AD8657 Analog Devices, AD8657 Datasheet - Page 20

• Manufacturer Part Number: AD8657
• Description: Precision, Micropower 18 V CMOS RRIO Op Amp
• Manufacturer: Analog Devices

Specifications of AD8657

• Vcc-vee: 2.7V to 18V
• Isy Per Amplifier: 18µA
• Packages: SOP,CSP
• -3db Bandwidth: 200kHz
• Slew Rate: 0.07V/µs
• Vos: 350µV
• Ib: 5pA
• # Opamps Per Pkg: 2
• Input Noise (nv/rthz): 50nV/rtHz

AD8657
EMI REJECTION RATIO
Circuit performance is often adversely affected by high frequency
electromagnetic interference (EMI). In the event where signal
strength is low and transmission lines are long, an op amp must
accurately amplify the input signals. However, all op amp pins—
the noninverting input, inverting input, positive supply, negative
supply, and output pins—are susceptible to EMI signals. These
high frequency signals are coupled into an op amp by various
means such as conduction, near field radiation, or far field radi-
ation. For instance, wires and PCB traces can act as antennas and
pick up high frequency EMI signals.
Precision op amps, such as the AD8657, do not amplify EMI or
RF signals because of their relatively low bandwidth. However,
due to the nonlinearities of the input devices, op amps can rectify
these out-of-band signals. When these high frequency signals
are rectified, they appear as a dc offset at the output.
To describe the ability of the AD8657 to perform as intended in
the presence of an electromagnetic energy, the electromagnetic
interference rejection ratio (EMIRR) of the noninverting pin is
specified in Table 2, Table 3, and Table 4 of the Specifications
section. A mathematical method of measuring EMIRR is
defined as follows:
4 mA TO 20 mA PROCESS CONTROL CURRENT
LOOP TRANSMITTER
The 2-wire current transmitters are often used in distributed
control systems and process control applications to transmit
analog signals between sensors and process controllers. Figure 73
shows a 4 mA to 20 mA current loop transmitter.
The transmitter powers directly from the control loop power
supply, and the current in the loop carries signal from 4 mA to
20 mA. Thus, 4 mA establishes the baseline current budget within
which the circuit must operate. Using the AD8657 is an excellent
EMIRR = 20 log (V
140
120
100
80
60
40
20
10M
V
V
IN
SY
= 100mV
= 2.7V TO 18V
Figure 72. EMIRR vs. Frequency
100M
IN_PEAK
PEAK
FREQUENCY (Hz)
/ΔV
OS
)
1G
10G
Rev. A | Page 20 of 24
choice due to its low supply current of 33 μA per amplifier over
temperature and supply voltage. The current transmitter controls
the current flowing in the loop, where a zero-scale input signal
is represented by 4 mA of current and a full-scale input signal
is represented by 20 mA. The transmitter also floats from the
control loop power supply, V
receiver. The loop current is measured at the load resistor, R
at the receiver side.
With a zero-scale input, a current of V
R. This creates a current flowing through the sense resistor,
I
details):
With a full-scale input voltage, current flowing through R is
increased by the full-scale change in V
increase in the current flowing through the sense resistor.
Therefore
When R >> R
receiver side is almost equivalent to I
Figure 73 is designed for a full-scale input voltage of 5 V. At 0 V
of input, loop current is 3.5 mA, and at a full scale of 5 V, the
loop current is 21 mA. This allows software calibration to fine
tune the current loop to the 4 mA to 20 mA range.
The AD8657 and ADR125 both consume only 160 µA quiescent
current, making 3.34 mA current available to power additional
signal conditioning circuitry or to power a bridge circuit.
0V TO 5V
NOTES
1. R1 + R2 = R´.
SENSE
I
I
I
, determined by the following equation (see Figure 73 for
V
SENSE, MIN
SENSE, DELTA
SENSE, MAX
IN
R
R
200kΩ
68kΩ
NULL
1MΩ
SPAN
1%
2kΩ
1%
1%
1%
Figure 73. 4 mA to 20 mA Current Loop Transmitter
R1
R2
V
= (V
= I
REF
SENSE
= (Full-Scale Change in V
SENSE, MIN
10µF
R3
1.2kΩ
REF
, the current through the load resistor at the
C2
× R)/(R
0.1µF
1/2
AD8657
C3
390pF
+ I
C1
SENSE, DELTA
V
DD
ADR125
NULL
OUT
, while signal ground is in the
GND
× R
V
IN
SENSE
3.3kΩ
SENSE
REF
IN
R4
C4
0.1µF
/R
/R
IN
.
)
R
SPAN
× R)/(R
SENSE
NULL
100Ω
C5
10µF
1%
D1
Q1
. This creates an
flows through
20mA
SPAN
4mA
TO
× R
V
18V
R
100Ω
DD
SENSE
L
L
,
)