Rail-to-rail opamp output swing with Vdd=5V, Vss=-5V, inverting amp configuration

Hello,
I wanna design a rail-to-rail input and output opamp and I wanna be sure I understand what it really does, so I will give a few examples. First of all, from what I understood, a regular opamp can't really reach the maximum swing, but a rail to rail opamp can do, even surpass it, but for simplicity let's say it is exactly the supply voltage.

So for Vdd = 5 V and for a Vss = -5V and I have my opamp as an inverting amplifier like in the photo I've attached, if my input signal is 1V and R2 =5k, R1=1k, then my output will be exactly 5 V, right? For an usual opamp it could be something like... 4.9 or 4.95... Is this correct?

Thank you!

You have to be careful using statements like exactly. Output stages in RR/IO opamps have strict limitations, how close they can get to the rail is limited by the load current, you have to be aware that the load current includes the current associated with the feedback network as well. There are normally a pair of graphs in the data sheets which show Vsupply-Vout versus load current for both supplies, since there are differences in PNP/NPN or P and N channel characteristics (this is sometimes  partially overcome at the expense of greater complexity in the output stage) The load can also influence the open loop gain and slew rate. You also have to be very aware of the input stage topology, of which there are a number for RR Input devices. Some rely on charge pumps, which generate some output noise, others on dual differential input stages, which result in changes in offset voltages and currents somewhere in the  common mode input operating characteristics. In inverting configurations this is not normally an issue.Some allow input voltages on the input greater than the rail, but you will get non linear performance if your gain tries to take the output beyond the rail at the output, which obviously can't happen.These devices continue to improve, cheers,Richard

Well, thank you for the thorough explanation, can you tell me how should I choose my currents through my circuit? Especially on my output "branch"? 
Thank you!

Look at the data sheet for the device you are using, there will be a graph showing how close the output will go to the rail versus output current. Determine what is an acceptable value, and use Ohms law to determine the minimum resistance you can have on the output for the given current at 5V. This should include the feedback resistor current plus the load current. You will find in all probability that the  resistor values are quite large.Cheers,Richard

One more comment, you mean -5V if the input is 1V, I missed that, you have an inverting amplifier,cheers,Richard

Yes, sorry, it seems like I've missed that detail

One last thing, can someone explain to me what is the benefit of a rail - to - rail op amp? 

With the advent of 3.3 and 5V only systems, the limited output swing of conventional op-amps presents quite a challenge, since the dynamic range is severely limited, you may have less than a volt of useful swing. Consider buffering an ADC input, where you need near rail to rail swing on the ADC input. You can generate higher voltage rails by various means for supplying the op-amps, but this adds to the complexity, and  you have to  prevent conduction of the protection diodes on the the lower voltage parts input pins. As I have already pointed out, there are quite a few additional issues that must be considered using RR/IO devices, I only touched on the subject, there is quite a bit on the web, most IC manufacturers have notes on the subject, see below for a few.See this note, and this one
Cheers,Richard

Hi again,
I tried designing a rail to rail op amp using cadence virtuoso and I think I succeeded. But I have a little problem for some simulations, for example:
If I have a 1.65 DC offset and a 50uV sinusoid and my output is between 3.2 and 0.1.. this means that I have an open loop gain for like >50 dB for sure. But if I try making a simulation it says it's 4dB and I followed a simulation tutorial like this one: https://youtu.be/L8wJhENPZNc
Does anyone know another simulation for the open loop gain?Thank you!

made me more informative with your work,i am feeling delighted with your work,best wishes from OREO TV

The youtube video shown is for a voltage follower, so expect the gain to be around 0dB.What model do you have for your op amp. A simple one will have a single pole, usually represented internally by an RC network. Additionally,  the input and output stages are not modelled, so there is no need to bias the input stages correctly, or load the output.A better model will include things like common mode DC characteristics, and the saturation voltage on the output stage, which means that you have to bias the inputs correctly, and have a reasonably high output resistive load on the output. In other words, you have to look at the data sheets and do a design as you would in the real world.I suggest that you look up"open loop gain op amp spice", or something of the sort, and proceed from there.Cheers,Richard