AD8554ARUZ Analog Devices Inc, AD8554ARUZ Datasheet - Page 20

• Manufacturer Part Number: AD8554ARUZ
• Description: IC OPAMP CHOPPER R-R 14TSSOP
• Manufacturer: Analog Devices Inc

Specifications of AD8554ARUZ

• Slew Rate: 0.4 V/µs
• Amplifier Type: Chopper (Zero-Drift)
• Number Of Circuits: 4
• Output Type: Rail-to-Rail
• Gain Bandwidth Product: 1.5MHz
• Current - Input Bias: 10pA
• Voltage - Input Offset: 1000µV
• Current - Supply: 850µA
• Current - Output / Channel: 30mA
• Voltage - Supply, Single/dual (±): 2.7 V ~ 5.5 V
• Operating Temperature: -40°C ~ 125°C
• Mounting Type: Surface Mount
• Package / Case: 14-TSSOP
• Op Amp Type: Zero Drift
• No. Of Amplifiers: 4
• Bandwidth: 1.5MHz
• Supply Voltage Range: 2.7V To 5.5V
• Amplifier Case Style: TSSOP
• No. Of Pins: 14
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• -3db Bandwidth: -
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant, Lead free / RoHS Compliant

AD8551/AD8552/AD8554
APPLICATIONS
A 5 V PRECISION STRAIN GAGE CIRCUIT
The extremely low offset voltage of the AD8552 makes it an
ideal amplifier for any application requiring accuracy with high
gains, such as a weigh scale or strain gage. Figure 65 shows a
configuration for a single-supply, precision, strain gage
measurement system.
A REF192 provides a 2.5 V precision reference voltage for A2.
The A2 amplifier boosts this voltage to provide a 4.0 V refer-
ence for the top of the strain gage resistor bridge. Q1 provides
the current drive for the 350 Ω bridge network. A1 is used to
amplify the output of the bridge with the full-scale output
voltage equal to
where R
Using the values given in Figure 65, the output voltage linearly
varies from 0 V with no strain to 4.0 V under full strain.
3 V INSTRUMENTATION AMPLIFIER
The high common-mode rejection, high open-loop gain, and
operation down to 3 V of supply voltage makes the AD855x an
excellent choice of op amp for discrete single-supply instrumen-
tation amplifiers. The common-mode rejection ratio of the
AD855x is greater than 120 dB, but the CMRR of the system is
also a function of the external resistor tolerances. The gain of
the difference amplifier shown in Figure 66 is given as
EQUIVALENT
V
2
OUT
×
LOAD
CELL
350Ω
B
2N2222
(
R +
is the resistance of the load cell.
R
4.0V
1
=
OR
B
Q1
Figure 65. A 5 V Precision Strain Gage Amplifier
V
R
1
⎛
⎜
⎜
⎝
2
NOTES
1. USE 0.1% TOLERANCE RESISTORS.
)
R
3
FULL-SCALE
R
+
4
40mV
R
1kΩ
AD8552-B
4
12.0kΩ
⎞
⎟
⎟
⎠
⎛
⎜ ⎜
⎝
1
+
R
R
1
2
5V
17.4kΩ
17.4kΩ
A2
⎞
⎟ ⎟
⎠
R
R
−
1
3
A1
V
AD8552-A
2.5V
2
100Ω
⎛
⎜ ⎜
⎝
100Ω
R
R
R
R
20kΩ
2
4
6
2
1
⎞
⎟ ⎟
⎠
REF192
4
V
0V TO 4.0V
2
OUT
3
(17)
(18)
Rev. D | Page 20 of 24
In an ideal difference amplifier, the ratio of the resistors are set
exactly equal to
Which sets the output voltage of the system to
Due to finite component tolerance, the ratio between the four
resistors is not exactly equal, and any mismatch results in a
reduction of common-mode rejection from the system.
Referring to Figure 66, the exact common-mode rejection ratio
can be expressed as
In the three-op amp, instrumentation amplifier configuration
shown in Figure 67, the output difference amplifier is set to
unity gain with all four resistors equal in value. If the tolerance
of the resistors used in the circuit is given as δ, the worst-case
CMRR of the instrumentation amplifier is
Consequently, using 1% tolerance resistors results in a worst-
case system CMRR of 0.02, or 34 dB. Therefore, either high
precision resistors or an additional trimming resistor, as shown
in Figure 67, should be used to achieve high common-mode
rejection. The value of this trimming resistor should be equal
to the value of R multiplied by its tolerance. For example, using
10 kΩ resistors with 1% tolerance requires a series trimming
resistor equal to 100 Ω.
V2
V1
V
CMRR
CMRR
A
OUT
Figure 67. A Discrete Instrumentation Amplifier Configuration
V
R
G
=
Figure 66. Using the AD855x as a Difference Amplifier
= A
R
V2
V1
R
MIN
=
2
1
V
=
R
(V1 − V2)
IF
AD8554-A
AD8554-B
=
1
R
R
R
R
R
R
R
2
4
3
2
4
3
4
1
R
δ
=
+
R
R
1
1
R
3
R
R
V
2
OUT
2
1
4
R
, THEN V
−
2
R
R
= 1 +
2
4
4
R
+
2
R
R
R
R
OUT
2R
R
R
3
G
2
TRIM
R
(V1 – V2)
=
R
AD8551/
AD8552/
AD8554
3
R
2
R
R
2
1
× (V1 – V2)
R
AD8554-C
V
OUT
V
OUT
(19)
(20)
(21)
(22)