Download Lecture 10

10. Design of Op. Amp. Circuits for
special applications
10.1
10.2
10.3
10.4
10.5
Negative Resistance Circuit
Sign Changer
Differential Voltage-Current Converters
Voltage-Current Converters
Analog Computation Circuits
1
10.1
Negative Resistance Circuit
RF
RA
Negative
Resistance
+
v
-
vo
i
R

R 
voRA
 v  vo   1  F   v
RA  RF
RA 

R  v v
R 
v  vo v 
i
   1  F     1  1  F 
R
R 
RA  R R 
RA 
 RR 
v
    A 
i
 RF 
v  v 
Negative Resistance generator !
EE3601-10
Electronics Circuit Design
2
Example
• Draw the negative resistance Op. Amp. circuit and find the value of the
negative resistance in terms of the circuit components, RF , RA and R
• Design the circuit to have a negative resistance of –1kW, and a specified
minimum dc path resistance at each op. amp. input of 10kW.
RF

R 
vR
v   v   o A  v  vo   1  F   v
RA  RF
RA 

R  v v
R 
v  vo v 
i
   1  F     1  1  F 
R
R 
RA  R R 
RA 
 RR 
v
    A 
i
 RF 
RA
+
v
vo
i
R
-
 RR 
v
  A   1kW, But dc path at non  inv input  R  10kW
i
 RF 
R
 1k
 A 
 0.1
RF  10k
dc path at inv. input RF // RA 
RF RA
RF  RA
R
 RA  1.1  10k  11kW, RF  A  110kW
0.1
EE3601-10
Electronics Circuit Design
 10k 
1
RA
RA

RF
RA
1  0.1
3
10.2
Sign Changer
R
vi
R
vO
+
-
R1
a
when the switch is at ""
v   v   vi no current in both R also in R1 
 vo  v   vi  vo  vi 
when the swich is at ""
v   v   0 no current in R1 
Sign Changer !
vo  0 0  vi

 vo  vi
R
R
EE3601-10
Electronics Circuit Design
4
Example
• Draw the Sign Changer Op. Amp. circuit and show how the output sign
changes according to the switch position
• Design the circuit which has an input resistance of 1MW.
when the switch is at ""
v   v   vi no current in both R also in R1 
 vo  v   vi  vo  vi 
R
vi
when the swich is at ""
R
-
v   v   0 no current in R1 
vO
+
R1
a
vo  0 0  vi

 vo  vi
R
R
when the switch is at "" Rin  a  R1  a  1MW  a or R1  1MW is OK
and vO  vi
when the swich is at "" Rin  R  1MW
and vO  vi
EE3601-10
Electronics Circuit Design
5
10.3
Differential Voltage-Current Converter
R2
v
vi
R1
vO
R1
RL
IL
R2
V  Vi V  Vo
V  V V  Vo

 IL  0  

R1
R2
R1
R2

V  Vi
V V
 IL  
R1
R1
IL 
V  V  V  Vi  V1  V


R1
R1
Differential voltage to current converter !
EE3601-10
Electronics Circuit Design
6
Example
• Draw the Differential voltage to current converter Op. Amp. circuit and find the
value of the load current IL in terms of the circuit components, R1 , V1 and V
• Design the circuit to have a full-swing load current IL of ±1mA for a specified full
input of ± 1mV at the circuit of the op. amp.
V  Vi V  Vo
V  V V  Vo

 IL  0  

R1
R2
R1
R2

0mV
v
R1
vi
+1mV
R1
V  Vi
V V
 IL  
R1
R1
IL 
R2
vO
RL
V  V  V  Vi  V1  V


R1
R1
R2
IL
+1mA
V V
 1mV
 1mV
IL  1
 1mA 
 R1 
 1kW
R1
R1
 1mA
We can take RL  R2  R1  1kW
EE3601-10
Electronics Circuit Design
7
10.4
Voltage-Current Converter
R
IL
R
V1
R1
V1
V  V  V1
but ILR1  V (no drop at R)
 V  V1  ILR1
IL 
V1
R1
Voltage to current converter !
EE3601-10
Electronics Circuit Design
8
Example
• Draw the Voltage to current converter Op. Amp. circuit and find the value of the
load current IL in terms of the circuit components, R1 , V1 and V
• Design the circuit to drive a dc motor with load current IL of 20mA from digital
drive voltage of 5V connected to the input of the op. amp. Circuit. (Note that op.
amp. power supply should be ±Vcc > ±5V)
R
V  V  V1
but ILR1  V (no drop at R)
 V  V1  ILR1
IL 
V1
R1
R
5V
20mA
IL
dc motor
5V
R1
V
5V
IL  1  20mA  R1 
 250W
R1
20mA
we can take R  R1  250W,
(Note input current int o the circuit is nearly zero from 5V digital source)
EE3601-10
Electronics Circuit Design
9
Summary of Design Equations
Special Application Circuits
1. Negative Resistance
Generator
RA
 RR 
v
  A 
i
 RF 
+
v
2. Sign Changer
RF
vo
i
R
-
vi
R
-
R
vO
R
R1
RL
R2
R1
4. Voltage to current converter
R1
IL
vO
+
switch at "" vo  v1 a switch at "" vo  v1
3. Differential voltage to
R2
current converter
v
vi
R
V V
IL  1
R1
V1
EE3601-10
Electronics Circuit Design
IL
V1
R1
V
IL  1
R1
10
10.5
Analog Computation Circuits
• Integration and Multiple Summing
Q
d3x
Q1  
3
dt
C
R
Z
R4
2
dt
dx
dt
C
R
RC=1
R1
R3
Q2 
R
RC=1
R2
d2x
R
Q3  x
X
C
RC=1
C
Y
R
iR
ic
vc
vv+
Y
1
X dt  Y  integral of X
CR

 d2 x  R


Z
 dt2  R
4


R
R  dx  R  d2 x  R
x        2   Z 

R1
R2  dt  R3  dt  R4
R
R


Y



x

Y
R1
R2
 dx  R


dt

 R3
 X1V1  Y1V2  X2V3  Y2V4  Output equation11
EE3601-10
Electronics Circuit Design
Example
• Design the Op. Amp. Circuit, that can operate the following differential equation
and whose lowest resistance at any input terminal of the Op. Amp should be 10kW
d3x
dx
d2x

4x


8
 12
dt
dt3
dt2
(a) Draw Integrators with RC=1 where the first input is the highest order in given
differential equation. Note the output of each Integrator.
Q
d3x
Q1  
dt3
R
d2x
Q2 
dt2
C
R
C
dx
dt
V1  x
C
R
RC=1
RC=1
RC=1
(b) Rearrange the given equation using available output of each Integrator.
d3x
dx
d2x
 4x 
 8 2  12
dt
dt3
dt
 d2x 
dx
  6 2
 4(-x) 
 8  dt2 
dt


Here the constant “12” is taken as a (+2V
battery) with a coefficient of 6.
Available outputs from the Integrators
EE3601-10
Electronics Circuit Design
12
(c) Find Z from the modified equation and design the Multiple-Input Op. Amp. to
get Vo same value as the highest order and draw the designed circuit:
Assume Vo 
d3x
 d2x 
dx
  6 2  X  9, Y  10, Z  9  10  1   2


4(-x)


8
3
2


dt
dt
 dt 
Case
Z
RY
RX
Ri
(n-inv)
Rj
(inv)
II
Z<0

- RF / Z
RF / Xi
RF / Yj
Z0
R
R
R
R
R
From table, RY  , RX   F , R1  F , R2  F , Ra  F , Rb  F
2
1
8
4
6
R
Lowest resistance is R2  F  10k,
8
R
80k
R1  F  80k, R2 
 10k,
1
8
80k
80k
Ra 
 20k, Rb 
 13.33k
4
6
 RF  80k
RF=80k
20k
2V
-x
dx/dt
-(d x/dt2)
Rx
2
EE3601-10
Electronics Circuit Design
VO
13.33k
80k
10k
40k
dx
d2 x
Vo  4x 
8
 12
dt
dt2
13
(d) Connect the Multiple-Input Op. Amp. Inputs with the outputs of the Integrators
showing the output equals to the highest order of the given equation and draw the
designed circuit:
Q
d3x
Q1  
dt3
d2x
Q2 
dt2
C
R
dx
dt
V1  x
C
R
C
R
Input signal
RC=1
RC=1
-x
2V
-(d2x/dt2)
RF=80k
20k
VO
13.33k
dx/dt
Rx
RC=1
80k
10k
40k
EE3601-10
Electronics Circuit Design
Output signal
dx
d2x
d3x
Vo  4x 
 8 2  12  3
dt
dt
dt
14
(e) form a loop connecting the output Vo back to the first input (highest
order of the equation) of the Integrators and get the Analog Computation
Circuit that can operate the given differential equation
Q
d3x
Q1  
dt3
d2x
Q2 
dt2
C
R
dx
dt
V1  x
C
R
C
R
Input signal
RC=1
RC=1
-x
2V
Rx
d3 x
VO
13.33k
dx/dt
-(d2x/dt2)
RF=80k
20k
80k
10k
40k
Output signal
EE3601-10
Electronics Circuit Design
Designed Analog
Computation
Circuit
RC=1
dx
d2 x
Vo 
 4x 
8
 12
3
2
dt
dt
dt
or
d3 x
dt3
8
d2 x
dt2

dx
 4x  12
dt
15
• Initial Condition of the equation
(f) The initial condition of the equation is electronically generated by
a voltage source connected at the output of the given term.
switch closes at t  0
2
switch closes at t  0
2
d x
d x
Q1   2  5 or
5
dt
dt 2
Q
d3x
dx
Q1 
 3
dt
Q1  
3
dt
5V
d2x
dt
t=0
R
C
"0"
RC=1
Q3  x  0
Q2 
2
3V
dx
dt
t=0
R
Q3  x
t=0
C
R
"0"
switch closes at t  0
C
"0"
RC=1
EE3601-10
Electronics Circuit Design
RC=1
16
Summary of Design Equations
Analog Computation Circuit
(a) Draw Integrators with RC=1 where the first input is the highest degree in given
differential equation. Note the output of each Integrator.
(b) Rearrange the given equation using available output of each Integrator.
(c) Find Z from the modified equation and design the Multiple-Input Op. Amp. to
get Vo same value as the highest order and draw the designed circuit:
(d) Connect the Multiple-Input Op. Amp. Inputs with the outputs of the Integrators
showing the output equals to the highest order of the given equation and draw the
designed circuit:
(e) form a loop connecting the output Vo back to the first input (highest
order of the equation) of the Integrators and get the Analog Computation
Circuit that can operate the given differential equation
(f) The initial condition of the equation is electronically generated by
a voltage source connected at the output of the given term.
EE3601-10
Electronics Circuit Design
17