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參數資料
| 型號： | AD8554AR |
| 廠商： | ANALOG DEVICES INC |
| 元件分類： | 運動控制電子 |
| 英文描述： | Zero-Drift, Single-Supply, Rail-to-Rail Input/Output Operational Amplifiers |
| 中文描述： | QUAD OP-AMP, 10 uV OFFSET-MAX, 1.5 MHz BAND WIDTH, PDSO14 |
| 封裝： | MS-012AB, SOIC-14 |
| 文件頁數： | 16/20頁 |
| 文件大小： | 262K |
| 代理商： | AD8554AR |
AD8551/AD8552/AD8554
–16–
REV. 0
V2
V1
V
OUT
R
1
R
3
R
4
R
2
AD855x
IFR
3
R
4
=R
1
R
2
, THEN V
OUT
=R
1
R
2
3
(V1
2
V2)
Figure 58. Using the AD855x as a Difference Amplifier
In an ideal difference amplifier, the ratio of the resistors are set
exactly equal to:
A
R
R
R
R
V
=
=
2
1
4
3
(19)
Which sets the output voltage of the system to:
V
Due to finite component tolerance the ratio between the four
resistors will not be exactly equal, and any mismatch results in a
reduction of common-mode rejection from the system. Referring
to Figure 58, the exact common-mode rejection ratio can be ex-
pressed as:
+
4
2
2
A V
V
OUT
=
)
1
2
(20)
CMRR
R R
R R
4
R R
R R
R R
2
=
+
4
3
3
(21)
In the 3 op amp instrumentation amplifier configuration shown
in Figure 59, the output difference amplifier is set to unity gain
with all four resistors equal in value. If the tolerance of the resis-
tors used in the circuit is given as
δ
, the worst-case CMRR of
the instrumentation amplifier will be:
CMRR
MIN
=
1
2
δ
(22)
V
OUT
R
R
R
R
AD8554-C
V2
R
R
V1
R
G
AD8554-B
AD8554-A
V
OUT
= 1 +R
G
(V1
2
V2)
R
TRIM
Figure 59. A Discrete Instrumentation Amplifier
Configuration
Thus, using 1% tolerance resistors would result 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
59, 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 k
resistors
with 1% tolerance would require a series trimming resistor equal to
100
.
A High Accuracy Thermocouple Amplifier
Figure 60 shows a K-type thermocouple amplifier configuration
with cold-junction compensation. Even from a +5 V supply, the
AD8551 can provide enough accuracy to achieve a resolution
of better than 0.02
°
C from 0
°
C to 500
°
C. D1 is used as a
temperature measuring device to correct the cold-junction error
from the thermocouple and should be placed as close as possible
to the two terminating junctions. With the thermocouple mea-
suring tip immersed in a zero-degree ice bath, R
6
should be
adjusted until the output is at 0 V.
Using the values shown in Figure 60, the output voltage will
track temperature at 10 mV/
°
C. For a wider range of tempera-
ture measurement, R
9
can be decreased to 62 k
. This will
create a 5 mV/
°
C change at the output, allowing measurements
of up to 1000
°
C.
AD8551
3
2
8
4
0V TO 5.00V
(0
8
C TO 500
8
C)
+5V
0.1
m
F
+
10
m
F
R
9
124k
V
R
8
453
V
R
5
40.2k
V
R
1
10.7k
V
R
2
2.74k
V
REF02EZ
0.1
m
F
+12V
2
6
4
+
+
–
–
D1
1N4148
R
53.6
V
R
4
5.62k
V
+5.000V
K-TYPE
THERMOCOUPLE
40.7
m
V/
8
C
R
6
200
V
1
Figure 60. A Precision K-Type Thermocouple Amplifier
with Cold-Junction Compensation
Precision Current Meter
Because of its low input bias current and superb offset voltage at
single supply voltages, the AD855x is an excellent amplifier for
precision current monitoring. Its rail-to-rail input allows the
amplifier to be used as either a high-side or low-side current
monitor. Using both amplifiers in the AD8552 provides a simple
method to monitor both current supply and return paths for
load or fault detection.
Figure 61 shows a high-side current monitor configuration. Here,
the input common-mode voltage of the amplifier will be at or near
the positive supply voltage. The amplifier’s rail-to-rail input provides
a precise measurement even with the input common-mode voltage at
the supply voltage. The CMOS input structure does not draw any
input bias current, ensuring a minimum of measurement error.
The 0.1
resistor creates a voltage drop to the noninverting
input of the AD855x. The amplifier’s output is corrected until
this voltage appears at the inverting input. This creates a current
through R
1
, which in turn flows through R
2
. The Monitor Output
is given by:
Monitor Output
R
R
R
I
SENSE
L
=
×
×
2
1
(23)
Using the components shown in Figure 61, the Monitor Output
transfer function is 2.5V/A.
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