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Is it possible to make a high-impedance, precision measurement of a 10V DC signal (dynamic range requirement is only ±1V with a DC-1kHz BW) using amplifiers and an Sigma-Delta ADC supplied only by +3.3V (or +5V)?

I can't use an instrumentation amplifier without at least a +12V/-2V supply. A voltage divider would be a trade-off between input impedance and Johnson noise.

Is there another way to do this that I am missing? A way to subtract the 10V bias without needing higher voltages in my circuit?

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  • \$\begingroup\$ Measurement with what? Is this going into an ADC or an analog circuit? \$\endgroup\$ – Transistor Oct 19 '18 at 22:37
  • \$\begingroup\$ A voltage divider followed by an opamp buffer? What specifically do you need the high impedance for? \$\endgroup\$ – David Oct 19 '18 at 22:43
  • \$\begingroup\$ Measurement is with an ADC. There is a variable contact resistance in the system so I do not want my meter to draw any current. Presently I am using an INA129 with +/-15V supplies from a DC-DC converter. \$\endgroup\$ – Mike Oct 19 '18 at 22:50
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    \$\begingroup\$ I think that you might be looking at a null-meter kind of setup. i.e. you have a 10V known source and measure the difference. You only need the range of the expected maximum difference (±1V in your case). Or you can ever have a reference adjustable from 9 to 11V. This is the "classic" way of making precise measurements in metrology. \$\endgroup\$ – Undertalk Oct 20 '18 at 1:34
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    \$\begingroup\$ Watch out for the sampling surge currents. 10pf at 100,000 at 10 volts will appear as (10pF * 10v = 100picoCoulomb ) * 100,000Hz or 1e-10 Coul * 10^+5 Hz = 1e-5 amperes. That, 10uA, times 24Kohm, causes 0.24 volts error. \$\endgroup\$ – analogsystemsrf Oct 20 '18 at 3:38

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