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How does the output (say, a sine wave) of a single supply inverting op amp compare with output of a single supply non-inverting, since it cannot go below 0v? Is the phase shifted? I'm trying to understand what happens, since I don't have access to an oscilloscope.

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An inverting or non-inverting opamp amplifier doesn't really matter that much, the main issue is the supply voltage. The supply voltage determines the output voltage range of the amplifier. Obviously when using a single-supply the output cannot go negative.

Let me illustrate that here:

enter image description here

The blue curve shows a pure sinewave which can have positive and negative values. You could get this when using an amplifier with a positive and a negative supply rail.

The red curve shows what happens when we try to make the same (blue) curve but in the situation where only a single (positive) supply is available. The sinewave becomes distorted, all its values below 0 V are "clipped".

To solve this we need to "lift" the voltage with a constant value. I used 1 V in this example. Adding a 1 V DC bias means the sinewave is "lifted" so that all its values become positive, see the magenta curve. For reference I re-plotted the red curve into the same figure as well.

If that 1 V DC offset is a problem it can often be removed using an AC coupling capacitor.

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If you only have a positive supply voltage, your ouput obviously can not go to the negative range. The phase is not shifted (at least not more or less than with a dual supply). If you have a positiv input voltage and your ouput would theoretically go to a negative voltage, then the ouput will just stay a 0 volts (and be in satuation). If, on the otherhand, you have negative differential input voltage, then your output will just be the positive voltage.

So, an inverting amplifier with single supply only makes sense, when the positive input is at a lower level than the negative input (or if you want the ouput to go into saturation, for some reason).

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    \$\begingroup\$ You can add an offset to the input signal, do your inversion, then remove the offset. That lets you use a single supply but still have positive and negative values in your output. \$\endgroup\$ – JRE Mar 11 at 9:08
  • \$\begingroup\$ That is true! But the positive input still must be on a lower potential than the negative input. But yeah, you could clamp the postive input to e.g. ground and then put a voltage swinging around -U_OS to the negative input. A capacitor at the output will remove the offset. But this obviously only works for non DC-inputs. \$\endgroup\$ – jusaca Mar 11 at 9:25

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