For electronic hardware personnel, the following conditions apply: the more accurate the data measurement, the better the effect. If you want good results, then the first thing to do is to improve the tool. If you want to measure data more accurately, you need to perform stable and stable data measurement first, and you must need an accurate power supply.
For example, to capture the analog quantity of a resistive pressure sensor, if you want to measure accurately, you need to provide a reliable power supply. It can be a voltage source that detects changes in current or a current source that detects changes in voltage. Compared with the latter, it is of course easier to provide a constant current source, and then detect the voltage change value caused by the sensor resistance change.
Next, I will introduce the design analysis of high-precision AC constant current source.
AC constant current source
For this circuit, where R5 = R6, R1 = R2, the input terminal is a 1V reference voltage input.
It can be seen from the figure that V1 = (1VREF + V5)/2. From the virtual short circuit of the operational amplifier, V2 = V1, and the current flowing through R1 is equal to V2/R1.
According to the virtual isolation of the operational amplifier, the current flowing into the reverse input terminal of op1 is 0, so the current flowing through R1 is equal to the current flowing through R2, where (V3-V2)/R2 = V2 / R1. And R1 = R2, so V3 = 2V2 = 2V1 = 1VREF + V5.
See again op2 of the operational amplifier, V5 = V4. Therefore, the current flowing through R4 is (V3-V4)/ R4 = (1VREF + V5-V4)/ R4, and V4 = V5, so the current flowing through R4 is a constant 1VREF/R4.
Knowing from the virtual terminal of the op amp, the current flowing into the positive input terminal of op2 is about 0, so the current flowing through R4 is equal to the output current I0, that is, the constant current I0 = 1VREF/R4.