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I'm trying to make a PCB board to measure Electrocardiogram (ECG) signal. I have some doubts about the DC offset of the 2 input signals (\$V_{IN-}\$, \$V_{IN+}\$) with reference to analog ground. I would appreciate it if someone can help.

TI's solution

I'm referencing the circuit above from an application report of TI. It uses 5V single supply. The small differential signal is first amplified with an instrumentation amplifier, with an integrator (A4) feeding to its output reference terminal to cancel out low frequency artifacts like different DC offset on electrodes. Then the signal is amplified again with a higher gain by an inverting amplifier (A3). At the bottom left, it is the driven right leg circuit. The common-mode voltage is picked up (A1) and amplified (A2). And then injected into the right leg using an inverting amplifier (I haven't fully understood the driven right leg circuit, which is probably the root of my problem).

The problem is that the 2 differential signals (\$V_{IN-}\$, \$V_{IN+}\$) need to be above 0.98V (according to the datasheet of INA118, which is the chip I use) with reference to the analog circuit common to ensure that the instrumentation amp operates in the linear region. I'm not sure if this is always satisfied. Should I add a DC offset to the 2 differential signals to make the instrumentation amplifier work properly? However, this would change the common-mode voltage and I don't know if the driven right leg circuit would still work as designed.

Below is part of the schematic for my simulation. The 2 differential signals (VG1 and VG2) have the the same ground as the circuit. In this case, I have to set the DC voltage of 2 sources to 1/2 VCC to get reasonable output. Am I wrong to set the source in this way?

my schematic

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The Vref = 2.5 compares to output of INA with same to drive REF to balance the input and output in support with RLD feedback to null the body galvanic electrode offsets.

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  • \$\begingroup\$ Is it because A2 forces a DC offset of (2.5-\$V_{CM}\$) throughout the body so the Vref=2.5 for the input? I ignored the A2 at first, so I thought the input was floating. \$\endgroup\$
    – Jack Black
    Commented Jul 18, 2022 at 10:11
  • \$\begingroup\$ A2 provides the common mode negative feedback with integration gain to null error for DC while Vref = 2.5 provides the input offset to input and output to bias the common mode in the difference amp. using REF (see internal schematic on datasheet) Both are necessary for a single supply. \$\endgroup\$ Commented Jul 18, 2022 at 14:22

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