This SPICE simulation shows the output voltage going below ground.
Is this correct. The op amp is single ended and I would of thought the waveform
would be clipped at ground and not show the negative swing. Is this correct ?

 

Maybe the presence of V3 is a bug, you are not using it,
remove it from schematic....?

Regards, Dana.

 

Is this correct ?

No. The reason your output is going below ground is that capacitor C2 has removed the DC component from the signal at the output of U1, leaving only the AC component.

 

Look directly at the opamp output and you will see that it doesn't go below ground.

Edit: Note that in your circuit R5 is in parallel with R6 so you could combine them into one resistor of the appropriate value.

 

@crutschow didn't notice that, thanks. The voltage divider sets the bias and the R6 is supposed to be a return current to ground. R6 through off the bias a bit. Looking the output of the op amp... the ac is riding on the bias so no negative swings. But after the capacitor that negative voltage (in real life) should be clipped at ground, correct?.

@OBWOB49 yes, you are correct but the waveform, if I looked at it on a scope would only go to ground. Maybe it's just a SPICE thing?

@ dana - I removed the extra source, but it didn't make a difference.

 

But after the capacitor that negative voltage (in real life) should be clipped at ground, correct?.

Not correct.
Why would it clip the negative voltage any more than it would clip the positive voltage.
Ground has no polarity.

The capacitor removes the dc bias at the output of the op amp so the capacitor output has a DC average of zero volts.
For that to happen, the output waveform must equally go above and below ground, as you observed.
It's definitely not a "SPICE thing" and you will see the exact same waveform with an oscilloscope in real life.

 

For some reason I was thinking the ground of the load resistor would cause the ac signal to clip the same as an op amp who's negative
supply terminal was grounded (and ac signal not biased).
But I think I understand how it works. The full sinewave is present since the op amp is biased. Running it through the cap just shifts the average around ground (0 volts). Thanks

 

But I think I understand how it works. The full sinewave is present since the op amp is biased. Running it through the cap just shifts the average around ground (0 volts).

Correct.