circuits 201
Scaled Design Example 11-21, T&R, 5th ed, p 544
More realistic values for components
Need to play games with elements to scale The ratio formulas for TV help permit this scaling
It certainly is possible to demand a design TV which is unrealizable with sensible component values
Like a pole at 10-3 Hz
100nF + -
10ΚΩ 10ΚΩ
+ - 25nF
10ΚΩ 30ΚΩ
MAE140 Linear Circuits 202
Second-order Stage Design
Circuit stages to yield
02ζω=R 1=L KC
−= 20
11
ω
KC 12=2
0ω≤K
+ -
120−
ω
K1Ω Ω
1Η Ω02ζω
F20
1
ω + -
Ω02
1ζω
H20
1
ω
F1HK1
€
TV(s) =
Ks 2 + 2ςω
0s+ω
0
2
MAE140 Linear Circuits 203
Circuit Synthesis
Given a stable transfer function TV(s), realize it via a cct using first-order and second-order stages
We are limited to stable transfer functions to keep within the linear range of the OpAmps There is an exception
When the unstable TV(s) is part of a stable feedback system
Come to MAE143B to find out
Transistor cct design is conceptually similar
cbsassssTV++
++= 2
2)( γβα
basssTV ++
=βα)(