Electronic – Current source improves linearity on collector BJT

A current source is the dual of a voltage source. An ideal voltage source has zero output impedance, so that the voltage doesn't drop under load. It shouldn't be shorted, because in theory there would flow an infinite current.
An ideal current source has infinite output impedance. This means that the load's impedance is negligible and won't influence the current flowing. Like voltage sources shouldn't be shorted, current sources shouldn't be left open. An open current source will still try to source the set current, and the theoretical current source will go to infinite voltage.

edit (following your comment)
Here you can read impedance as resistance. If the current source would have a limited resistance changes in load would change the current, because the total resistance would change. You don't want that. So if the current source's resistance is infinite the load can be ignored and the resistance always remains the same (infinite). Therefore the current will as well.

A practical current source may be constructed as follows:

One diode has the same voltage drop as the base-emitter junction, so the other diode sets the transistor's emitter to about 0.7V. A fixed voltage across a fixed resistor gives a fixed emitter current, which is about the same as the collector current if the transistor's \$H_{FE}\$ is high enough. (Strictly speaking this is a current sink rather than a current source, but the principle remains the same.)

Another current sink uses an opamp as control element:

The main thing you need to know about opamps in this configuration is that they will try to keep the voltage on both inputs equal. So suppose you set \$V_{SET}\$ to 1V, then the opamp will try to make the - input also 1V. It does so by inserting current into the transistor's base. This will cause a current through the load \$I_{LOAD}\$ which is (almost) equal to \$I_{SET}\$. And \$I_{SET}\$ is constant to get the 1V across \$R_{SET}\$, according to Ohm's Law:

\$ I_{SET} = \dfrac{V_{SET}}{R_{SET}} \$

Since \$V_{SET}\$ and \$R_{SET}\$ are constant, so will \$I_{SET}\$ be. QED.

The base emitter (diode) junction is forward biased just like the two diodes you added and so, in effect, you get 3 lots of variations in junction voltage (for a given current) with temperature. The answer is no, I'm afraid not. Here's how a typical diode alters its voltage, for a given current, against temperature AND, the same is true of the base-emitter junction: -

Three different operating currents produce surprisingly similar slopes that tell you that a diode's forward voltage drop largely reduces at 2mV for a one degC rise in temperature. Resistors don't do this of course and you also have to consider that a diode has got a very low dynamic resistance once biased at some arbitrary operating point of a few hundred micro-amps upwards. This dynamic resistance is lower than a typical emitter resistor: -

Look at the right hand portion of the diagram - I've drawn two horizontal lines at 10mA and 15mA with the corresponding forward voltage drops. The difference (deltas) allows you to calculate the dynamic resistance = 0.1 volts / 5mA = 20 ohms i.e. probably less than the emitter resistor you might choose BUT, it changes with current so you get more gain than you bargained for (gain harder to define) and high signal non-linearity (distortion).

Setting the operating point of the collector at about half the supply voltage is useful to be able to obtain the maximum swing of signal (in terms of Vp-p) at the collector i.e. one side of the output signal doesn't clip much earlier than the other. There are subtleties here but that's the basic rule to maximize output amplitude and no, neither does this affect temperature stability.

Either use negative feedback (with care) or use an emitter resistor to lower the gain of the common emitter amplifier.