Schematic
I'll start with the schematic:
simulate this circuit – Schematic created using CircuitLab
We know that a BJT can operate over a fairly wide range of collector currents while it's \$\beta\$ remains fairly flat (stable.) But temperature definitely impacts \$\beta\$ and so does part-to-part variation as well as drifting over time, too. You'd really like to have a design that is fairly immune to variations of \$\beta\$ in a specific BJT. \$V_{_\text{BE}}\$ is also highly temperature-dependent. So temperature is something to worry over. I'm mostly going to gloss over the details. But keep in mind that I'm still thinking about it when designing this even for simple educational needs.
(There's a process to figure all this out called sensitivity analysis so that you can quantify just how immune a circuit is due to some quantified impact of any concern you have. But we'll avoid all that for now.)
\$I_{_\text{Q}}\$
We can get a voltage gain, \$A=5\$, around most any reasonable quiescent current. But together with \$V_{_\text{CC}}\$, this choice directly impacts the output impedance. Low \$I_{_\text{Q}}\$ means higher output impedance. And usually, we seek lower output impedance.
Also, BJTs are designed to work over maybe 3 or perhaps 4 orders of magnitude current (their \$\beta\$ is relatively flat over a range like this, but outside of it may change much more drastically -- plus there are other odd behaviors that intrude and you may wish to avoid.) You can usually tell what collector currents a BJT 'likes' by skimming its datasheet. But for most small signal BJTs it's better to stay in this range: \$10\:\mu\text{A}\le I_{_\text{Q}}\le 10\:\text{mA}\$.
MIT has a datasheet for the 2N2222A. Find this:
The fact they bother to provide both a min and max there tells me this is a good number. So values near that will be good. They don't say much for anything below about \$100\:\mu\text{A}\$, so it may be good to avoid such low values. And the higher values are like more for switching, not analog amplification. This package just could not take the heat of those much larger collector currents, otherwise.
So I'm setting \$I_{_\text{Q}}=2\:\text{mA}\$.
Headroom Voltages
We will want to set the quiescent \$V_{_\text{E}}\$ to something more than \$500\:\text{mV}\$ to mitigate temperature variations on the operating point. More is better, less is worse. I won't go through the quantitative calcs here. But suffice it that you want some distance here.
And anything you reserve there has to be \$A=5\$ times larger for the voltage drop across the collector resistor.
More, you have to also account for the maximum allowable peak (not peak-to-peak, but peak) output voltage swing, \$V_{_\text{PKout}}\$. And you also need to reserve out something for the absolute minimum allowable \$V_{_\text{CEmin}}\$ to keep the BJT out of saturation (which means distortion or clipping.) That must be at least \$1\:\text{V}\$ but here again more is better.
Working through the details you should start with something like this:
$$V_{_\text{CC}}-I_{_\text{Q}}\cdot R_{_\text{C}}-I_{_\text{Q}}\cdot R_{_\text{E}}-V_{_\text{CEmin}}-V_{_\text{PKout}}-\frac1{A}V_{_\text{PKout}}=0\:\text{V}$$
Now, in writing that I assumed the current the collector resistor and the emitter resistor are identical. They aren't. But \$\beta\$ is usually high enough that we don't care about the difference. It's a waste of time worrying over it.
Also, I subtracted out \$V_{_\text{PKout}}\$ once (that's for when the output signal drives the collector voltage downwards towards the emitter on the down-going part of the cycle) and then \$\frac1{A}V_{_\text{PKout}}\$ again because of the fact that when the collector is moving down the emitter is moving up. And the amount it moves up will be \$\frac1{A}\$ of how far the collector voltage moves down. (They are pinching towards each other.)
Solving the above for \$V_{_\text{PKout}}\$ (recognizing that \$V_{_\text{E}}=I_{_\text{Q}}\cdot R_{_\text{E}}\$ and that \$A\cdot V_{_\text{E}}=I_{_\text{Q}}\cdot R_{_\text{C}}\$):
$$V_{_\text{PKout}}=\frac{A}{A+1}\left(V_{_\text{CC}}-V_{_\text{CEmin}}\right)-A\cdot V_{_\text{E}}$$
For a voltage gain of \$A=5\$ and \$I_{_\text{Q}}=2\:\text{mA}\$, and if we squeeze things really tight so that \$V_{_\text{CEmin}}=1\:\text{V}\$ and allow for \$V_{_\text{E}}=1\:\text{V}\$, then \$V_{_\text{PKout}}=2.5\:\text{V}\$. So that's \$5\:\text{V}\$ peak-to-peak and the input then should not exceed \$V_{_\text{PKin}}=500\:\text{mV}\$.
Resistors
This is pretty easy now. The first pair is just \$R_{_\text{E}}=\frac{1\:\text{V}}{2\:\text{mA}}=500\:\Omega\$ and \$R_{_\text{C}}=A\cdot R_{_\text{E}}=2.5\:\text{k}\Omega\$.
(At \$2\:\text{mA}\$ the dynamic emitter resistance, which is a slope and not a real resistor of any kind, would be \$13\:\Omega\$ at room temp. Comparing this to \$R_{_\text{E}}=500\:\Omega\$ means it's not a big concern for our hoped-for voltage gain but it may slightly affect the precise value. That's not usually an issue.)
Now all we need to do is bias the darned thing.
The biasing pair of resistors need to have about 10% of the quiescent current in them. Since we usually can expect the base current of the BJT to be less than 1% of the quiescent current, this means that the base current won't seriously impact the biasing point.
So let's calculate them. Note that \$R_1\$ carries our quiescent 10% biasing current plus the base current. So 11% of \$I_{_\text{Q}}\$. But \$R_2\$ only needs to carry 10% of \$I_{_\text{Q}}\$, since the base has by then subtracted its bit.
We also know that the base voltage will be about \$700\:\text{mV}\$ above \$V_{_\text{E}}=1\:\text{V}\$, so \$V_{_\text{B}}=1.7\:\text{V}\$.
Thus: \$R_1=\frac{V_{_\text{CC}}-V_{_\text{B}}}{0.11\,\cdot\, I_{_\text{Q}}}\approx 37.73\:\text{k}\Omega\$ and \$R_2=\frac{V_{_\text{B}}}{0.10\,\cdot\, I_{_\text{Q}}}= 8.5\:\text{k}\Omega\$.
Obviously, you need to select values you can get for all this and most of these aren't directly available. So let's to a quick hack.
Let's set \$R_{_\text{E}}=560\:\Omega\$ and \$R_{_\text{C}}=2.7\:\text{k}\Omega\$. We bump them both up which will slightly reduce \$I_{_\text{Q}}=\frac{1\:\text{V}}{560\:\Omega}=1.786\:\text{mA}\$ and will slightly lower our \$A=\frac{2.7\:\text{k}\Omega}{560\:\Omega}=4.8\$ gain.
Then \$R_1\approx 42\:\text{k}\Omega\$ and \$R_2\approx 9.5\:\text{k}\Omega\$. That looks like a candidate for increasing, as well, so that \$R_1= 39\:\text{k}\Omega\$ and \$R_2= 9.1\:\text{k}\Omega\$, which are available. This will slightly muss with our quiescent point again. But not a lot. So let's see:
I've included the operating point in text on the schematic. You can see that the quiecent current is very close to the \$2\:\text{mA}\$ we designed for. And the gain does appear to be about \$4.8\$. And no clipping or any visible distortion there.
Must be an accident. :)
One can also perform runs at higher and lower temperatures to see just how much things change. I did that. But it wasn't much -- a few tens of microamp changes in the quiescent current. So I didn't bother posting all that up.
Known quantities:
