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ADE1202 数据表(PDF) 27 Page - Analog Devices |
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ADE1202 数据表(HTML) 27 Page - Analog Devices |
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27 / 42 page ![]() Data Sheet ADE1202 Rev. 0 | Page 27 of 42 The coefficients gi, where i = 0, 1, 2, … 15, are the coefficients of the generating polynomial defined by the CRC-16-CCITT algorithm as follows: G(x) = x16 + x12 + x5 + 1 (9) g0 = g5 = g12 = 1 (10) All other gi coefficients are equal to 0. FB(j) = aj − 1 XOR b15(j − 1) (11) b0(j) = FB(j) AND g0 (12) bi(j) = FB(j) AND gi XOR bi − 1(j − 1), i = 1, 2, 3, … 15 (13) Equation 11, Equation 12, and Equation 13 must be repeated for j = 1, 2, … 16. The value written into the SPI communication CRC contains Bit bi(16), where i = 0, 1, … 15. PROTECTING THE INTEGRITY OF CONFIGURATION REGISTERS Configuration registers are either user accessible registers (R/W registers listed in Table 17) or internal registers that are not user accessible. On power-up, the user accessible configuration registers can be written without restriction. When the registers are configured, write 0xADE1 to the LOCK register to send configuration information from the isolated side to the nonisolated side. This action also disables write access to the configuration registers from the SPI port to protect the integrity of the configuration. When the protection is enabled, read back the LOCK register to ensure that Bit 0 (LOCK) was set to 1. When the LOCK register is read, Bit 0 (LOCK) shows the protection status. If the LOCK bit is 0, the protection is disabled. If the LOCK bit is 1, the protection is enabled. The lock function does not affect the ADDR_RELOAD bit, the LOCK register, and the INT_STATUS register, which can all be written when LOCK = 1. To disable the register protection, write 0xADE0 to the LOCK register. To change any configuration registers, disable the protection, change the value of the register, and then reenable the protection. VERSION The REVID bits (Bits[8:5]) in the CTRL register identify the version of the IC. INSULATION WEAR OUT The lifetime of insulation caused by wear out is determined by the isolation thickness, material properties, and the voltage stress applied. It is important to verify that the product lifetime is adequate at the application working voltage. The working voltage supported by an isolator for wear out may not be the same as the working voltage supported for tracking. The working voltage applicable to tracking is specified in most standards. Testing and modeling show that the primary driver of long-term degradation is displacement current in the polyimide insulation causing incremental damage. The stress on the insulation can be broken down into broad categories, such as dc stress, which causes very little wear out because there is no displacement current, and an ac component time varying voltage stress, which causes wear out. The ratings in certification documents are typically based on 60 Hz sinusoidal stress because this value reflects isolation from the line voltage. However, many practical applications have combinations of 60 Hz ac and dc across the barrier, as shown in Equation 14. Because only the ac portion of the stress causes wear out, the equation can be rearranged to solve for the ac rms voltage, as shown in Equation 15. For insulation wear out with the polyimide materials used in the ADE1202, the ac rms voltage determines the product lifetime. 22 RMS AC RMS DC V V V = + (14) or 22 = − AC RMS RMS DC V VV (15) where: VRMS is the total rms working voltage. VAC RMS is the time varying portion of the working voltage. VDC is the dc offset of the working voltage. Calculation and Use of Parameters Example The following example frequently arises in power conversion applications. Assume that the line voltage on one side of the isolation is 240 V ac rms and a 400 V dc bus voltage is present on the other side of the isolation barrier. The isolator material is polyimide. To establish the critical voltages in determining the creepage, clearance, and lifetime of a device, see Figure 47 and the following equations. TIME VAC RMS VRMS VDC VPEAK Figure 47. Critical Voltage Example Calculate the working voltage across the barrier from Equation 16 with the following equations: 22 RMS AC RMS DC V V V = + (16) 22 240 400 RMS V = + (17) In this example, VRMS = 466 V. |
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