Radio Frequency amplifiers quite often use common base circuits and I believe the main reason is because "miller capacitance" does not have the same detrimental effect.
On a normal common-emitter configuration, the input is fed to the base and the output is at the collector but, internally the capacitance between collector and base acts as negative feedback and can reduce gain.
To combat this, if the base is held at a fixed potential to ground (as per in its normally biased state), and this is supplemented by decent capacitance to ground, the miller capacitance is effectively shunted to ground. The down-side is that feeding an input to the emitter requires a stronger drive voltage because the emitter input impedance is lower.
It is lower because the emitter fed input has to also "handle" collector currents as well as the little bit of emitter-to-base current normally needed for amplification.
Consider also the differential amplifier: -
This is another circuit that uses the emitter as an input. On this occasion the input to the emitter comes from the emitter (outputting) of the other BJT. In fact, both emitters are simultaneously inputs and outputs.
The terminology can be confusing for a newbie, actually. The term "virtual short circuit" refers to the fact that in an opamp circuit with negative feedback the circuit is arranged in a way that (ideally) makes the voltage across the two opamp inputs zero.
Since one of the properties of a short circuit between two points is that the voltage across those points is zero, the people who invented that terminology considered (I guess) an intuitive thing to call what happens between the input terminals of the opamp a "virtual short". They called it "virtual" because it lacks the other property of a real (ideal) short: to gobble up any amount of current without problems! Alas, that's no small difference! They could have called the thing in a less confusing manner ("the principle of voltage balancing"!?!), but "the virtual short principle" sounds cooler, probably! Who knows?!
So, when we say that between the two inputs there is a virtual short, it's just an easy and conventional way to say that the circuit strives to balance the voltages at the inputs, i.e. it tries to make them and keep them equal.
Note that the existence of the "virtual short" is a property of the circuit, not of the opamp (although it exploits the ideally infinite gain of the opamp), whereas the fact that no current flows into the inputs is a property of the opamp (ideally).
EDIT (prompted by a comment)
I'll try to be clearer about what I said above. The virtual short is exclusively due to two key factors combined together: very high gain + negative feedback.
Let's do some math do convince ourselves. Let's call \$V^+\$ and \$V^-\$ the voltages at the non-inverting and the inverting inputs of the opamp, respectively, and \$V_o\$ the output voltage. A real opamp, in this respect, is a differential amplifier, i.e. \$V_o = A(V^+-V^-)\$, where \$A\$ is the open-loop gain of the opamp.
Inverting that relationship you get \$V^+-V^-=V_o/A\$. Thus, for finite \$V_o\$ and infinite \$A\$, you get that the difference between the inputs becomes zero.
Where did the negative feedback play a role? Nowhere, till now!!! The catch is that a real opamp needs negative feebdback to keep its output from saturating, in which case the simple linear model of the opamp (i.e. that gain formula) would no longer apply, except outside a very small interval of input voltages (assuming a classic non-inverting configuration where \$V^+\$ is the input voltage and \$V^-\$ is a fraction of the output).
Apply negative feedback and you'll get a zero differential voltage at the inputs over a meaningful range of input voltages.
Best Answer
My recommendation: Forget the link which gaves you the above information, which is false resp. misleading. (By the way: This link leads you to other "explanations" which also are wrong). Hence, you should not blindly trust any information available in the internet.
The text says that the "25 mV value being the internal voltage drop across the depletion layer of the forward biased pn diode junction". That`s pure nonsense.
This value of 25 mV is the so-called "temperature voltage VT" which depends on the environment temperature and appears in the exponent of the e-function describing the relation between the controlling base-emitter voltage and the emitter current.
And what about the "resistance" re, which appears in the above figure outside the transistor?. In fact, it is NOT a resistance - it is the inverse of the transconductance gm=1/re - and some people prefer the use of re instead of gm. Note that the transconductance gm=d(Ic)/d(Vbe) is nothing else than the SLOPE of the transfer curve Ic=f(Vbe) - measured in the selected DC operating point.
More than that, it can be easily shown that the slope d(Ic)/d(Vbe) is identical to gm=Ic/Vt (VT: temp. voltage); this gives you the relation between gm=1/re and VT.
Hence, gm is the most important parameter which determines gain. It relates input voltage and output current (therefore, it is called "mutual" transconductance gm). This can be seen in the known gain formulas (common emitter):
(a) without feedback: Gain=-gm*Rc
(b) With feedback (emitter resistor Re): Gain=-gmRc/(1+gmRe)
(Sometimes you can read: (a) -Rc/re and (b)-Rc/(Re+re) ).
EDIT: Differential input resistance at the base node (without feedback resistor Re):
The input characteristic of the BJT is also exponential with the slope
1/rbe=d(Ib)/d(Vbe)=(1/beta)[d(Ic)/d(Vbe)]=gm/beta.
Hence: rbe=beta/gm (or: rbe=beta*re).
Final comment: I think, this post is a typical example for the confusion which can be caused by using such "artificial" terms like re which have no physical meaning.