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I'm trying to make a impedance meter using STM32 and I want to ask that question to make sure it work ( at least the theory).

So, Can opamps work with signal that may go higher or lower than supply voltage (single positive supply) as long as it's output remains in Vsupply region ?

Sorry for my bad English and thanks !

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    \$\begingroup\$ That's in the datasheet of the opamp. \$\endgroup\$ Dec 9 '17 at 15:51
  • \$\begingroup\$ There are lots of ways to make an opamp. How much of an excursion beyond the supply rail are you dealing with? \$\endgroup\$ Dec 9 '17 at 16:44
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    \$\begingroup\$ Usually, none of the op-amp pins can be exposed to voltages outside of the supply range. Sometimes the voltages cannot even be close to the supply limits. But you may be able to work around this using voltage dividers. \$\endgroup\$
    – mkeith
    Dec 9 '17 at 17:52
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Some single supply op-amps have a common-mode range that extends slightly below the negative supply rail.

At least one TI op-amp has a built-in charge pump and the common mode range extends 200mV below the negative supply rail and 200mV above the positive supply rail. So do some R-R input op-amps (at the expense of a "kink" in the input response- they're really two front ends that have to pass off at some input common mode voltage).

But generally for the most selection and best price/performance in op-amps you wan to keep both inputs within the supply rails, and often well within the supply rails (no closer than a volt or two for some op-amps).

Input common mode range (or sometimes Input Voltage Range) is the data sheet spec you need to look at.

For example, the OP-07 has a specified range with +/-15V supplies that is +/-13V.

The LM324 has a range from 0V to 2V below the positive supply (depending a bit on temperature range etc.) and is extremely cheap.

The MCP6001 works with inputs as much as 300mV outside the power supply rails (and is fairly cheap).

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  • \$\begingroup\$ Sorry for the late response. My net was down for a few weeks. \$\endgroup\$
    – Long Pham
    Jan 7 '18 at 3:31
  • \$\begingroup\$ I'm curious about this because I saw an Instructable about making Arduino inverting converter. There is a shematic in which an inverting amplifier used as feedback path to the arduino. \$\endgroup\$
    – Long Pham
    Jan 7 '18 at 3:41
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The answer is "no" and "yes".

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) Voltage follower. (b) Voltage divider and follower. (c) Inverting amplifier.

  • (a) Since the IN signal is connected directly to the op-amp non-inverting input it is not allowed to exceed the input voltage limits. Usually that means keeping within the supply rail voltages.
  • (b) By adding a divider in front the amplifier can now handle very large voltages - eleven times higher in this case.
  • (c) In the inverting mode the signal can be attenuated by increasing the input resistor. Gain in this case is -0.1 so the input could go to ten times the supply rail voltage for a rail-to-rail op-amp.

Note that (a) doesn't load the input signal due to the very high input impedance of the op-amp. (b) and (c) do load the IN signal. Whether this is important to you or not depends on the preceding circuit.

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  • \$\begingroup\$ Supposed that Vee = 0V and it is a unity gain inverting amplifier, will it be able to reverse a negative input signal ? \$\endgroup\$
    – Long Pham
    Jan 7 '18 at 3:50
  • \$\begingroup\$ When commenting on my answer please use the node names given on my schematic or "inverting input" / "non-inverting input". I have no idea what you mean by "Vee". \$\endgroup\$
    – Transistor
    Jan 7 '18 at 10:18
  • \$\begingroup\$ Oh, I mean op amp negative supply terminal \$\endgroup\$
    – Long Pham
    Jan 7 '18 at 14:51
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    \$\begingroup\$ Theoretically, yes. In practice it depend on how well the opamp works as the inputs get close to negative supply. \$\endgroup\$
    – Transistor
    Jan 7 '18 at 14:55

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