AMR sensor circuit explanation - Page 1
Forum section: electronics / beginners

author: nForce
author_rank: Frequent Contributor
timestamp: October 17, 2018, 11:47:09 am
content: I don't understand this circuit in attachment. What does that integrator do in feedback?Can someone explain slow and in detail. I know what is an AMR sensor and Wheatstone bridge I just don't understand the outside circuitry.

author: station240
author_rank: Unknown
timestamp: October 17, 2018, 02:10:45 pm
content: There is an explanation of how this exact circuit works here.http://archives.sensorsmag.com/articles/0399/0399_18/index.htmThe integrator in the feedback path, is there to remove DC offsets caused by the circuit itself (especially the opamps), so the reading is relative to 0.I'm sure someone else can better explain the rest of the circuit, it's really late where I am.

author: nForce
author_rank: Frequent Contributor
timestamp: October 18, 2018, 04:09:29 pm
content: Thanks, but why is the integrator connected to the non-inverting input of the other op amp? Shouldn't be to the inverting, because we want negative feedback?Do you maybe know what does that not gate with the capacitor do?Thanks again.

author: Benta
author_rank: Super Contributor
timestamp: October 18, 2018, 04:20:11 pm
content: Quote from: nForce on October 18, 2018, 04:09:29 pmThe integrator has negative gain, which is why it has to go to the non-inverting input. Otherwise you'd have negative x negative = positive feedback.

author: nForce
author_rank: Frequent Contributor
timestamp: October 19, 2018, 10:59:44 am
content: Thanks Benta, do you know:Quote

author: David Hess
author_rank: Super Contributor
timestamp: October 20, 2018, 01:05:53 am
content: The integrator has a low pass response but placing it within the feedback loop creates a high pass response; the integrator operates as a servo amplifier continuously adjusting the input signal to produce a zero output. This configuration takes maximum advantage of low input bias current operational amplifiers to allow for the longest time constants limited only by the integration capacitor and resistor.This is commonly done with instrumentation amplifiers because it avoids a more complex high pass filtering network on the input which would likely compromise common mode performance and defeat the high input impedance of an instrumentation amplifier. Placing a high pass filter on the output would allow the DC coupled instrumentation amplifier saturate if a high input offset was present. Note that usually the feedback path from the integrator goes to the instrumentation amplifier's reference input instead of its positive input but the ultimate result is the same; they might have needed to do it that way to prevent the input stage itself of the instrumentation amplifier from saturating.The whole circuit is a chopper amplifier with a differential input. The 200Hz signal chops the transducer's differential signal which is then amplified while common mode noise is rejected by the instrumentation amplifier. Only the AC component matters so the integrator was added to prevent saturation of the instrumentation amplifier as described above allowing much higher gain to be used even in the presence of excessive DC offset. The demodulator when reconstructs the original signal.