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03-1 Operational Amplifiers *
* 03-1
These are common components in conditioning circuits.
There are two inputs, v+ and vi, two power supplies, +Vs and Vs, and
an output, vo .
vo = min f+Vs; max fVs; (v+ v )Agg, were A is the op-amp gain.
Ignoring saturation, vo = A (v+ v ) .
Ideal Op-Amp Properties
Infinite input impedance.
Infinite gain. (A = 1)
Zero output impedance.
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03-2 *
* 03-2
Where to Find Ideal Op-Amps
An electronics textbook.
However, in certain circuits a real op-amp performs almost the same as
an ideal op-amp would.
Simplifying Assumptions
Current into inputs is zero.
When used in a negative feedback configuration, v+ = v .
Op-Amp Circuits to be Covered
Non-inverting amplifier.
Inverting amplifier.
Summing amplifier.
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03-3 Non-Inverting Amplifier *
* 03-3
vo = RA__+_RB_____Rvi.
A
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03-4 Versatility of Inverting Amplifier *
* 03-4
Use of Non-Inverting Amplifier in Conditioning Circuits
vo = RA__+_RB_____Rvi.
A
Traditional Use, Voltage Amplifier
Input is vi, output is vo .
Hc(vi) = RA__+_RB_____Rvi
A
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03-5 Non-Inverting Amplifier Example Problem *
* 03-5
Design a system with output vo = H (x), where process variable x is
water level, x 2 [0 m; 1 m], and H (x) = 10x V__m.
Note: most example problems will not be as complete as the archetypical
problem covered earlier.
Solution:
Use same float-and-cable system as in previous example problem.
Use 100 k three-terminal variable resistor with 1 V voltage source across
fixed terminals:
Ht(x) = 1x V___m.
Problem will be solved two ways:
First way, we know what kind of conditioning circuit is needed.
Second way, we have to determine algebraicly the type of conditioning
circuit needed.
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03-6 *
* 03-6
First Way: Use Non-Inverting Amplifier
Obviously, all that is needed is an amplifier with a gain of 10. A non-
inverting amplifier will do.
Then:
H(x) = Hc(Ht(x)) = A 1x V___m
10x V___m= A 1x V___m
A = 10
So choose resistors such that (RA + RB )=RA = 10.
For example, RA = 10 k and RB = 90 k.
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03-7 *
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Second Way: Derive Conditioning-Circuit Function
Pretend we don't know that a simple amplifier is needed.
H(x) = Hc(Ht(x))
We need to solve for Hc.
Let y = Ht(x) = 1x V___m.
Then x = H1t (y) = 1y m___V.
Substituting:
i m j
H 1y ____V = Hc(y)
i m j i m j
Hc(y) = H 1y ____V = 10 1y ____V
= 10y
Therefore our conditioning circuit needs to multiply y, a voltage, by a
constant.
A non-inverting amplifier will do just that.
(The remainder of the solution is identical to the first way.)
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03-8 Inverting Amplifier *
* 03-8
vo = RB___RvA .
A
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03-9 *
* 03-9
RB
Output vo = _____vA .
RA
Traditional Use, Voltage Amplifier
Input is vA and output is vo .
Hc(vA ) = RB___RvA = A1 vA
A
where A1 = RB___R.
A
Resistance to Voltage Converter
Input is RB , output is vo .
Hc(RB ) = RB___RvA = A2 RB
A
where A2 = vA =RA .
Here, vA is a fixed voltage, buried in the constant A2 .
03-9 EE 4770 Lecture Transparency. Formatted 9:45, 22 January 1999 from lsli03*
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03-10 *
* 03-10
Inverted Resistance to Voltage Converter
Input is RA , output is vo .
Hc(RA ) = RB___RvA = A3 =RA
A
where A3 = RB vA .
vA is a fixed voltage here also.
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03-11 *
* 03-11
Current to Voltage Converter
This circuit is similar to the inverting amplifier.
Input is iA , output is vo .
Hc(iA ) = RB iA = A4 iA .
where A4 = RB .
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03-12 Inverting Amplifier Example Problem *
* 03-12
Design a system with output vo = H (x), where process variable x is
water level, x 2 [0 m; 1 m], and H (x) = 10x V__m.
This is the same as the non-inverting amplifier problem.
Solution:
Use same float-and-cable system as in previous example problem.
Use a 100 k two__-terminal variable resistor:
Ht(x) = 100x k____m.
The variable resistor will be the "input" to the inverting amplifier.
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03-13 *
* 03-13
Conditioning-Circuit Function Derivation
So far: H(x) = 10x V__m(given) and Ht(x) = 100x k___m(choice of trans-
ducer).
Solve for Hc in:
Hc(Ht(x)) = H(x)
Let y = Ht(x) = 100x k____m.
Then x = 0:01y _m___k.
Substituting:
i m j
Hc(y) = H 0:01y _____k
i m j V
= 10 0:01y _____k ____m
= 0:1y _V___k
For the inverting amplifier used as a resistance-to-voltage converter:
Hc(RB ) = A2 RB .
RB ! y and A2 ! 0:1 _V___k.
Choose RA and vA so that the following equation is satisfied:
0:1 _V___k= vA___R
A
For example, RA = 60 k and vA = 6 V.
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03-14 Another Inverting Amplifier Sample Problem *
* 03-14
Design a system with output vo = H (x), where process variable x is
pressure in a sealed cylinder, x 2 [100 kPa ; 1000 kPa ], and H (x) =
____x_______ 2
V. The cylinder has an area of 100 cm . The piston can
100 kPa
reach a maximum height of 10 cm at which point the pressure will be
100 kPa . The cylinder contents is held at a constant temperature.
Plan: Deduce pressure by measuring the position of the piston.
Ideal gas law: P S = n
03-15 *
* 03-15
Solution Plan:
Compute position of piston, y, in terms of pressure, x.
Measure position of piston with variable resistor.
Find conversion circuit to produce H(x).
Transducer(s)
Two transducers are being used:
- Pressure-to-position. (The piston.) Use notation y = Ht1 (x).
- Position-to-resistance. Use notation z = Ht2 (y).
Pressure to Position
Recall: P S = 105 kPa cm 3.
Here P ! x
: : :and S ! y100 cm 2.
So: xy100 cm 2 = 105 kPa cm 3.
5 kPa cm 3 103 kPa cm
Or: y = 10______________x100=cm__2___________x= Ht1 (x).
Position to Resistance
Use a 5 k variable resistor.
Connect it such that Ht2 (y) = ___y____105cmk.
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03-16 *
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Conversion Circuit Function
Desired output: H(x) = _____x_____100VkPa.
Hc(Ht2 (Ht1 (x))) = H(x)
Let z = Ht2 (Ht1 (x)) = 5 105 _kPa__x.
Then: x = 5 105 _kPa__z.
Substituting:
Hc(Ht2 (Ht1 (x))) = H(x)
Hc(z) = H(5 105 _kPa__z)
Hc(z) = 5_zk V
Conversion Circuit Choice
Use inverting amplifier as inverted-resistance-to-voltage converter.
Hc(RA ) = _A3__R, where A3 = RB vA .
A
RA ! z and A3 ! 5 k V.
Choose RB and vA so that 5 k V = RB vA .
_________________________________________
_ R = 500 and v = 10 V _
For example, ___B______________A______________________._
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03-17 Summing Amplifier *
* 03-17
Xn v
vo = RB _Ai___.
i=1 RAi
Applications
Adding response of several transducers.
Adding "a constant" to the output of a transducer.
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03-18 *
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Gain/Offset Circuit
Frequently used conditioning circuit.
Uses one inverting amplifier and one summing amplifier.
RB vB vC RD RA
vo = __________ RE ______________ .
RD RA vB RC
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03-19 *
* 03-19
Gain/Offset Circuit
vo = _RB_vB___R RE vC_RD_RA_____ .
D RA vB RC
vo = A5 (RE O5 ) .
Hc(RE ) = A5 (RE O5 ) .
In this form,
- RE is the input,
- A5 = _RB_vB___Rdetermines the gain,
D RA
- and O5 = vC_RD_RA_____vdetermines the offset.
B RC
Note that offset can be changed without affecting gain.
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03-20 Gain/Offset Circuit Example *
* 03-20
Design a system with output vo = H (x), where process variable x is
water level, x 2 [0 m; 1 m], and H (x) = 10x V__m.
This is identical to an earlier problem.
However, it will be solved using a different transducer.
Transducer function will account for the small deviation from perfection.
Transducer:
Ht(x) = e1 _x__m5 k + e2 ,
where e1 = 0:91 and e2 = 37 .
(If e1 = 1 and e2 = 0 then the transducer would be perfect.)
The conditioning circuit should be designed to give the proper out-
put, taking into account e1 and e2 .
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03-21 *
* 03-21
Proceeding in the usual manner:
Let y = Ht(x) = e1 _x__m5 k + e2 .
Then x = ____m____e(y e2 ).
1 5 k
Hc(Ht(x)) = H(x)
Hc(y) = 10 V___m__m____e(y e2 )
1 5 k
= __2_V___e(y e2 )
1 k
Looks like a gain/offset circuit.
A5 ! __2_V___eand O5 ! e2 .
1 k
Choose component values so that following are simultaneously satisfied:
__2_V___ = _RB_vB___ and e = vC_RD_RA_____
e1 k RD RA 2 vB RC
Choose reasonable values for RA , RD , vB , and vC .
vB = 5 V and RD = 5 k.
Possible reasons: a 5 V supply is available.
Current through transducer (RE ) will be 1 mA , not too large or
small for many cases.
RA = 10 k and vC = 5 V.
Solving equations then yields:
RB = 22:0 k.
RC = 1:35 M.
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| David M. Koppelman - koppel@ee.lsu.edu | Modified 22 Jan 1999 9:46 (15:46 UTC) |