库卡 KPS-600 伺服系统

4线模式下，激励电流必须设置为250 µA，3线模式下设置为100 µA。 对于4线模式，假设RTD值为3.92 kΩ。 来自AIN0的激励电流流经RREF + RRTD + RRETURN = 3.92 kΩ + 3.92 kΩ + 3 kΩ = 10.84 kΩ。 因此，AIN0处的电压等于250 µA × 10.84 kΩ = 2.71 V。AD7124-4指定激励电流输出端的输出顺从电压为AVDD − 0.35 V，即3.3 V – 0.35 V = 2.95 V。 因为2.95 V > 2.71 V，所以即使对于大RTD电阻，250 µA激励电流也能正常工作。

The excitation current must be set to 250 µA for 4-wire mode and 100 µA for 3-wire mode. For 4-wire mode, assume an RTD value of 3.92 kΩ. The excitation current from AIN0 flows through RREF + RRTD + RRETURN = 3.92 kΩ + 3.92 kΩ = 10.84 kΩ. Ω + 3 kΩ = 10.84 kΩ. Therefore, the voltage at AIN0 is equal to 250 µA × 10.84 kΩ = 2.71 V. The AD7124-4 specifies the output compliance voltage at the excitation current output as AVDD - 0.35 V, i.e., 3.3 V - 0.35 V = 2.95 V. Since 2.95 V > 2.71 V, the 250 µA excitation current works even for large RTD resistors.
 

 在3线模式下，来自AIN4的引脚补偿激励电流也会流经3 kΩ返回电阻，在AIN0处产生一个附加电压：250 μA × 3 kΩ = 0.75 V。因此，AIN0处的总电压等于2.71 V + 0.75 V = 3.46 V，这违反了裕量要求。 所以，在3线模式下，各激励电流必须降至100 μA以提供足够的裕量。 

In 3-wire mode, the pin-compensated excitation current from AIN4 also flows through the 3 kΩ return resistor, generating an additional voltage at AIN0: 250 μA × 3 kΩ = 0.75 V. Therefore, the total voltage at AIN0 equals 2.71 V + 0.75 V = 3.46 V, which violates the margin requirement. Therefore, in 3-wire mode, the individual excitation currents must be reduced to 100 μA to provide sufficient margin.

考虑30 V电压连接在AIN+和AIN-之间的情况。从AIN+朝里看，30V电压看到R1(3kΩ)，之后是内部ESD保护二极管，再后面是从AIN3朝外看到的3kΩ电阻与从AIN4朝外看到的3 kΩ电阻并联。忽略内部ESD保护二极管，AIN+与AIN-之间的总电阻为3 kΩ + 3 kΩ|l3 kΩ=4.5 kΩ。因此，流经AD7124-4的电流限值为30V÷4.5 kΩ=6.7 mA。

Consider the case where the 30 V voltage is connected between AIN+ and AIN-. Looking in from AIN+, the 30 V voltage sees R1 (3 kΩ), followed by the internal ESD protection diode, and then followed by the 3 kΩ resistor seen from AIN3 facing out in parallel with the 3 kΩ resistor seen from AIN4 facing out. Ignoring the internal ESD protection diode, the total resistance between AIN+ and AIN- is 3 kΩ + 3 kΩ|l3 kΩ = 4.5 kΩ. Therefore, the current limit flowing through the AD7124-4 is 30V ÷ 4.5 kΩ = 6.7 mA.