I had to transform a 10Vpp (without DC level) signal into a 5Vpp with a virtual ground at 2.5V DC. I've come with this simple circuit that works both in simulation and real life. But I would like to know if there is something that I've missed in this design:

The op amp is configured as non-inverted with gain. R8 appears disconnected in the capture, but it is connected to the negative input.

Related with this:

What happens if I insert a 10Vpp (±5V) to an op amp connected to 5V and ground (as power source) using an AC coupling capacitor? Will the negative part of the signal damage the amplifier?

 

I don't think you need the op amp. You should be able to do it with a capacitor and resistors.

 

I don't think you need the op amp. You should be able to do it with a capacitor and resistors.

Be aware that the amplifier's output is what I need to adapt. That op amp is the end of an amplification circuit whose signal is 10Vpp (±5V). The next stage is a CMOS counter with 5V power source. So I thought it would be better to adapt the signal using R14, R15, C12 and D1.

 

With a capacitor, the performance will depend on the input frequency.
If you want to operate down to DC, I think you can do it with resistors and diode clamp and eliminate the capacitor and op amp.

Edit: I reread you comments. Keep the op amp. All you need to do is attenuate the signal and level shift. No capacitor is needed.
What is confusing is your drawing. Turn your circuit around, op amp inputs on the left, outputs on the right.

 

Remove the cap and and R15; then place another diode from the junction of R14 and D1 to +5V with the cathode at +5V. D1 limits the lower swing to -.6V and D2 limits the swing to 5.6V.

 

Here is what you are looking for:

 

This will also work, as long as your op amp output goes all the way to +5V (rail-to-rail output). Otherwise, the logic '1' may be too low. Mr Chip's circuit will have the same problem, only worse.

 

OP did say signal is 5V peak-to-peak.
If you want a reliable circuit that works for all voltage ranges, use a comparator.

 

OP did say signal is 5V peak-to-peak.
If you want a reliable circuit that works for all voltage ranges, use a comparator.

Ten volts p-p. Typo, of course.

 

What happens if I insert a 10Vpp (±5V) to an op amp connected to 5V and ground (as power source) using an AC coupling capacitor? Will the negative part of the signal damage the amplifier?

It could. There are ways to protect the input.

 

Ten volts p-p. Typo, of course.

Thanks for the correction.