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CIRCUITS WITH RESISTIVE Serhan YAMAÇLI 15th October 2007
Presentation Outline 1. Current-to-voltage Converters 2. Voltage-to-current Converters 3. Current Amplifiers 4. Difference Amplifiers 5. Instrumentation Amplifiers 6. Instrumentation Applications 7. Transducer Bridge Amplifiers
Amplifiers
Current amplifiers
Voltage-mode amplifiers Voltage
Current
Current
(Section 3)
Voltage
(Last lecture) Transimpedance amplifiers Current
Voltage
(Section 1)
Transconductance amplifiers Voltage
Current
(Section 2)
1. CURRENT-TO-VOLTAGE CONVERTERS • A current-to-voltage converter is a circuit in which current is the input and voltage is the output . • Voltage-to-current converters are also called transresistance amplifiers. • The gain of a voltage-to-current converter is the ratio of the output voltage to input current and given in A/V. Current
Voltage
Figure 1.1 Transresistance amplifier block diagram
Vout A= I in
(1.1)
1. CURRENT-TO-VOLTAGE CONVERTERS
Figure 1.2 Basic I-V converter
vO − 0 iI + =0 R
(1.2)
vO = − RiI
(1.3)
Magnitude of the gain is also called sensitivity.
1. CURRENT-TO-VOLTAGE CONVERTERS • If a sensitivity of 1V/μA is aimed, R must be 1MΩ. • The element may also be a frequency dependent element, then the input-output of the circuit can be expressed as:
vO = − Z 0 ( s)iI
(1.4)
• The circuit is called a transimpedance amplifier.
• For this type of amplifier, opamp eliminates the input and output loadings.
1. CURRENT-TO-VOLTAGE CONVERTERS • If the opamp is non-ideal, then the circuit parameters can be given as
A = −R
1 1+1/ T
Ri =
rd || ( R + ro ) 1+ T
Ro ≅
ro 1+ T
(1.5)
where
T=
αrd rd + R + ro
(1.6)
1. CURRENT-TO-VOLTAGE CONVERTERS SPICE SIMULATIONS OF I-V CONVERTER The circuit is simulated using LM324 opamp macromodel. 1
20V
2
40mA (-19.271m,19.267)
10V
20mA
0V
0A
-10V
-20mA
(19.948m,-19.937) -20V
>> -40mA -30mA 1
V(U3A:OUT)
2
-20mA I(I1)
-10mA
0A
10mA
20mA
30mA
I_I1
Figure 1.3 DC characteristics of the transresistance amplifier Note: A macromodel is an equivalent circuit that models both linear and nonlinear characteristics of a circuit/element [2, 3]. Note 2: LM324 supply voltages are taken as
± 15V [4]
1. CURRENT-TO-VOLTAGE CONVERTERS SPICE SIMULATIONS OF I-V CONVERTER 1.0
(1.2669M,704.663m)
0.5
0
1.0Hz
V(U3A:OUT)
10Hz
I(I1)
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
10MHz
Frequency
Figure 1.4 AC characteristics of the transresistance amplifier The -3dB cut-off frequency of this amplifier is 1.26MHz.
100MHz
1. CURRENT-TO-VOLTAGE CONVERTERS SPICE SIMULATIONS OF I-V CONVERTER 1
10V
2
10mA
5V
5mA
0V
0A
-5V
-5mA
-10V
>> -10mA 0s 1
0.5ms V(U3A:OUT) 2
I(I1)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
Time
Figure 1.5 Transient characteristics of the transresistance amplifier R=1k
4.0ms
1. CURRENT-TO-VOLTAGE CONVERTERS HIGH-SENSITIVITY I-V CONVERTERS • High sensitivity values require high resistance values thus they may not be realizable as integrated circuit (IC) form. • The circuit of Figure 1.6 shows a high-sensitivity I-V converter.
Fig.1.6 High-sensitivity I-V converter
1. CURRENT-TO-VOLTAGE CONVERTERS HIGH-SENSITIVITY I-V CONVERTERS
− v1 v1 vO − v1 − + =0 R R1 R2 v1 = − RiI
(1.7)
(1.8)
Combining (1.7) and (1.8) gives
vO = − kRiI R2 R2 k = 1+ + R1 R
(1.9)
(1.10)
1. CURRENT-TO-VOLTAGE CONVERTERS SPICE SIMULATIONS OF HIGH SENSITIVITY I-V CONVERTER 1
20V
2
10mA .4502m,19.253)
0V
0A
(6.6414m,-19.876) -20V
>> -10mA -6.8mA -6.0mA 1 V(U3A:OUT)
2
-4.0mA I(I1)
-2.0mA
0A
2.0mA
4.0mA
6.0mA
I_I1
Figure 1.7 DC characteristics of the high-sensitivity I-V converter
8.0mA
1. CURRENT-TO-VOLTAGE CONVERTERS SPICE SIMULATIONS OF HIGH SENSITIVITY I-V CONVERTER
3.0
(503.501K,2.1926)
2.0
1.0
0 1.0Hz V(U3A:OUT)
10Hz I(I1)
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
10MHz
Frequency
Figure 1.8 AC characteristics of the high-sensitivity I-V converter The -3dB cut-off frequency of this circuit is found to be 503kHz.
100MHz
1. CURRENT-TO-VOLTAGE CONVERTERS SPICE SIMULATIONS OF HIGH SENSITIVITY I-V CONVERTER
1
20V
2
10mA
10V
5mA
0V
0A
-10V
-5mA
-20V
>> -10mA 0s 1
0.5ms V(U3A:OUT) 2
I(I1)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
Time
Figure 1.9 Transient response of high-sensitivity I-V converter
4.0ms
1. CURRENT-TO-VOLTAGE CONVERTERS PHOTODETECTOR AMPLIFIERS • I-V converters are primarily used in photodetector circuits.
Figure 1.10. Photoconductive (a), and photovoltaic detectors • The incident light amplitude varies the photodiode current and then it is converted to voltage for further processing. For example in fiberoptic detectors.
2. VOLTAGE-TO-CURRENT CONVERTERS • A voltage-to-current converter (V-I converter), also called transcoductance amplifier, gets a voltage input and generates a current output proportional to the input voltage.
Voltage
Current
Figure 2.1 V-I converter block diagram Ideal case
Practical case
iO = AvI
(2.1)
1 iO = AvI − vL RO
(2.2)
Voltage of load
2. VOLTAGE-TO-CURRENT CONVERTERS • For true V-I conversion
RO → ∞
(2.3)
• (2.3) means an open-circuit. This may cause the circuit not to operate properly since the output current may not have a path. • Voltage compiance is the range of permissible values of vL for which the circuit works properly. • Floating-type V-I converter: Both terminals of the load is uncommitted. • Grounded-type V-I converter: One of the terminals of the load is grounded.
2. VOLTAGE-TO-CURRENT CONVERTERS FLOATING TYPE V-I CONVERTORS
Figure 2.2. Floating-type V-I converters
RiO = vI
(2.4)
vI iO = R
(2.5)
2. VOLTAGE-TO-CURRENT CONVERTERS FLOATING TYPE V-I CONVERTORS From circuit of Figure 2.2(a):
vO = vI + vL
(2.6)
Opamp output swing voltage region:
VOL < vO < VOH
(2.7)
Combining (2.6) and (2.7) yields
(VOL − vI ) < vL < (VOH − vI ) Voltage compliance of the circuit
(2.8)
2. VOLTAGE-TO-CURRENT CONVERTERS FLOATING TYPE V-I CONVERTORS For Figure 2.2(b)
vI − 0 iO = R
(2.9)
The circuit swings to the voltage vO = −vL Voltage compliance is then
VOL < vL < VOH The voltage compliance is greater than the previous circuit.
(2.10)
2. VOLTAGE-TO-CURRENT CONVERTERS FLOATING TYPE V-I CONVERTORS • These circuits are said to be bidirectional since the input voltage can be either positive or negative, no matter the polarity, the circuits work properly. • If the load is a capacitor, the the input-output relation of the circuit would be
iO = sCv I
(2.11)
which represents a lossless integrator.
• Integrators are primarily used in waveform generators, V-F, F-V converters, and ADCs.
2. VOLTAGE-TO-CURRENT CONVERTERS PRACTICAL V-I CONVERTOR LIMITATIONS
vI − vD + vL + roiO − αvD = 0 (2.12) vD vI − vD iO + − =0 rd R
Figure 2.3 V-I converter with practical opamp model
(2.13)
α − R / rd 1 A= R 1 + α + rO / R + rO / rd
(2.14)
RO = ( R || rd )(1 + α ) + ro
(2.15)
2. VOLTAGE-TO-CURRENT CONVERTERS SIMULATIONS OF FLOATING TYPE V-I CONVERTORS 8.0mA
(7.1693,7.1377m)
4.0mA
0A
-4.0mA
(-7.4603,-7.4542m)
-8.0mA
-15V
-I(RL)
-10V
-5V
0V
5V
10V
V_V3
Figure 2.4 DC characteristics of the floating V-I converter
15V
2. VOLTAGE-TO-CURRENT CONVERTERS SIMULATIONS OF FLOATING TYPE V-I CONVERTORS 1.0mA
(538.270K,706.845u)
0.5mA
0A
1.0Hz -I(RL)
10Hz
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
Frequency
Figure 2.5 AC characteristics of the floating V-I converter The -3dB cut-off frequency of the circuit is 538kHz.
10MHz
2. VOLTAGE-TO-CURRENT CONVERTERS SIMULATIONS OF FLOATING TYPE V-I CONVERTORS 1
8.0mA
2
10V
4.0mA
5V
0A
0V
-4.0mA
-5V
-8.0mA
>> -10V 0s 1
-I(RL)
2
0.5ms V(V4:+)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
Time
Figure 2.6 Transient characteristics of the floating V-I converter
4.0ms
2. VOLTAGE-TO-CURRENT CONVERTERS GROUNDED V-I CONVERTORS
Figure 2.7 Howland current pump and its Norton equivalent circuit
2. VOLTAGE-TO-CURRENT CONVERTERS GROUNDED V-I CONVERTORS Output resistance of the Howland current pump is
R2 RO = R2 / R1 − R4 / R3
(2.16)
In order to have an infinite output resistance (the ideal current source),
R4 R2 = R3 R1 Balanced bridge
(2.17)
2. VOLTAGE-TO-CURRENT CONVERTERS GROUNDED V-I CONVERTORS Voltage compliance of the circuit is
vL ≤
R1 Vsat R1 + R2
(2.18)
Saturation voltage of the operational amplifier
2. VOLTAGE-TO-CURRENT CONVERTERS EFFECTS OF RESISTANCE MISMATCHES The balanced bridge can not be implemented in practice. The non-ideality can be modelled by imbalance factor ε
R4 R2 = (1 − ε ) R3 R1
(2.19)
The output resistance of the circuit is found as
R1 Ro = ε
(2.20)
2. VOLTAGE-TO-CURRENT CONVERTERS SIMULATIONS OF GROUNDED V-I CONVERTORS 10mA (7.0263,7.0210m)
5mA
0A
-5mA (-7.4474,-7.4396m)
-10mA -15V
-I(R5)
-10V
-5V
0V
5V
10V
V_V3
Figure 2.8 DC characteristics of the grounded-type V-I converter
15V
2. VOLTAGE-TO-CURRENT CONVERTERS SIMULATIONS OF GROUNDED V-I CONVERTORS 1.0mA
0.8mA
(176.198K,716.320u)
0.6mA
0.4mA
0.2mA 1.0Hz -I(R5)
10Hz
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
Frequency
Figure 2.9 AC characteristics of the grounded-type V-I converter The -3dB cut-off frequency of the circuit is 178kHz
10MHz
2. VOLTAGE-TO-CURRENT CONVERTERS SIMULATIONS OF GROUNDED V-I CONVERTORS 1
10V
2
10mA
5V
5mA
0V
0A
-5V
-5mA
-10V
>> -10mA 0s 1
V(R1:1)
0.5ms 2 I(R5)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
4.0ms
Time
Figure 2.10 Transient characteristics of the grounded-type V-I converter • Note the inverse direction of the current.
2. VOLTAGE-TO-CURRENT CONVERTERS EFFECTS OF FINITE GAIN OF OPAMP If the open-loop gain of opamp is α, then the output resistance of the Howland current pump is
α RO = ( R1 || R2 )1 + 1 + R2 / R1
RO decreased from infinite to this finite number.
(2.21)
2. VOLTAGE-TO-CURRENT CONVERTERS EFFECTS OF FINITE GAIN OF OPAMP Balance condition:
R4 R2 A + R2 B = R3 R1
(2.22)
Transfer characteristics:
R2 / R1 iO = vI R2 B
Figure 2.11 Improved Howland circuit
(2.23)
The advantage of the circuit is to provide power saving.
3. CURRENT AMPLIFIERS Opamps can be used as current amplifiers. The transfer characteristics of a practical current amplifier:
1 iO = AiI − vL RO
Output current
Gain
Output resistance
(3.1)
Load voltage
Ideally, iO must be independent of vL, that is
RO → ∞
(3.2)
3. CURRENT AMPLIFIERS FLOATING TYPE CURRENT AMPLIFIER
Figure 3.1 Floating type current amplifier
3. CURRENT AMPLIFIERS FLOATING TYPE CURRENT AMPLIFIER For infinite-gain opamp:
iO +
If the open-loop gain of opamp is α
iI R2 iI = 0 R1
(3.3)
R2 R1
(3.4)
If
A = 1+
iO = AiI
R2 / R1 A = 1+ 1+1/ α
(3.6)
RO = R1 (1 + α )
(3.7)
Voltage compliance is: (3.5)
Note that output resistance approcahes infinity.
− (VOH + i2 R1 ) ≤ vL ≤ −(VOL + i2 R1 ) (3.8)
3. CURRENT AMPLIFIERS GROUNDED TYPE CURRENT AMPLIFIER Input-output relation:
iO = AiS −
1 vL RO
(3.9)
Where the gain is
A=−
R2 R1
(3.10)
Output resistance is: Figure 3.2 Grounded type current amplifier
R2 RO = − R S R1
(3.11)
If A=-1, the circuit is called as current reverser or current mirror. Source resistance
3. CURRENT AMPLIFIERS SIMULATIONS OF FLOATING TYPE CURRENT AMPLIFIER
20mA (-14.050m,14.060m)
10mA
0A
-10mA (14.820m,-
-20mA
-16mA
I(I1)
-
-12mA
I(RL)
-8mA
-4mA
0A
4mA
8mA
12mA
I_I1
Figure 3.3 DC transfer characteristics of grounded current amplifier
16mA
3. CURRENT AMPLIFIERS SIMULATIONS OF FLOATING TYPE CURRENT AMPLIFIER 1.0mA
(1.1538M,712.435u)
0.5mA
0A 1.0Hz I(I2)
-I(RL)
10Hz
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
Frequency
Figure 3.4 AC transfer characteristics of grounded current amplifier
10MHz
3. CURRENT AMPLIFIERS SIMULATIONS OF FLOATING TYPE CURRENT AMPLIFIER
1.0mA
0.5mA
0A
-0.5mA
-1.0mA
0s
I(I2)
I(RL)
0.5ms
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
Time
Figure 3.5 Transient characteristics of grounded current amplifier
4.0ms
4. DIFFERENCE AMPLIFIERS A difference amplifier is an amplifier that has one output and two inputs and satisfies: • Amplification of the difference voltage at the input terminals • Rejection of the common voltage at the input terminals
Figure 4.1 Difference amplifier (a), differential and common inputs (b)
4. DIFFERENCE AMPLIFIERS The bridge balance must be satisfied to operate:
R4 R2 = R3 R1
(4.1)
R2 vO = (v2 − v1 ) R1
Output voltage
(4.2)
Differential input voltage
4. DIFFERENCE AMPLIFIERS Differential mode component:
vDM = v2 − v1
(4.3)
Common mode component:
vCM
v1 + v2 = 2
(4.4)
Input voltages can be expressed in of differential mode and common mode inputs (Figure 4.1b):
v1 = vCM −
vDM 2
(4.5)
v2 = vCM +
vDM 2
(4.6)
4. DIFFERENCE AMPLIFIERS The common mode and differential mode input resistances can be exxpressed as
Figure 4.2 Differential mode and common mode input resistances
Rid = 2R1
(4.7)
R1 + R2 Ric = 2
(4.8)
4. DIFFERENCE AMPLIFIERS EFFECTS OF RESISTANCE MISMATCHES A difference amplifier is insensitive to common mode for infinite gain and perfectly balanced resistances. If the opamp is ideal but the resistances are mismatched:
R2 ' = R2 (1 − ε )
(4.9)
imbalance factor
vO = vDM + ACM vCM
Figure 4.3 Differential mode and common mode input resistances
R2 R1 + 2 R2 ε 1 − = R1 R1 + R2 2 R2 ACM = ε R1 + R2
(4.10) (4.11)
(4.12)
4. DIFFERENCE AMPLIFIERS EFFECTS OF RESISTANCE MISMATCHES • is called the differential mode gain. • Acm is called the common mode gain.
CMRR = Acm
(4.13)
Common mode rejection ratio
CMRRdB = 20 log10
CMRR in dB
Acm
(4.14)
4. DIFFERENCE AMPLIFIERS EFFECTS OF RESISTANCE MISMATCHES If ε<<1, then
R2 εR2 ≅ / Acm R1 R1 + R2 1 + R2 / R1 ≅ 20 log10 Acm dB ε
(4.15)
(4.16)
• For a fixed imbalance factor, CMRR increases with increasing R2/R1.
4. DIFFERENCE AMPLIFIERS SIMULATIONS OF DIFFERENTIAL AMPLIFIER INA105 from BURR-BROWN • INA105 is a monolithic differential amplifier that employs very good matched resistors (2/10000 sensitivity).
Figure 4.4 INA 105 monolithic differential amplifier [5]
4. DIFFERENCE AMPLIFIERS SIMULATIONS OF DIFFERENTIAL AMPLIFIER INA105 from BURR-BROWN 5.0V
(-1.1491,1.0851) (-429.825m,429.950m)
0V
(3.5702,-3.5518)
-5.0V -5.0V V(V1:+)
-4.0V V(U1:OUT)
-3.0V
-2.0V
-1.0V
0.0V
1.0V
2.0V
3.0V
4.0V
5.0V
V_V1
Figure 4.5 DC sweep of INA 105 monolithic differential amplifier • Note that, in the linear region, if the input voltage is 429.825mV (applied to inverting terminal), output voltage is -429.950mV.
4. DIFFERENCE AMPLIFIERS SIMULATIONS OF DIFFERENTIAL AMPLIFIER INA105 from BURR-BROWN 1.0V
(1.4610M,719.235m)
0.5V
0V 1.0Hz
V(V1:+)
10Hz
V(U1:OUT)
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
Frequency
Figure 4.6 AC sweep of INA 105 monolithic differential amplifier • Note that, -3dB cut-off frequency is 1.45MHz.
10MHz
4. DIFFERENCE AMPLIFIERS SIMULATIONS OF DIFFERENTIAL AMPLIFIER INA105 from BURR-BROWN 400mV
200mV
0V
-200mV
-400mV
0s
V(V5:+)
0.5ms V(U1:OUT)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
Time
Figure 4.7 Transient response of INA 105 monolithic differential amplifier
4.0ms
4. DIFFERENCE AMPLIFIERS DIFFERENCE AMPLIFIER CALIBRATION
Figure 4.8 Difference amplifier calibration circuit • Rpot is adjusted toachieve calibrated difference amplifier.
4. DIFFERENCE AMPLIFIERS DIFFERENCE AMPLIFIER WITH VARIABLE GAIN
The CM gain can be given as:
ACM
Figure 4.9 Differential amplifier with variable gain
2 R2 R2 1 + = R1 RG
(4.17)
• Note that gain is inveresely proportional with RG.
4. DIFFERENCE AMPLIFIERS DIFFERENCE AMPLIFIER WITH VARIABLE GAIN
The CM gain can be given as:
ACM
R2 RG = R1 R3
(4.18)
A linearly tuneable gain is achieved. Figure 4.10 Differential amplifier with linearly adjustible gain
4. DIFFERENCE AMPLIFIERS GROUND-LOOP INTERFERENCE PROBLEM
Figure 4.11 An inverting amplifier that is subject to ground-loop interference Zg is the distributed impedance of the ground line. Amplifier sees vi and vg in series, so;
vO = −
R2 (vi + v g ) R1
(4.19)
Figure 4.11 A difference amplifier that eliminates cross-talk for common ground terminal connections
R2 vO = − vi R1
(4.20)
5. INSTRUMENTATION AMPLIFIERS An instrumentation amplifier is a difference amplifier that satisfies:
3. Extremely high (ideally infinite) common mode and differential mode input resistances 4. Very low (ideally zero) output impedance 5. Accurate and stable gain, typically between 1V/V and 1000V/V 6. Extremely high common mode rejection ratio •
Instrumentation amplifiers are used in the areas in which the differential mode input signal level is very low such as transducer output in an industrial application or biomedical engineering.
•
The 2nd, 3rd and 4th expressions may be satisfied by a difference amplifier but extremely high input impedance can not be satisfied.
5. INSTRUMENTATION AMPLIFIERS TRIPLE OPAMP INSTRUMENTATION AMPLIFIER From the first stage:
2R vO1 − vO 2 = 1 + 3 (v1 − v2 ) RG
(5.1)
From the second stage:
vO =
R2 (vO 2 − vO1 ) R1
(5.2)
Combining these two equations yields:
First stage
Second stage
Figure 5.1 Triple opamp instrumentation amplifier configuration
2 R3 R2 vO = AI . AII = 1 + RG R1 (5.3)
5. INSTRUMENTATION AMPLIFIERS DUAL OPAMP INSTRUMENTATION AMPLIFIERS
Figure 5.2 Dual opamp instrumentation amplifier For OA1
R3 v3 = (1 + )v1 R4
For OA2 (5.5)
R2 R2 vO = − v3 + 1 + v2 R1 R1
(5.6)
5. INSTRUMENTATION AMPLIFIERS DUAL OPAMP INSTRUMENTATION AMPLIFIERS Combining these equations yields:
R2 1 + R3 / R4 vO = 1 + v2 − v1 1 + R1 / R2 R1 If
R3 R1 = R4 R2
(5.7)
(5.8)
then, the input-output relationship becomes:
R vO = 1 + 2 (v2 − v1 ) R1
(5.9)
5. INSTRUMENTATION AMPLIFIERS DUAL OPAMP INSTRUMENTATION AMPLIFIERS WITH VARIABLE GAIN
Figure 5.3 Variable gain instrumentation amplifier
A = 1+
R2 2 R2 + R1 RG
(5.10)
5. INSTRUMENTATION AMPLIFIERS DUAL OPAMP INSTRUMENTATION AMPLIFIER
• Although dual opamp instrumantation amplifiers take the advantage of lower number of active and ive devices, it suffers from highfrequency performance degredation. • The reason for the performance decrement is: v1 and v2 input voltages meet at different times since v1 has to through OA1 to catch v2.
5. INSTRUMENTATION AMPLIFIERS MONOLITHIC INSTRUMENTATION AMPLIFIER
• The monolithic instrumentation amplifiers allows better optimization of CMRR, lineraity and noise. • Also, monolithic laser trimmed resistances allows the impalance factor to be minimized. • The SPICE simulations of a monolithic instrumentation amplifer from Burr-Brown, namely INA-101 are carried. • INA-101 allows the adjustement of the gain by an external resistor.
5. INSTRUMENTATION AMPLIFIERS SPICE SIMULATIONS OF A MONOLITHIC INSTRUMENTATION AMPLIFIER
40kΩ A = 1+ RG
Gain
Figure 5.4 Simulated instrumentation amplifier INA 101 [6]
(5.4)
5. INSTRUMENTATION AMPLIFIERS SPICE SIMULATIONS OF A MONOLITHIC INSTRUMENTATION AMPLIFIER 20V
(3.2984,9.889) 10V
0V
(-3.3246,-9.912) -10V
-20V
-15V
V(U1:+)
V(RL:2)
-10V
-5V
0V
5V
10V
V_V1
Figure 5.5 Simulated DC characteristics of INA 101 instrumentation amplifier (RG=20k, A=3V/V)
15V
5. INSTRUMENTATION AMPLIFIERS SPICE SIMULATIONS OF A MONOLITHIC INSTRUMENTATION AMPLIFIER 3.0V
(385.478K,2.1408)
2.0V
1.0V
0V
1.0Hz
V(V4:+)
10Hz
V(RL:2)
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
Frequency
Figure 5.6 Simulated AC characteristics of INA 101 instrumentation amplifier (RG=20k, A=3V/V) • Note that –dB cut-off frequency is 338kHz.
10MHz
5. INSTRUMENTATION AMPLIFIERS SPICE SIMULATIONS OF A MONOLITHIC INSTRUMENTATION AMPLIFIER 4.0V
2.0V
0V
-2.0V
-4.0V
0s
V(U1:+)
0.5ms V(RL:2)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
4.0ms
Time
Figure 5.7 Simulated transient characteristics of INA 101 instrumentation amplifier (RG=20k, A=3V/V)
5. INSTRUMENTATION AMPLIFIERS FLYING CAPACITOR TECHNIQUES • •
A capacitor is flipped back and forth between source and amplifier to achieve high common mode rejection ratio (CMRR). Circuit ignores common mode input hence provides high common mode rejection ratio.
R vO = 1 + 2 (v2 − v1 ) R1
(5.5)
In fact, a noninverting amplifier Figure 5.8 Employing switched capacitor to achieve high commpn mode rejection ratio
6. INSTRUMENTATION APPLICATIONS ACTIVE GUARD DRIVE • In the situations where source and amplifier are far apart the signal is trnasmitted by shielded wires such as coaxial lines. • The signal is transmitted in a double ended form for achieving the same noise pick-up. • The common noise is considered as common mode signal and rejected by the difference amplifier. • For this reason, double ended transmission is also named as balanced transmission. THE DISADVANTAGE: • The distributed capacitance of the cable decreases the CMRR. • Consider the circuit of Figure 6.1.
6. INSTRUMENTATION APPLICATIONS ACTIVE GUARD DRIVE First coaxial line model
Second coaxial line model Figure 6.1 The distributed capacitance of the balanced line • Since Rs1.C1 and Rs2.C2 time constants will not be the same, this unbalance will decrease common mode rejection ratio because the amplifier will see a CM input.
6. INSTRUMENTATION APPLICATIONS ACTIVE GUARD DRIVE CMRR due to imbalance is:
1 CMRRdB = 20 log10 2πfRdmCcm where
Rdm = Rs1 − Rs 2
Ccm
C1 + C2 = 2
• For example, if f=60Hz, Rdm=1kΩ and Ccm=1nF, CMRR=68.5dB.
(6.1)
(6.2)
(6.3)
6. INSTRUMENTATION APPLICATIONS ACTIVE GUARD DRIVE
Figure 6.2 Reducing Ccm by using an active guard • Vcm is fed to shield of the balanced line, thus reducing CM input at the input of amplifier.
6. INSTRUMENTATION APPLICATIONS DIGITALLY PROGRAMMABLE GAIN • In automatic instrumentation such as data acquistion systems, the programming of instrumentation amplifier electronically may be needed. • The gain of the first stage is
Routside AI = 1 + Rinside
(6.4)
Routside = 2R1
(6.5)
Rinside = 2( R2 + R3 + .... + Rn ) + Rn +1 Figure 6.3 Digitally programmable IA
(6.6)
6. INSTRUMENTATION APPLICATIONS OUTPUT OFFSETTING • In some applications, like a voltage-to-frequency converters, the output may need to have an offset value. • The need comes from the fact that: the input of the succeeding circuit mya have only one polarity.
vO = A(v2 − v2 ) + (1 + R2 / R1 )[ R1 /( R1 + R2 )]Vref (6.7)
vO = A(v2 − v1 ) + Vref Figure 6.4 IA with offset control
(6.8)
6. INSTRUMENTATION APPLICATIONS CURRENT OUTPUT INSTRUMENTATION AMPLIIFIERS
iO =
1 + 2 R3 / RG (v2 − v1 ) R1 (6.9)
Figure 6.5 Current output IA • The current output is generated using Howland current pump at the output stage
6. INSTRUMENTATION APPLICATIONS CURRENT OUTPUT INSTRUMENTATION AMPLIIFIERS
iO =
1 1 (v2 − v1 ) − vL R RO (6.10)
R2 / R1 RO = R3 R5 / R4 − ( R2 + R3 ) / R1 Figure 6.6 Current output IA with dual opamps
(6.11)
| vL |≤ Vsat − 2 v2 − v1 (6.12)
6. INSTRUMENTATION APPLICATIONS SIMULATIONS OF CURRENT OUTPUT INSTRUMENTATION AMPLIIFIERS 1
5.0V
2
20uA (3.6075,14.240u)
10uA
0V
0A
-10uA (-3.8037,-14.978u)
-5.0V
>>
-20uA -5.0V 1
-4.0V V(V2:+) 2
I(RL)
-3.0V
-2.0V
-1.0V
0.0V
1.0V
2.0V
3.0V
V_V2
Figure 6.7 DC characteristics of current output IA with LM324
4.0V
5.0V
6. INSTRUMENTATION APPLICATIONS SIMULATIONS OF CURRENT OUTPUT INSTRUMENTATION AMPLIIFIERS 3.0uA
2.0uA
1.0uA
0A
1.0Hz
I(RL)
10Hz
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
Frequency
Figure 6.8 AC characteristics of current output IA with LM324
10MHz
6. INSTRUMENTATION APPLICATIONS SIMULATIONS OF CURRENT OUTPUT INSTRUMENTATION AMPLIIFIERS 1
1.0V
2
4.0uA
0.5V
2.0uA
0V
0A
-0.5V
-2.0uA
-1.0V
>> -4.0uA 0s 1
V(V5:+)
0.5ms 2 I(RL)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
Time
Figure 6.9 Transient characteristics of current output IA with LM324
4.0ms
6. INSTRUMENTATION APPLICATIONS CURRENT INPUT INSTRUMENATION AMPLIFIER
vO 2 = vCM − iI R3 vO1 = vCM + R3iI vO =
R2 (vO 2 − vO1 ) R1
vO = − Figure 6.10 Current input instrumentation amplifier
2 R2 R3iI R1
(6.13) (6.14) (6.15)
(6.16)
7. TRANSDUCER BRIDGE AMPLIFIERS
• Resistive transducer: A device whose resistance varies by a physical paramater such as temperature, light frequency/amplitude, pressure. • Resistive transducer transducer are made a part of a circuit to generate an electrical signal proportional to the physical parameter. • Parameter’s magnitude and the electrical signal are wanted to be in a linear relationship.
7. TRANSDUCER BRIDGE AMPLIFIERS TRANSDUCER RESISTANCE DEVIATION • Transducer resistances are expressed as
R + ∆R Resistance at a reference condition
(7.1) Deviation
It can be also expressed as
R = R (1 + δ )
∆R δ= R
Fractional deviation
(7.2)
7. TRANSDUCER BRIDGE AMPLIFIERS THE TRANSDUCER BRIDGE • ΔR is converted to a ΔV by means of a voltage divider.
Figure 7.1 Transducer bridge with instrumentation amplifier • This voltage dividers are referred to as bridge legs.
7. TRANSDUCER BRIDGE AMPLIFIERS THE TRANSDUCER BRIDGE
v1 =
VREF R (1 + δ ) R VREF δ = VREF + R1 + R (1 + δ ) R1 + R 2 + R1 / R + R / R1 + (1 + R / R1 )δ
R v2 = VREF R1 + R
(7.4)
δ vO = A(v1 − v2 ) = AVREF 1 + R1 / R + (1 + R / R1 )(1 + δ ) If
(7.3)
(7.5)
δ << 1 vO ≅
AVREF AVREF δ= δ 2 + R1 / R + R / R1 4
(7.6) If R1=R
7. TRANSDUCER BRIDGE AMPLIFIERS THE TRANSDUCER BRIDGE
Figure7.2 Bridge calibration of a transducer bridge Due to the tolerances of bridge resistances and IA’s reference voltage, a calibration is needed. R3 is varied to achieve vO=0V at ΔR=0.
7. TRANSDUCER BRIDGE AMPLIFIERS STRAIN GAUGE BRIDGES A resistor with resistivity ρ, cross-sectional area S and lenght l has a resistance of:
ρl R= S
(7.7)
l ' = l + ∆l
(7.8)
S ' = S − ∆S
(7.9)
ρ (l + ∆l ) R= S − ∆S
(7.10)
If the wire is strained,
7. TRANSDUCER BRIDGE AMPLIFIERS STRAIN GAUGE BRIDGES The volume of the resistor is not changed, so
(l + ∆l )( S − ∆S ) = Sl
(7.11)
∆l ∆l ∆l ∆R = R 2 + ≅ 2 R l l l
(7.12)
Unstrained resistance
Fractional elognation
7. TRANSDUCER BRIDGE AMPLIFIERS STRAIN GAUGE BRIDGES
Figure 7.3 Strain-gauge bridge and instrumentation amplifier Load cell
7. TRANSDUCER BRIDGE AMPLIFIERS STRAIN GAUGE BRIDGES
v1 = VB ( R + ∆R) /( R + ∆R + R − ∆R) = VB ( R + ∆R ) / 2 R
(7.13)
v2 = VB ( R − ∆R ) / 2 R
(7.14)
v1 − v2 = VB ∆R / R = VBδ
(7.15)
vO = AVREF δ
(7.16)
• Note that, voltage deviation is perfectly linear with fractional elongation. • Also, in the circuit R3 adjusts sensitivity, R2 nulls the voltage in the absence of strain.
7. TRANSDUCER BRIDGE AMPLIFIERS BRIDGE AMPLIFIER WITH SINGLE OPAMP AMPLIFIER
vO =
R2 δ VREF R R1 / R + (1 + R1 / R2 )(1 + δ ) (7.17) For
Figure 7.4 Single opamp bridge amplifier
vO ≅
δ << 1
R2 δ VREF R 1 + R1 / R + R1 / R2 (7.18)
7. TRANSDUCER BRIDGE AMPLIFIERS BRIDGE LINEARIZATION • Bridge circuits except strain-gauge has a nonlinear voltage response with the fractional elognation. • The linear response may be achieved using a current source to drive the bridge. Current source (V-I converter)
Figure 7.5 Bridge linearization using a current source
7. TRANSDUCER BRIDGE AMPLIFIERS BRIDGE LINEARIZATION
VREF IB = R1
(7.19)
v1 = VREF + R (1 + δ ) I B / 2
(7.20)
v2 = VREF + RI B / 2
(7.21)
v1 − v2 = RδI B / 2
(7.22)
vO =
ARVREF δ 2R1
(7.23)
7. TRANSDUCER BRIDGE AMPLIFIERS BRIDGE LINEARIZATION WITH SINGLE TRANSDUCER
Figure 7.6 Bridge linearization using single transducer
− R(1 + δ ) v1 = VREF R1
vO =
(7.24)
− R2 R v1 − VREF 2 R1 R1
From (7.24) and (7.25),
vO = VREF δ
R2 R1
(7.26)
(7.25)
7. TRANSDUCER BRIDGE AMPLIFIERS SIMULATIONS OF BRIDGE LINEARIZATION WITH SINGLE TRANSDUCER
• The bridge transducer witgh single transducer element has been simulated in OrCad SPICE. • The output voltage is plotted versus the transducer resistance value. • In rder to plot the variation, a parametric sweep in OrCad SPICE is utilized [7, 8, 9]. • Supply voltages and reference voltage are taken as 15V. • R1=R=10k are taken. • LM324 opamp macromodels are used.
7. TRANSDUCER BRIDGE AMPLIFIERS SIMULATIONS OF BRIDGE LINEARIZATION WITH SINGLE TRANSDUCER 0V
-5V
-10V
-15V
0
0.1K V(U2A:OUT)
0.2K
0.3K
0.4K
0.5K
0.6K
0.7K
0.8K
0.9K
1.0K
RL
Figure 7.7 Variation of output voltage via the change of transducer resistance Note that, vo is negative since, fractinal elongation is negative (i.e. transducer resistance is below 1k.)
1.1K
7. REFERENCES •
S. Franco, Design with operational amplifiers and anategrated circuits, McGraw Hill, USA, 2001.
•
S. Kılınç, M. Saygıner, U. Çam and H. Kuntman, Simple and accurate macromodel for current operational amplifier (COA), Proceedings of ELECO 2005: The 4th International Conference on Electrical and Electronics Engineering, (Electronics), pp.1-5, 7-11 December 2005, Bursa, Turkey.
•
H. Kuntman: Simple and accurate nonlinear OTA macromodel for simulation of CMOS OTAC active filters, International Journal of Electronics, Vol.77, No.6, pp.993-1006, 1994.
•
LM324 datasheet, National Semiconductors, http://cache.national.com/ds/LM/LM124.pdf, accessed on 10th Oct. 2007.
•
INA105 datasheet, Texas Instruments, http://focus.ti.com/lit/ds/symlink/ina105.pdf, accessed on 10th Oct. 2007.
•
INA101 datasheet, Burr-Brown, www.hardware.dibe.unige.it/DataSheets/INA101.pdf, accessed on 12th Oct. 2007.
•
Pspice Reference Guide, ece-classweb.ucsd.edu/spring06/ece139/PspiceRef.pdf, accessed on 13th Oct. 2007.
•
DC Simulation and analysis, http://ecen4303.okstate.edu/ecen1322/lab2ar.rtf, accessed on 13th Oct. 2007.
•
Pspice Notes v.3.0, http://wwwferp.ucsd.edu/najmabadi/CLASS/COMMON/PSPICE/PSpice_Notes_v3.0.doc, accessed 13th Oct. 2007.
Presentation Outline 1. Current-to-voltage Converters 2. Voltage-to-current Converters 3. Current Amplifiers 4. Difference Amplifiers 5. Instrumentation Amplifiers 6. Instrumentation Applications 7. Transducer Bridge Amplifiers
Amplifiers
Current amplifiers
Voltage-mode amplifiers Voltage
Current
Current
(Section 3)
Voltage
(Last lecture) Transimpedance amplifiers Current
Voltage
(Section 1)
Transconductance amplifiers Voltage
Current
(Section 2)
1. CURRENT-TO-VOLTAGE CONVERTERS • A current-to-voltage converter is a circuit in which current is the input and voltage is the output . • Voltage-to-current converters are also called transresistance amplifiers. • The gain of a voltage-to-current converter is the ratio of the output voltage to input current and given in A/V. Current
Voltage
Figure 1.1 Transresistance amplifier block diagram
Vout A= I in
(1.1)
1. CURRENT-TO-VOLTAGE CONVERTERS
Figure 1.2 Basic I-V converter
vO − 0 iI + =0 R
(1.2)
vO = − RiI
(1.3)
Magnitude of the gain is also called sensitivity.
1. CURRENT-TO-VOLTAGE CONVERTERS • If a sensitivity of 1V/μA is aimed, R must be 1MΩ. • The element may also be a frequency dependent element, then the input-output of the circuit can be expressed as:
vO = − Z 0 ( s)iI
(1.4)
• The circuit is called a transimpedance amplifier.
• For this type of amplifier, opamp eliminates the input and output loadings.
1. CURRENT-TO-VOLTAGE CONVERTERS • If the opamp is non-ideal, then the circuit parameters can be given as
A = −R
1 1+1/ T
Ri =
rd || ( R + ro ) 1+ T
Ro ≅
ro 1+ T
(1.5)
where
T=
αrd rd + R + ro
(1.6)
1. CURRENT-TO-VOLTAGE CONVERTERS SPICE SIMULATIONS OF I-V CONVERTER The circuit is simulated using LM324 opamp macromodel. 1
20V
2
40mA (-19.271m,19.267)
10V
20mA
0V
0A
-10V
-20mA
(19.948m,-19.937) -20V
>> -40mA -30mA 1
V(U3A:OUT)
2
-20mA I(I1)
-10mA
0A
10mA
20mA
30mA
I_I1
Figure 1.3 DC characteristics of the transresistance amplifier Note: A macromodel is an equivalent circuit that models both linear and nonlinear characteristics of a circuit/element [2, 3]. Note 2: LM324 supply voltages are taken as
± 15V [4]
1. CURRENT-TO-VOLTAGE CONVERTERS SPICE SIMULATIONS OF I-V CONVERTER 1.0
(1.2669M,704.663m)
0.5
0
1.0Hz
V(U3A:OUT)
10Hz
I(I1)
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
10MHz
Frequency
Figure 1.4 AC characteristics of the transresistance amplifier The -3dB cut-off frequency of this amplifier is 1.26MHz.
100MHz
1. CURRENT-TO-VOLTAGE CONVERTERS SPICE SIMULATIONS OF I-V CONVERTER 1
10V
2
10mA
5V
5mA
0V
0A
-5V
-5mA
-10V
>> -10mA 0s 1
0.5ms V(U3A:OUT) 2
I(I1)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
Time
Figure 1.5 Transient characteristics of the transresistance amplifier R=1k
4.0ms
1. CURRENT-TO-VOLTAGE CONVERTERS HIGH-SENSITIVITY I-V CONVERTERS • High sensitivity values require high resistance values thus they may not be realizable as integrated circuit (IC) form. • The circuit of Figure 1.6 shows a high-sensitivity I-V converter.
Fig.1.6 High-sensitivity I-V converter
1. CURRENT-TO-VOLTAGE CONVERTERS HIGH-SENSITIVITY I-V CONVERTERS
− v1 v1 vO − v1 − + =0 R R1 R2 v1 = − RiI
(1.7)
(1.8)
Combining (1.7) and (1.8) gives
vO = − kRiI R2 R2 k = 1+ + R1 R
(1.9)
(1.10)
1. CURRENT-TO-VOLTAGE CONVERTERS SPICE SIMULATIONS OF HIGH SENSITIVITY I-V CONVERTER 1
20V
2
10mA .4502m,19.253)
0V
0A
(6.6414m,-19.876) -20V
>> -10mA -6.8mA -6.0mA 1 V(U3A:OUT)
2
-4.0mA I(I1)
-2.0mA
0A
2.0mA
4.0mA
6.0mA
I_I1
Figure 1.7 DC characteristics of the high-sensitivity I-V converter
8.0mA
1. CURRENT-TO-VOLTAGE CONVERTERS SPICE SIMULATIONS OF HIGH SENSITIVITY I-V CONVERTER
3.0
(503.501K,2.1926)
2.0
1.0
0 1.0Hz V(U3A:OUT)
10Hz I(I1)
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
10MHz
Frequency
Figure 1.8 AC characteristics of the high-sensitivity I-V converter The -3dB cut-off frequency of this circuit is found to be 503kHz.
100MHz
1. CURRENT-TO-VOLTAGE CONVERTERS SPICE SIMULATIONS OF HIGH SENSITIVITY I-V CONVERTER
1
20V
2
10mA
10V
5mA
0V
0A
-10V
-5mA
-20V
>> -10mA 0s 1
0.5ms V(U3A:OUT) 2
I(I1)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
Time
Figure 1.9 Transient response of high-sensitivity I-V converter
4.0ms
1. CURRENT-TO-VOLTAGE CONVERTERS PHOTODETECTOR AMPLIFIERS • I-V converters are primarily used in photodetector circuits.
Figure 1.10. Photoconductive (a), and photovoltaic detectors • The incident light amplitude varies the photodiode current and then it is converted to voltage for further processing. For example in fiberoptic detectors.
2. VOLTAGE-TO-CURRENT CONVERTERS • A voltage-to-current converter (V-I converter), also called transcoductance amplifier, gets a voltage input and generates a current output proportional to the input voltage.
Voltage
Current
Figure 2.1 V-I converter block diagram Ideal case
Practical case
iO = AvI
(2.1)
1 iO = AvI − vL RO
(2.2)
Voltage of load
2. VOLTAGE-TO-CURRENT CONVERTERS • For true V-I conversion
RO → ∞
(2.3)
• (2.3) means an open-circuit. This may cause the circuit not to operate properly since the output current may not have a path. • Voltage compiance is the range of permissible values of vL for which the circuit works properly. • Floating-type V-I converter: Both terminals of the load is uncommitted. • Grounded-type V-I converter: One of the terminals of the load is grounded.
2. VOLTAGE-TO-CURRENT CONVERTERS FLOATING TYPE V-I CONVERTORS
Figure 2.2. Floating-type V-I converters
RiO = vI
(2.4)
vI iO = R
(2.5)
2. VOLTAGE-TO-CURRENT CONVERTERS FLOATING TYPE V-I CONVERTORS From circuit of Figure 2.2(a):
vO = vI + vL
(2.6)
Opamp output swing voltage region:
VOL < vO < VOH
(2.7)
Combining (2.6) and (2.7) yields
(VOL − vI ) < vL < (VOH − vI ) Voltage compliance of the circuit
(2.8)
2. VOLTAGE-TO-CURRENT CONVERTERS FLOATING TYPE V-I CONVERTORS For Figure 2.2(b)
vI − 0 iO = R
(2.9)
The circuit swings to the voltage vO = −vL Voltage compliance is then
VOL < vL < VOH The voltage compliance is greater than the previous circuit.
(2.10)
2. VOLTAGE-TO-CURRENT CONVERTERS FLOATING TYPE V-I CONVERTORS • These circuits are said to be bidirectional since the input voltage can be either positive or negative, no matter the polarity, the circuits work properly. • If the load is a capacitor, the the input-output relation of the circuit would be
iO = sCv I
(2.11)
which represents a lossless integrator.
• Integrators are primarily used in waveform generators, V-F, F-V converters, and ADCs.
2. VOLTAGE-TO-CURRENT CONVERTERS PRACTICAL V-I CONVERTOR LIMITATIONS
vI − vD + vL + roiO − αvD = 0 (2.12) vD vI − vD iO + − =0 rd R
Figure 2.3 V-I converter with practical opamp model
(2.13)
α − R / rd 1 A= R 1 + α + rO / R + rO / rd
(2.14)
RO = ( R || rd )(1 + α ) + ro
(2.15)
2. VOLTAGE-TO-CURRENT CONVERTERS SIMULATIONS OF FLOATING TYPE V-I CONVERTORS 8.0mA
(7.1693,7.1377m)
4.0mA
0A
-4.0mA
(-7.4603,-7.4542m)
-8.0mA
-15V
-I(RL)
-10V
-5V
0V
5V
10V
V_V3
Figure 2.4 DC characteristics of the floating V-I converter
15V
2. VOLTAGE-TO-CURRENT CONVERTERS SIMULATIONS OF FLOATING TYPE V-I CONVERTORS 1.0mA
(538.270K,706.845u)
0.5mA
0A
1.0Hz -I(RL)
10Hz
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
Frequency
Figure 2.5 AC characteristics of the floating V-I converter The -3dB cut-off frequency of the circuit is 538kHz.
10MHz
2. VOLTAGE-TO-CURRENT CONVERTERS SIMULATIONS OF FLOATING TYPE V-I CONVERTORS 1
8.0mA
2
10V
4.0mA
5V
0A
0V
-4.0mA
-5V
-8.0mA
>> -10V 0s 1
-I(RL)
2
0.5ms V(V4:+)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
Time
Figure 2.6 Transient characteristics of the floating V-I converter
4.0ms
2. VOLTAGE-TO-CURRENT CONVERTERS GROUNDED V-I CONVERTORS
Figure 2.7 Howland current pump and its Norton equivalent circuit
2. VOLTAGE-TO-CURRENT CONVERTERS GROUNDED V-I CONVERTORS Output resistance of the Howland current pump is
R2 RO = R2 / R1 − R4 / R3
(2.16)
In order to have an infinite output resistance (the ideal current source),
R4 R2 = R3 R1 Balanced bridge
(2.17)
2. VOLTAGE-TO-CURRENT CONVERTERS GROUNDED V-I CONVERTORS Voltage compliance of the circuit is
vL ≤
R1 Vsat R1 + R2
(2.18)
Saturation voltage of the operational amplifier
2. VOLTAGE-TO-CURRENT CONVERTERS EFFECTS OF RESISTANCE MISMATCHES The balanced bridge can not be implemented in practice. The non-ideality can be modelled by imbalance factor ε
R4 R2 = (1 − ε ) R3 R1
(2.19)
The output resistance of the circuit is found as
R1 Ro = ε
(2.20)
2. VOLTAGE-TO-CURRENT CONVERTERS SIMULATIONS OF GROUNDED V-I CONVERTORS 10mA (7.0263,7.0210m)
5mA
0A
-5mA (-7.4474,-7.4396m)
-10mA -15V
-I(R5)
-10V
-5V
0V
5V
10V
V_V3
Figure 2.8 DC characteristics of the grounded-type V-I converter
15V
2. VOLTAGE-TO-CURRENT CONVERTERS SIMULATIONS OF GROUNDED V-I CONVERTORS 1.0mA
0.8mA
(176.198K,716.320u)
0.6mA
0.4mA
0.2mA 1.0Hz -I(R5)
10Hz
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
Frequency
Figure 2.9 AC characteristics of the grounded-type V-I converter The -3dB cut-off frequency of the circuit is 178kHz
10MHz
2. VOLTAGE-TO-CURRENT CONVERTERS SIMULATIONS OF GROUNDED V-I CONVERTORS 1
10V
2
10mA
5V
5mA
0V
0A
-5V
-5mA
-10V
>> -10mA 0s 1
V(R1:1)
0.5ms 2 I(R5)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
4.0ms
Time
Figure 2.10 Transient characteristics of the grounded-type V-I converter • Note the inverse direction of the current.
2. VOLTAGE-TO-CURRENT CONVERTERS EFFECTS OF FINITE GAIN OF OPAMP If the open-loop gain of opamp is α, then the output resistance of the Howland current pump is
α RO = ( R1 || R2 )1 + 1 + R2 / R1
RO decreased from infinite to this finite number.
(2.21)
2. VOLTAGE-TO-CURRENT CONVERTERS EFFECTS OF FINITE GAIN OF OPAMP Balance condition:
R4 R2 A + R2 B = R3 R1
(2.22)
Transfer characteristics:
R2 / R1 iO = vI R2 B
Figure 2.11 Improved Howland circuit
(2.23)
The advantage of the circuit is to provide power saving.
3. CURRENT AMPLIFIERS Opamps can be used as current amplifiers. The transfer characteristics of a practical current amplifier:
1 iO = AiI − vL RO
Output current
Gain
Output resistance
(3.1)
Load voltage
Ideally, iO must be independent of vL, that is
RO → ∞
(3.2)
3. CURRENT AMPLIFIERS FLOATING TYPE CURRENT AMPLIFIER
Figure 3.1 Floating type current amplifier
3. CURRENT AMPLIFIERS FLOATING TYPE CURRENT AMPLIFIER For infinite-gain opamp:
iO +
If the open-loop gain of opamp is α
iI R2 iI = 0 R1
(3.3)
R2 R1
(3.4)
If
A = 1+
iO = AiI
R2 / R1 A = 1+ 1+1/ α
(3.6)
RO = R1 (1 + α )
(3.7)
Voltage compliance is: (3.5)
Note that output resistance approcahes infinity.
− (VOH + i2 R1 ) ≤ vL ≤ −(VOL + i2 R1 ) (3.8)
3. CURRENT AMPLIFIERS GROUNDED TYPE CURRENT AMPLIFIER Input-output relation:
iO = AiS −
1 vL RO
(3.9)
Where the gain is
A=−
R2 R1
(3.10)
Output resistance is: Figure 3.2 Grounded type current amplifier
R2 RO = − R S R1
(3.11)
If A=-1, the circuit is called as current reverser or current mirror. Source resistance
3. CURRENT AMPLIFIERS SIMULATIONS OF FLOATING TYPE CURRENT AMPLIFIER
20mA (-14.050m,14.060m)
10mA
0A
-10mA (14.820m,-
-20mA
-16mA
I(I1)
-
-12mA
I(RL)
-8mA
-4mA
0A
4mA
8mA
12mA
I_I1
Figure 3.3 DC transfer characteristics of grounded current amplifier
16mA
3. CURRENT AMPLIFIERS SIMULATIONS OF FLOATING TYPE CURRENT AMPLIFIER 1.0mA
(1.1538M,712.435u)
0.5mA
0A 1.0Hz I(I2)
-I(RL)
10Hz
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
Frequency
Figure 3.4 AC transfer characteristics of grounded current amplifier
10MHz
3. CURRENT AMPLIFIERS SIMULATIONS OF FLOATING TYPE CURRENT AMPLIFIER
1.0mA
0.5mA
0A
-0.5mA
-1.0mA
0s
I(I2)
I(RL)
0.5ms
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
Time
Figure 3.5 Transient characteristics of grounded current amplifier
4.0ms
4. DIFFERENCE AMPLIFIERS A difference amplifier is an amplifier that has one output and two inputs and satisfies: • Amplification of the difference voltage at the input terminals • Rejection of the common voltage at the input terminals
Figure 4.1 Difference amplifier (a), differential and common inputs (b)
4. DIFFERENCE AMPLIFIERS The bridge balance must be satisfied to operate:
R4 R2 = R3 R1
(4.1)
R2 vO = (v2 − v1 ) R1
Output voltage
(4.2)
Differential input voltage
4. DIFFERENCE AMPLIFIERS Differential mode component:
vDM = v2 − v1
(4.3)
Common mode component:
vCM
v1 + v2 = 2
(4.4)
Input voltages can be expressed in of differential mode and common mode inputs (Figure 4.1b):
v1 = vCM −
vDM 2
(4.5)
v2 = vCM +
vDM 2
(4.6)
4. DIFFERENCE AMPLIFIERS The common mode and differential mode input resistances can be exxpressed as
Figure 4.2 Differential mode and common mode input resistances
Rid = 2R1
(4.7)
R1 + R2 Ric = 2
(4.8)
4. DIFFERENCE AMPLIFIERS EFFECTS OF RESISTANCE MISMATCHES A difference amplifier is insensitive to common mode for infinite gain and perfectly balanced resistances. If the opamp is ideal but the resistances are mismatched:
R2 ' = R2 (1 − ε )
(4.9)
imbalance factor
vO = vDM + ACM vCM
Figure 4.3 Differential mode and common mode input resistances
R2 R1 + 2 R2 ε 1 − = R1 R1 + R2 2 R2 ACM = ε R1 + R2
(4.10) (4.11)
(4.12)
4. DIFFERENCE AMPLIFIERS EFFECTS OF RESISTANCE MISMATCHES • is called the differential mode gain. • Acm is called the common mode gain.
CMRR = Acm
(4.13)
Common mode rejection ratio
CMRRdB = 20 log10
CMRR in dB
Acm
(4.14)
4. DIFFERENCE AMPLIFIERS EFFECTS OF RESISTANCE MISMATCHES If ε<<1, then
R2 εR2 ≅ / Acm R1 R1 + R2 1 + R2 / R1 ≅ 20 log10 Acm dB ε
(4.15)
(4.16)
• For a fixed imbalance factor, CMRR increases with increasing R2/R1.
4. DIFFERENCE AMPLIFIERS SIMULATIONS OF DIFFERENTIAL AMPLIFIER INA105 from BURR-BROWN • INA105 is a monolithic differential amplifier that employs very good matched resistors (2/10000 sensitivity).
Figure 4.4 INA 105 monolithic differential amplifier [5]
4. DIFFERENCE AMPLIFIERS SIMULATIONS OF DIFFERENTIAL AMPLIFIER INA105 from BURR-BROWN 5.0V
(-1.1491,1.0851) (-429.825m,429.950m)
0V
(3.5702,-3.5518)
-5.0V -5.0V V(V1:+)
-4.0V V(U1:OUT)
-3.0V
-2.0V
-1.0V
0.0V
1.0V
2.0V
3.0V
4.0V
5.0V
V_V1
Figure 4.5 DC sweep of INA 105 monolithic differential amplifier • Note that, in the linear region, if the input voltage is 429.825mV (applied to inverting terminal), output voltage is -429.950mV.
4. DIFFERENCE AMPLIFIERS SIMULATIONS OF DIFFERENTIAL AMPLIFIER INA105 from BURR-BROWN 1.0V
(1.4610M,719.235m)
0.5V
0V 1.0Hz
V(V1:+)
10Hz
V(U1:OUT)
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
Frequency
Figure 4.6 AC sweep of INA 105 monolithic differential amplifier • Note that, -3dB cut-off frequency is 1.45MHz.
10MHz
4. DIFFERENCE AMPLIFIERS SIMULATIONS OF DIFFERENTIAL AMPLIFIER INA105 from BURR-BROWN 400mV
200mV
0V
-200mV
-400mV
0s
V(V5:+)
0.5ms V(U1:OUT)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
Time
Figure 4.7 Transient response of INA 105 monolithic differential amplifier
4.0ms
4. DIFFERENCE AMPLIFIERS DIFFERENCE AMPLIFIER CALIBRATION
Figure 4.8 Difference amplifier calibration circuit • Rpot is adjusted toachieve calibrated difference amplifier.
4. DIFFERENCE AMPLIFIERS DIFFERENCE AMPLIFIER WITH VARIABLE GAIN
The CM gain can be given as:
ACM
Figure 4.9 Differential amplifier with variable gain
2 R2 R2 1 + = R1 RG
(4.17)
• Note that gain is inveresely proportional with RG.
4. DIFFERENCE AMPLIFIERS DIFFERENCE AMPLIFIER WITH VARIABLE GAIN
The CM gain can be given as:
ACM
R2 RG = R1 R3
(4.18)
A linearly tuneable gain is achieved. Figure 4.10 Differential amplifier with linearly adjustible gain
4. DIFFERENCE AMPLIFIERS GROUND-LOOP INTERFERENCE PROBLEM
Figure 4.11 An inverting amplifier that is subject to ground-loop interference Zg is the distributed impedance of the ground line. Amplifier sees vi and vg in series, so;
vO = −
R2 (vi + v g ) R1
(4.19)
Figure 4.11 A difference amplifier that eliminates cross-talk for common ground terminal connections
R2 vO = − vi R1
(4.20)
5. INSTRUMENTATION AMPLIFIERS An instrumentation amplifier is a difference amplifier that satisfies:
3. Extremely high (ideally infinite) common mode and differential mode input resistances 4. Very low (ideally zero) output impedance 5. Accurate and stable gain, typically between 1V/V and 1000V/V 6. Extremely high common mode rejection ratio •
Instrumentation amplifiers are used in the areas in which the differential mode input signal level is very low such as transducer output in an industrial application or biomedical engineering.
•
The 2nd, 3rd and 4th expressions may be satisfied by a difference amplifier but extremely high input impedance can not be satisfied.
5. INSTRUMENTATION AMPLIFIERS TRIPLE OPAMP INSTRUMENTATION AMPLIFIER From the first stage:
2R vO1 − vO 2 = 1 + 3 (v1 − v2 ) RG
(5.1)
From the second stage:
vO =
R2 (vO 2 − vO1 ) R1
(5.2)
Combining these two equations yields:
First stage
Second stage
Figure 5.1 Triple opamp instrumentation amplifier configuration
2 R3 R2 vO = AI . AII = 1 + RG R1 (5.3)
5. INSTRUMENTATION AMPLIFIERS DUAL OPAMP INSTRUMENTATION AMPLIFIERS
Figure 5.2 Dual opamp instrumentation amplifier For OA1
R3 v3 = (1 + )v1 R4
For OA2 (5.5)
R2 R2 vO = − v3 + 1 + v2 R1 R1
(5.6)
5. INSTRUMENTATION AMPLIFIERS DUAL OPAMP INSTRUMENTATION AMPLIFIERS Combining these equations yields:
R2 1 + R3 / R4 vO = 1 + v2 − v1 1 + R1 / R2 R1 If
R3 R1 = R4 R2
(5.7)
(5.8)
then, the input-output relationship becomes:
R vO = 1 + 2 (v2 − v1 ) R1
(5.9)
5. INSTRUMENTATION AMPLIFIERS DUAL OPAMP INSTRUMENTATION AMPLIFIERS WITH VARIABLE GAIN
Figure 5.3 Variable gain instrumentation amplifier
A = 1+
R2 2 R2 + R1 RG
(5.10)
5. INSTRUMENTATION AMPLIFIERS DUAL OPAMP INSTRUMENTATION AMPLIFIER
• Although dual opamp instrumantation amplifiers take the advantage of lower number of active and ive devices, it suffers from highfrequency performance degredation. • The reason for the performance decrement is: v1 and v2 input voltages meet at different times since v1 has to through OA1 to catch v2.
5. INSTRUMENTATION AMPLIFIERS MONOLITHIC INSTRUMENTATION AMPLIFIER
• The monolithic instrumentation amplifiers allows better optimization of CMRR, lineraity and noise. • Also, monolithic laser trimmed resistances allows the impalance factor to be minimized. • The SPICE simulations of a monolithic instrumentation amplifer from Burr-Brown, namely INA-101 are carried. • INA-101 allows the adjustement of the gain by an external resistor.
5. INSTRUMENTATION AMPLIFIERS SPICE SIMULATIONS OF A MONOLITHIC INSTRUMENTATION AMPLIFIER
40kΩ A = 1+ RG
Gain
Figure 5.4 Simulated instrumentation amplifier INA 101 [6]
(5.4)
5. INSTRUMENTATION AMPLIFIERS SPICE SIMULATIONS OF A MONOLITHIC INSTRUMENTATION AMPLIFIER 20V
(3.2984,9.889) 10V
0V
(-3.3246,-9.912) -10V
-20V
-15V
V(U1:+)
V(RL:2)
-10V
-5V
0V
5V
10V
V_V1
Figure 5.5 Simulated DC characteristics of INA 101 instrumentation amplifier (RG=20k, A=3V/V)
15V
5. INSTRUMENTATION AMPLIFIERS SPICE SIMULATIONS OF A MONOLITHIC INSTRUMENTATION AMPLIFIER 3.0V
(385.478K,2.1408)
2.0V
1.0V
0V
1.0Hz
V(V4:+)
10Hz
V(RL:2)
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
Frequency
Figure 5.6 Simulated AC characteristics of INA 101 instrumentation amplifier (RG=20k, A=3V/V) • Note that –dB cut-off frequency is 338kHz.
10MHz
5. INSTRUMENTATION AMPLIFIERS SPICE SIMULATIONS OF A MONOLITHIC INSTRUMENTATION AMPLIFIER 4.0V
2.0V
0V
-2.0V
-4.0V
0s
V(U1:+)
0.5ms V(RL:2)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
4.0ms
Time
Figure 5.7 Simulated transient characteristics of INA 101 instrumentation amplifier (RG=20k, A=3V/V)
5. INSTRUMENTATION AMPLIFIERS FLYING CAPACITOR TECHNIQUES • •
A capacitor is flipped back and forth between source and amplifier to achieve high common mode rejection ratio (CMRR). Circuit ignores common mode input hence provides high common mode rejection ratio.
R vO = 1 + 2 (v2 − v1 ) R1
(5.5)
In fact, a noninverting amplifier Figure 5.8 Employing switched capacitor to achieve high commpn mode rejection ratio
6. INSTRUMENTATION APPLICATIONS ACTIVE GUARD DRIVE • In the situations where source and amplifier are far apart the signal is trnasmitted by shielded wires such as coaxial lines. • The signal is transmitted in a double ended form for achieving the same noise pick-up. • The common noise is considered as common mode signal and rejected by the difference amplifier. • For this reason, double ended transmission is also named as balanced transmission. THE DISADVANTAGE: • The distributed capacitance of the cable decreases the CMRR. • Consider the circuit of Figure 6.1.
6. INSTRUMENTATION APPLICATIONS ACTIVE GUARD DRIVE First coaxial line model
Second coaxial line model Figure 6.1 The distributed capacitance of the balanced line • Since Rs1.C1 and Rs2.C2 time constants will not be the same, this unbalance will decrease common mode rejection ratio because the amplifier will see a CM input.
6. INSTRUMENTATION APPLICATIONS ACTIVE GUARD DRIVE CMRR due to imbalance is:
1 CMRRdB = 20 log10 2πfRdmCcm where
Rdm = Rs1 − Rs 2
Ccm
C1 + C2 = 2
• For example, if f=60Hz, Rdm=1kΩ and Ccm=1nF, CMRR=68.5dB.
(6.1)
(6.2)
(6.3)
6. INSTRUMENTATION APPLICATIONS ACTIVE GUARD DRIVE
Figure 6.2 Reducing Ccm by using an active guard • Vcm is fed to shield of the balanced line, thus reducing CM input at the input of amplifier.
6. INSTRUMENTATION APPLICATIONS DIGITALLY PROGRAMMABLE GAIN • In automatic instrumentation such as data acquistion systems, the programming of instrumentation amplifier electronically may be needed. • The gain of the first stage is
Routside AI = 1 + Rinside
(6.4)
Routside = 2R1
(6.5)
Rinside = 2( R2 + R3 + .... + Rn ) + Rn +1 Figure 6.3 Digitally programmable IA
(6.6)
6. INSTRUMENTATION APPLICATIONS OUTPUT OFFSETTING • In some applications, like a voltage-to-frequency converters, the output may need to have an offset value. • The need comes from the fact that: the input of the succeeding circuit mya have only one polarity.
vO = A(v2 − v2 ) + (1 + R2 / R1 )[ R1 /( R1 + R2 )]Vref (6.7)
vO = A(v2 − v1 ) + Vref Figure 6.4 IA with offset control
(6.8)
6. INSTRUMENTATION APPLICATIONS CURRENT OUTPUT INSTRUMENTATION AMPLIIFIERS
iO =
1 + 2 R3 / RG (v2 − v1 ) R1 (6.9)
Figure 6.5 Current output IA • The current output is generated using Howland current pump at the output stage
6. INSTRUMENTATION APPLICATIONS CURRENT OUTPUT INSTRUMENTATION AMPLIIFIERS
iO =
1 1 (v2 − v1 ) − vL R RO (6.10)
R2 / R1 RO = R3 R5 / R4 − ( R2 + R3 ) / R1 Figure 6.6 Current output IA with dual opamps
(6.11)
| vL |≤ Vsat − 2 v2 − v1 (6.12)
6. INSTRUMENTATION APPLICATIONS SIMULATIONS OF CURRENT OUTPUT INSTRUMENTATION AMPLIIFIERS 1
5.0V
2
20uA (3.6075,14.240u)
10uA
0V
0A
-10uA (-3.8037,-14.978u)
-5.0V
>>
-20uA -5.0V 1
-4.0V V(V2:+) 2
I(RL)
-3.0V
-2.0V
-1.0V
0.0V
1.0V
2.0V
3.0V
V_V2
Figure 6.7 DC characteristics of current output IA with LM324
4.0V
5.0V
6. INSTRUMENTATION APPLICATIONS SIMULATIONS OF CURRENT OUTPUT INSTRUMENTATION AMPLIIFIERS 3.0uA
2.0uA
1.0uA
0A
1.0Hz
I(RL)
10Hz
100Hz
1.0KHz
10KHz
100KHz
1.0MHz
Frequency
Figure 6.8 AC characteristics of current output IA with LM324
10MHz
6. INSTRUMENTATION APPLICATIONS SIMULATIONS OF CURRENT OUTPUT INSTRUMENTATION AMPLIIFIERS 1
1.0V
2
4.0uA
0.5V
2.0uA
0V
0A
-0.5V
-2.0uA
-1.0V
>> -4.0uA 0s 1
V(V5:+)
0.5ms 2 I(RL)
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
Time
Figure 6.9 Transient characteristics of current output IA with LM324
4.0ms
6. INSTRUMENTATION APPLICATIONS CURRENT INPUT INSTRUMENATION AMPLIFIER
vO 2 = vCM − iI R3 vO1 = vCM + R3iI vO =
R2 (vO 2 − vO1 ) R1
vO = − Figure 6.10 Current input instrumentation amplifier
2 R2 R3iI R1
(6.13) (6.14) (6.15)
(6.16)
7. TRANSDUCER BRIDGE AMPLIFIERS
• Resistive transducer: A device whose resistance varies by a physical paramater such as temperature, light frequency/amplitude, pressure. • Resistive transducer transducer are made a part of a circuit to generate an electrical signal proportional to the physical parameter. • Parameter’s magnitude and the electrical signal are wanted to be in a linear relationship.
7. TRANSDUCER BRIDGE AMPLIFIERS TRANSDUCER RESISTANCE DEVIATION • Transducer resistances are expressed as
R + ∆R Resistance at a reference condition
(7.1) Deviation
It can be also expressed as
R = R (1 + δ )
∆R δ= R
Fractional deviation
(7.2)
7. TRANSDUCER BRIDGE AMPLIFIERS THE TRANSDUCER BRIDGE • ΔR is converted to a ΔV by means of a voltage divider.
Figure 7.1 Transducer bridge with instrumentation amplifier • This voltage dividers are referred to as bridge legs.
7. TRANSDUCER BRIDGE AMPLIFIERS THE TRANSDUCER BRIDGE
v1 =
VREF R (1 + δ ) R VREF δ = VREF + R1 + R (1 + δ ) R1 + R 2 + R1 / R + R / R1 + (1 + R / R1 )δ
R v2 = VREF R1 + R
(7.4)
δ vO = A(v1 − v2 ) = AVREF 1 + R1 / R + (1 + R / R1 )(1 + δ ) If
(7.3)
(7.5)
δ << 1 vO ≅
AVREF AVREF δ= δ 2 + R1 / R + R / R1 4
(7.6) If R1=R
7. TRANSDUCER BRIDGE AMPLIFIERS THE TRANSDUCER BRIDGE
Figure7.2 Bridge calibration of a transducer bridge Due to the tolerances of bridge resistances and IA’s reference voltage, a calibration is needed. R3 is varied to achieve vO=0V at ΔR=0.
7. TRANSDUCER BRIDGE AMPLIFIERS STRAIN GAUGE BRIDGES A resistor with resistivity ρ, cross-sectional area S and lenght l has a resistance of:
ρl R= S
(7.7)
l ' = l + ∆l
(7.8)
S ' = S − ∆S
(7.9)
ρ (l + ∆l ) R= S − ∆S
(7.10)
If the wire is strained,
7. TRANSDUCER BRIDGE AMPLIFIERS STRAIN GAUGE BRIDGES The volume of the resistor is not changed, so
(l + ∆l )( S − ∆S ) = Sl
(7.11)
∆l ∆l ∆l ∆R = R 2 + ≅ 2 R l l l
(7.12)
Unstrained resistance
Fractional elognation
7. TRANSDUCER BRIDGE AMPLIFIERS STRAIN GAUGE BRIDGES
Figure 7.3 Strain-gauge bridge and instrumentation amplifier Load cell
7. TRANSDUCER BRIDGE AMPLIFIERS STRAIN GAUGE BRIDGES
v1 = VB ( R + ∆R) /( R + ∆R + R − ∆R) = VB ( R + ∆R ) / 2 R
(7.13)
v2 = VB ( R − ∆R ) / 2 R
(7.14)
v1 − v2 = VB ∆R / R = VBδ
(7.15)
vO = AVREF δ
(7.16)
• Note that, voltage deviation is perfectly linear with fractional elongation. • Also, in the circuit R3 adjusts sensitivity, R2 nulls the voltage in the absence of strain.
7. TRANSDUCER BRIDGE AMPLIFIERS BRIDGE AMPLIFIER WITH SINGLE OPAMP AMPLIFIER
vO =
R2 δ VREF R R1 / R + (1 + R1 / R2 )(1 + δ ) (7.17) For
Figure 7.4 Single opamp bridge amplifier
vO ≅
δ << 1
R2 δ VREF R 1 + R1 / R + R1 / R2 (7.18)
7. TRANSDUCER BRIDGE AMPLIFIERS BRIDGE LINEARIZATION • Bridge circuits except strain-gauge has a nonlinear voltage response with the fractional elognation. • The linear response may be achieved using a current source to drive the bridge. Current source (V-I converter)
Figure 7.5 Bridge linearization using a current source
7. TRANSDUCER BRIDGE AMPLIFIERS BRIDGE LINEARIZATION
VREF IB = R1
(7.19)
v1 = VREF + R (1 + δ ) I B / 2
(7.20)
v2 = VREF + RI B / 2
(7.21)
v1 − v2 = RδI B / 2
(7.22)
vO =
ARVREF δ 2R1
(7.23)
7. TRANSDUCER BRIDGE AMPLIFIERS BRIDGE LINEARIZATION WITH SINGLE TRANSDUCER
Figure 7.6 Bridge linearization using single transducer
− R(1 + δ ) v1 = VREF R1
vO =
(7.24)
− R2 R v1 − VREF 2 R1 R1
From (7.24) and (7.25),
vO = VREF δ
R2 R1
(7.26)
(7.25)
7. TRANSDUCER BRIDGE AMPLIFIERS SIMULATIONS OF BRIDGE LINEARIZATION WITH SINGLE TRANSDUCER
• The bridge transducer witgh single transducer element has been simulated in OrCad SPICE. • The output voltage is plotted versus the transducer resistance value. • In rder to plot the variation, a parametric sweep in OrCad SPICE is utilized [7, 8, 9]. • Supply voltages and reference voltage are taken as 15V. • R1=R=10k are taken. • LM324 opamp macromodels are used.
7. TRANSDUCER BRIDGE AMPLIFIERS SIMULATIONS OF BRIDGE LINEARIZATION WITH SINGLE TRANSDUCER 0V
-5V
-10V
-15V
0
0.1K V(U2A:OUT)
0.2K
0.3K
0.4K
0.5K
0.6K
0.7K
0.8K
0.9K
1.0K
RL
Figure 7.7 Variation of output voltage via the change of transducer resistance Note that, vo is negative since, fractinal elongation is negative (i.e. transducer resistance is below 1k.)
1.1K
7. REFERENCES •
S. Franco, Design with operational amplifiers and anategrated circuits, McGraw Hill, USA, 2001.
•
S. Kılınç, M. Saygıner, U. Çam and H. Kuntman, Simple and accurate macromodel for current operational amplifier (COA), Proceedings of ELECO 2005: The 4th International Conference on Electrical and Electronics Engineering, (Electronics), pp.1-5, 7-11 December 2005, Bursa, Turkey.
•
H. Kuntman: Simple and accurate nonlinear OTA macromodel for simulation of CMOS OTAC active filters, International Journal of Electronics, Vol.77, No.6, pp.993-1006, 1994.
•
LM324 datasheet, National Semiconductors, http://cache.national.com/ds/LM/LM124.pdf, accessed on 10th Oct. 2007.
•
INA105 datasheet, Texas Instruments, http://focus.ti.com/lit/ds/symlink/ina105.pdf, accessed on 10th Oct. 2007.
•
INA101 datasheet, Burr-Brown, www.hardware.dibe.unige.it/DataSheets/INA101.pdf, accessed on 12th Oct. 2007.
•
Pspice Reference Guide, ece-classweb.ucsd.edu/spring06/ece139/PspiceRef.pdf, accessed on 13th Oct. 2007.
•
DC Simulation and analysis, http://ecen4303.okstate.edu/ecen1322/lab2ar.rtf, accessed on 13th Oct. 2007.
•
Pspice Notes v.3.0, http://wwwferp.ucsd.edu/najmabadi/CLASS/COMMON/PSPICE/PSpice_Notes_v3.0.doc, accessed 13th Oct. 2007.
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