Friday 7 July 2017

Why has this configuration chosen for feedback?


I'm a newbie in electronics and Trying to understand and build this circuit. would somebody tell me the rule of R1 & C2 in this circuit and their magnitude effect on the controlling of process.why didnt feedback start just after the output of opamp?How could i find the transfer function between the voltage output of Hall effect sensor and the voltage output of MOSFET drain?


The schematic is the circuit of a magnetic levitating tool at:


http://www.bis0uhr.de/index.htm?http://www.bis0uhr.de/projekte/schwebekugel/english.php%99https://www.youtube.com/


Thanks so much


enter image description here



Answer



That inner loop ensures that there is negative feedback at high frequencies. Normally the negative feedback loop is formed between the magnetic field in the inductor and the hall-effect sensor.


Suppose the current through the inductor goes up. This increases the magnetic field. The hall effect sensor senses this. It outputs a positive voltage. This tends to turn MOSFET OFF, reducing the current, and thus reducing the magnetic field. It will settle to some equilibrium point.


Note that this feedback mechanism can only occur so fast. At very high frequencies the loop can't respond quickly enough and this can lead to instability.



So you add a small capacitor and series resistor into the inner loop which can respond quickly and ensure that the op-amp stays in a closed-loop configuration for high frequencies.


Whoever designed the circuit probably designed it guided by this intuition and didn't do calculations. They probably saw some oscillation and realized the problem, intuitively, and tried 100n and 100k, with a time-constant below the frequency of the observed oscillation. This is the kind of thing you might tailor on your breadboard until the oscillation is quelled, not derive an equation for.


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