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AD8657ACPZ-R7 数据表(PDF) 20 Page - Analog Devices |
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AD8657ACPZ-R7 数据表(HTML) 20 Page - Analog Devices |
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20 / 24 page ![]() AD8657 Rev. A | Page 20 of 24 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: 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 72. EMIRR vs. Frequency 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 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, while 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 flowing through the sense resistor, ISENSE, determined by the following equation (see Figure 73 for details): 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 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. RL 100 Ω VDD 18V C2 10µF C3 0.1µF C1 390pF C4 0.1µF R4 3.3k Ω Q1 D1 4mA TO 20mA R3 1.2k Ω RNULL 1M Ω 1% VREF RSPAN 200k Ω 1% VIN 0V TO 5V R1 68k Ω 1% R2 2k Ω 1% NOTES 1. R1 + R2 = R´. 1/2 AD8657 C5 10µF RSENSE 100 Ω 1% VOUT GND ADR125 VIN Figure 73. 4 mA to 20 mA Current Loop Transmitter |
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