AD8546/AD8548

Data Sheet

Rev. C | Page 20 of 24

Note that 100 kΩ resistors are used in series with the input of

the op amp. If smaller resistor values are used, the supply current

of the system increases much more. For more information about

using op amps as comparators, see the AN-849 Application Note,

Using Op Amps as Comparators.

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.

Op amps, such as the AD8546 and AD8548, 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 AD8546/AD8548 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:

EMIRR = 20 log (VIN_PEAK/ΔVOS)

20

40

60

80

100

120

140

10M

100M

1G

10G

FREQUENCY (Hz)

VIN = 100mVPEAK

VSY = 2.7V TO 18V

Figure 70. EMIRR vs. Frequency

4 mA TO 20 mA PROCESS CONTROL CURRENT

LOOP TRANSMITTER

A 2-wire current transmitter is often used in distributed control

systems and process control applications to transmit analog signals

between sensors and process controllers. Figure 71 shows a 4 mA

to 20 mA current loop transmitter.

RL

100Ω

VDD

18V

C2

10µF

C3

0.1µF

C1

390pF

C4

0.1µF

R4

3.3kΩ

Q1

D1

4mA

TO

20mA

R3

1.2kΩ

RNULL

1MΩ

1%

VREF

RSPAN

200kΩ

1%

VIN

0V TO 5V

R1

68kΩ

1%

R2

2kΩ

1%

NOTES

1. R1 + R2 = R´.

1/2

AD8546

C5

10µF

RSENSE

100Ω

1%

VOUT

GND

ADR125

VIN

Figure 71. 4 mA to 20 mA Current Loop Transmitter

The transmitter is powered 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.

The AD8546 is an excellent 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, VDD, whereas signal

ground is in the receiver. The loop current is measured at the load

resistor, RL, at the receiver side.

With a zero-scale input, a current of VREF/RNULL flows through

R. This creates a current, ISENSE, that flows through the sense

resistor, as determined by the following equation:

ISENSE,MIN = (VREF × R)/(RNULL × RSENSE)

With a full-scale input voltage, current flowing through R is

increased by the full-scale change in VIN/RSPAN. This creates an

increase in the current flowing through the sense resistor.

ISENSE,DELTA = (Full-Scale Change in VIN × R)/(RSPAN × RSENSE)

Therefore,

ISENSE,MAX = ISENSE,MIN + ISENSE,DELTA

When R >> RSENSE, the current through the load resistor at the

receiver side is almost equivalent to ISENSE.

Figure 71 shows a design for a full-scale input voltage of 5 V. At

0 V of input, the loop current is 3.5 mA, and at a full-scale input

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.

Together, the AD8546 and the ADR125 consume quiescent

current of only 160 µA, making 3.34 mA current available to

power additional signal conditioning circuitry or to power a

bridge circuit.