But I couldn't understand what meant by the output impedance of this
circuit. What is the meaning of the output impedance here and how can
we quantify it in terms of R hfe ect.?
You've got to think like a transistor but, to help you, try thinking like a forward biased diode. Let's say your anode voltage is connected to a perfect battery (strong voltage source) then, under very light load conditions on the cathode (say 0.1 mA), the cathode voltage might be 0.5 V lower than the anode: -
If you increased the current a little bit to 0.15 mA, the forward voltage dropped would be about 0.525V. Take the "before" and "after" numbers and work out what the dynamic resistance is: -
Dynamic resistance = change in volts / change in amps = 25 mV/ 0.05 mA = 500 ohms.
If you did this at a higher level of forward conduction (say at 1mA to 1.5mA) you'd get a forward volt change of 10 mV (640 mV to 650 mV). Now, the dynamic resistance has become 10 mV / 0.5 mA = 20 ohms.
This dynamic impedance represents the output impedance of the diode (at the cathode) when the anode is tied to a solid fixed supply voltage.
simulate this circuit – Schematic created using CircuitLab
If the diode's cathode were connected to a fixed 100 ohm resistor to ground, can you see that the dynamic impedance of the cathode in parallel with the fixed 100 ohms becomes the new output impedance: -
simulate this circuit
Think about converting the voltage source in series with the dynamic impedance of the diode to a current source like this: -
You see now that the dynamic impedance of the diode is parallel to the 100 ohm and this makes the net resistance smaller. You don't have to consider what value the voltage source is or what the current source amperage is because you reconvert back to the thevenin equivalent and you have the net output impedance.
Note - this added resistor is the emitter resistor when a BJT is considered
So, if the total current drawn from the cathode is ~0.1mA the "source" impedance of the cathode and the combined resistor is 500 ohms || 100 ohms = 83 ohms.
This drops to 20 ohms || 100 ohms (17 ohms) when ~1mA is being taken through the cathode. How does all this relate to a BJT you might be asking. Here's how...
The base emitter junction is a forward biased diode; the base is the anode and the emitter is the cathode but, the clever thing about BJTs is that although there may be "weak voltage" at the base that is easily affected by loading, the collector current doesn't allow this to happen and it is the collector current that replaces the base current as the source of current to the emitter (cathode). Thus, you can still regard the emitter has having the output impedance of a diode when that diode's anode is connected to a strong voltage source.
What I've said is a little oversimplified because there is still a bit of base current taken when providing current to the emitter but, this is usually at about 1% of the level of the pure diode scenario - this boils down to the BJT having a current gain of about 100.