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ECG

Other Parts Discussed in Thread: INA333, INA326, OPA333

Hi, 

I am not particularly clear on the use of the reference pin on the INA333.  For instance, in the attached TINA simulation of my ECG circuit using E.M Spinelli's circuit, it posits that the ref voltage is Vcc/2 and also in the Right-leg drive. However, as I'm using positive and negative supply voltages my reference is zero and I'm not clear on its use. Also, the results of my simulated circuit makes the output inverted and very low in mV.  Help would be greatly appreciated. 

Regards,

Jeffrey. Circuit functionality ECG.TSC

  • Hi Jeff,

    A few year back I did some work on an ECG presentation and on one slide discussed an instrumentation amplifier's REF pin function. The slide shows a different instrumentation amplifier (IA), the INA326, but it too is a 3-Op amp topology similar to the INA333. The REF function is the same for both of the IAs so the slide is applicable to the INA333 as well. Additionally, the ECG circuit I present, like yours, uses an op amp as an integrator between the IA output and its REF input.

    Here's the slide and the accompanying  explanation:

    The INA326 output level may be referenced to a voltage applied to the reference pin, pin 5. If 0V is applied to the non-inverting input, then the output will be referenced to zero volts and the swing can move upwards from 0V. If the reference pin is set to +2.5V, then the output can swing both above and below +2.5V, within the output swing bounds. This reference voltage is sometimes referred to as a pedestal voltage, because it raises the output up from ground (0V) and sets it to a specific dc level.

    The integrator shown in the schematic is referenced to +2.5V applied to the U2 op amp non-inverting input. At DC the integrator’s gain is very large and any deviation from +2.5V sensed at the inverting input – as the result of a common-mode DC voltage on the INA’s inputs - will result in a large DC voltage at the output. This DC voltage is then applied to the INA326 reference input in in such a manner that it drives (servos) the INA’s output back to +2.5V.

    The integrator gain falls off rapidly as the frequency increases. Thus, AC signals having a frequency above the integrator’s -3dB cutoff frequency have little effect on the reference voltage applied to the INA.

    The net result is a DC restorer circuit that compensates for a DC common-mode voltage, such as may be present with the electrodes. It also provides a high-pass transfer characteristic with a cutoff frequency that is a function of the integrator RC constant. This results in a circuit equivalent to a capacitive-coupled amplifier, but without any capacitors directly in the signal path. High quality, high capacitance capacitors can be large and costly and are avoided by applying this technique.


    Simulating the INA333 the circuit with an op amp integrator in the reference circuit has always proven to be difficult. I find with Vref in your circuit set to 0 V, that the circuit converges but the measured voltages in the circuit are about 0 V, as they should be. When Vref is made more positive it becomes difficult to achieve convergence. The INA333 and OPA333 have very complex models having hundreds of lines of model code.

    The TINA simulator software I use is the Industrial version, which has more options. One of which is an interactive mode. I start a normal dc simulation with Vref equal to 0 V. The simulator comes to a stable solution and converges after several tens of seconds. Then I put the simulator in interactive mode, click on the Vref supply symbol and change the reference voltage to the desired level. The circuit then readily finds solution and the correct node voltages are shown.

    You can see an example where I used this procedure and later set Vref to 2.0 V. The measured points in the circuit achieve the correct 2.0 V level established by Vref. 

    I hope this helps.

    Regards, Thomas

    Precision Analog Applications Engineering

  • Hi Thomas, 

    First of all, thank you for your reply. I really appreciate it

    I can really infer a great deal of understanding from your post. A lot of things didn't quite make sense while running the simulation but now, I get it. Thank you very much! 

    Regards, 

    Jeffrey

  • Hi Jeffery,

    Thank you for the feedback.I am glad the information was of use to you. I hope your ECG project progresses well.

    Regards, Thomas

    Precision Analog Applications Engineering