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A Query about servo integrators on the "Ref" pin of the INA333 and some other doubts.

Yash Chitalia
Prodigy 30 points
Other Parts Discussed in Thread: INA333

Hi,

I was looking at a long chain of threads talking about how the reference voltage of the INA333 is set by a voltage divider followed by a buffer and an Integrator circuit.

However, I really haven't understood the importance of the circuit, and whether it is mandatory/recommended to use it. Could someone explain this concept, direct me to a link where this factor has been explained?

Further, what is the recommended input resistance value at each input of the In-Amp for an ECG circuit? (I have attached an image of a circuit used in an older thread, and have highlighted my queries with red circles)

 

Also, I had certain doubts about a circuit, that was posted later on the same link. I have attached the image of that circuit below for reference.

1. The First stage in this circuit has an IN-AMP which has been referenced to ground, and furthermore, is single supply. So if I understand the circuit correctly, the output of the INAMP will be half wave rectified?

2. Also, now let us assume that the reference pin of the INAMP was given VCC/2 thereby shifting the entire ECG signal by +VCC/2 volts( i.e biasing the ECG signal at half supply). Will the filter stages that follow be referenced to VCC/2 volts in that case? I mean (referring to the figure below) will the GND points in circle 2, circle3 and circle 4 be replaced by VCC/2 points? Or will the Circle 4 GND point remain where it is (since it is the final stage of the circuit, and we want our final signal to be referenced to GND)?

 

Thanks a lot,

Yash

  • 0
    TI__Expert 7185 points

    The purpose of the integrator is to act as a high pass filter, driving Vout to the DC value of VREF, thereby removing all of the DC offsets incurred at the front end of the INA (due to VOS, IB*R, CMRR, and especially electrode offsets).

    The resistors that are shown in this circuit are not necessary for proper operation or use of the INA, they are simply there for suggested protection.  According to AAMI EC11: 1991 section 3.2.9 and EC13: 1992 section 3.2.9.2 the filter resistance added to the inputs of the instrumentation gain amplifier must be larger than 100k ohms to keep input currents that could be generated during a device fault or EOS condition to a safe level for the patient.  These requirements are specifically for clinical ECG systems but there may be more relaxed requirements depending on the end application of which I am not aware.

    With regard to the circuit you have referred to in the link--you are correct that if the INA is not referenced to a mid supply DC voltage there could be clipping at the output of the INA for signals that go below ground.  If each of the successive filter stages is DC biased at the non-inverting input at VCC/2, your overall signal will swing about this voltage.  Keep in mind that even if all of the circled grounds that you have shown were referenced to a DC voltage, these will have no effect on the overall DC bias level because the enitre path for the dynamic signal is AC and the DC voltage is therefore seen as an AC ground.  To summarize, in the circuit that you referenced in the link circle 1 (the reference voltage) is really the only one that needs to be reference to VCC/2.

    I hope this helps.

    Matt

     

  • 0 in reply to Matthew Hann
    Prodigy 10 points

    Hi Matt and TI,

    I understand that integrator is shown like a HPF on signal path. Since integrator works that way a 1-pole HPF the Boden are the same in both the passive RC HPF and integrator. So why to use integrator instead of simpler and cheaper passive RC HPF? And how to simulate the difference in TINA with INA12x etc. inamp?

    Edit: I think I've got it: It is recommend to drive low impedance to ref-pin with op amp buffer, therefore no cost to change it to integrator.