What about the design of high-precision AC constant current source?
For electronic hardware people, the following applies: the more precise the data measurement, the better. If you want good results, the first thing to do is improve your tools. 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 value of a resistive pressure sensor, a reliable power source is required if the measurement is to be accurate. It can be a voltage source that senses a change in current, or a current source that senses a change in voltage. Compared with the latter, it is easier of course to provide a constant current source and then detect the value of the voltage change caused by the change of the sensor resistance.
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.
V1 = (1VREF + V5)/2. From the virtual short circuit of the op-amp, V2 = V1, the current flowing through R1 is equal to V2/R1.
From the virtual isolation of the operational amplifier, the current flowing into the inverting input of op1 is zero, 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 op amp, V5 = V4. Therefore, the current through R4 is (V3-V4)/R4 = (1VREF + V5-V4)/R4, and V4 = V5, so the current through R4 is a constant 1VREF/R4.
Knowing from the virtual terminal of the operational amplifier, 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 is realized.
