Although in both the common base amplifier and in the common emitter amplifier (non-swamped: with a nonexistent or bypassed emitter resistor), the signal appears to be coupled to ground only across the BE diode. So intuitively, you might expect the same low impedance in both.
However, in the common base amplifier, there is a dynamic action going on which lowers the input impedance.
The action in the common base amplifier can be understood like this. Suppose that the input AC signal swings a little bit low. This increases the \$V_{BE}\$ voltage. The collector-emitter current is very sensitive to \$V_{BE}\$ so a small increase \$V_{BE}\$ causes a large increase in the current. But in this common base amplifier, the collector-emitter current goes through the input. So, the input pulls down the voltage slightly, and a lot more current pours out!
Since the current flowing out of the emitter reacts sensitively to the input, it means that the input stage behaves rather like a voltage source. The opposite (a current source) would not react to voltage changes at all: the same current would flow regardless of our attempts to shake the voltage this way and that.
And, of course, voltage sources are what? Low impedances!
So in summary, the common base amplifier has a kind of active voltage source behavior, according to how the collector-emitter current responds to changes in the applied input voltage, and this gives it a lower impedance than you would expect just from looking at it statically.
This behavior is absent in the common emitter, because the input interacts with the base current, not the collector emitter current. In the common emitter, we even get feedback-driven impedance raising action if we add an unbypassed emitter resistor.
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.
Best Answer
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.