1
\$\begingroup\$

TI makes the ADS1118 which is an ADC with a programmable gain amplifier (PGA). In one of their application examples (page 4), they indicate that small differential input voltages (1-2mV) must have a common mode bias. This seems unnecessary since common mode voltage will be rejected by the differential amplifier. Why is a common-mode bias required? Will this work if the common-mode bias is removed?

enter image description here

\$\endgroup\$

closed as unclear what you're asking by Andy aka, Voltage Spike, Sparky256, Chetan Bhargava, Bimpelrekkie Feb 11 '18 at 14:32

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    \$\begingroup\$ I never saw "1-2mV" mentioned on page 4 at all. Maybe you can point us in the right direction. Maybe I'm going blind? \$\endgroup\$ – Andy aka Feb 2 '18 at 18:53
  • 1
    \$\begingroup\$ I don’t see anything about that at all. Try copy and paste and embed into your question. \$\endgroup\$ – Andy aka Feb 2 '18 at 19:15
  • 1
    \$\begingroup\$ There is nothing on page 4 of the link in your question. I know what a TC symbol is. \$\endgroup\$ – Andy aka Feb 2 '18 at 19:23
  • 1
    \$\begingroup\$ Where does page 4 say or hint “they indicate that small differential input voltages (1-2mV) must have a common mode bias”. I can’t see it said nor even can I construe this as a hidden meaning. \$\endgroup\$ – Andy aka Feb 2 '18 at 19:36
  • 1
    \$\begingroup\$ Please highlight in para 3 where it mentions 1 or 2 mV signals require the resistors? \$\endgroup\$ – Andy aka Feb 2 '18 at 20:04
1
\$\begingroup\$

The inputs need to be biased at the 'midpoint', between the supply voltage (Vdd, 3.3V) and GND. The midpoint would be 1.165 volts. This places the signal into the middle of the output range of the Diff Amp. Otherwise, if the inputs to the DA were left floating, the output of the DA could ride along the top or bottom rail. The 1~2mv signal is what is being generated by the thermocouples and has nothing to do with biasing. (Refer to the picture attached.) Biasing the inputs of the DA allows the signal to reside at the midpoint, which allows it to swing either positive or negative. (Center waveform.) If the inputs were left floating, the output signal would ride along the top or bottom rail and would not be able to swing high or low properly. (Top and bottom waveforms.)

enter image description here

I hope that answers your question. Please ask and I will try to explain it in more detail if needed.

\$\endgroup\$
1
\$\begingroup\$

Generally when you're feeding a signal into a differential amplifier, you want the signal to be somewhere in the middle of the amplifier's range, as it's difficult to make the amplifier perform accurately when its inputs are very close to the rails. Also if your input is near the rails and you exceed either supply rail, then the amplifier will likely no longer be linear.

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.