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.
Real answer is you don't. Hfe varies by temperature, part to part variability, collector current and voltage. Since you can't know what Hfe is at any given time, you make a circuit which is robust to the variations of Hfe.
Start by using a typical value and determine the component values (resistors) and bias points in the circuit. Then check to see if you are in the operating envelope of the transistor (power dissipation, max currents, max breakdown voltages). Check the bias points to ensure you have adequate headroom for your AC circuit (ie you can achieve the gain you need at the signal level you need and you are still in the linear region).
If you are, then without changing the component values, change the Hfe for the worst value and check the operating envelope again to see if you are still OK. Check the bias points to ensure you still have adequate headroom for your AC circuit (ie you can achieve the gain you need at the signal level you need and you are still in the linear region).
If you are, then without changing the component values, change the Hfe for the best value and check the operating envelope again to see if you are still OK. Check the bias points to ensure you still have adequate headroom for your AC circuit (ie you can achieve the gain you need at the signal level you need and you are still in the linear region).
If you are, congratulations, you've accomplished something called robust engineering, and you can be sure if you get a random sample of (the same type) transistors, that the circuit will work properly.
If you fail along the way, then either find a different transistor that is better suited to the operating envelope, or revise the component values to better meet the circuits needs within the transistor's parameters.
If you are making a one-off circuit you don't need to do all this, just get typical values, calculate function and operating envelope, and make a one off. Then check to see if it's near your calculations. But don't assume because you made one that works, you can make many and won't have problems.
Best Answer
The answer is ... it depends what else you want to achieve.
Increasing all the resistors will increase the input impedance.
The limit starts to come in where the transistor loses gain at very low currents.
As R4 is dominating the circuit as drawn (assuming reasonable hFE like 100 or so), sorting R4 will give you the most bang for buck.
In order to increase the AC input impedance, without disturbing the DC bias, you could bootstrap R4, by splitting it into two parts, say two 5k resistors in series, and connecting a capacitor from Q1 emitter to the mid point. As Q1 emitter follows the base with a gain of 0.99ish, that will all but eliminate the signal current that gets conducted down R4.
With R4 bootstrapped, R3 now dominates. You could replace R3 and R4 potential divider by a single resistor from a point of the right voltage, bootstrapped to Q1 emitter. That eliminates R3 and R4 for signals.
Replacing Q1 with a darlington transistor for much higher hFE will reduce the R2 term.