First of all, I'm going to assume your complete circuit looks like this:
[BTW, you should post your complete circuit if you expect to get any meaningful answers.]
Secondly, the unity voltage gain of the common collector refers to AC, not DC.
From the image above, you can see that the output voltage will be \$V_Z-V_{BE}\$.
And \$V_{BE}\$ will have some variation with the collector current, but not too much: \$V_{BE}\propto ln(I_C)\$.
On the other hand, \$I_B\$ is not negligible, it could be up to 20mA (for the transistor's minimum \$h_{FE}\$ of 50), and you don't really show how you are biasing your zener, so it could be that the base is sucking more than you are providing and the voltage across the zener will drop, and this drop will be directly reflected at the output voltage of this circuit.
By the way, from the 2SD1047 datasheet, \$V_{BE}\$ at 1A will be about 0.7V, so your output should be about 4.3V (not 5V), and like I said, will vary a bit with \$I_C\$. At 1A, it will dissipate quite a bit: \$1A(20V-4.3V)\approx 16W\$. The transistor should be able to thermally handle it though, since its thermal resistance is only 1.25°C/W.
Answering questions:
@LvW Just out of curiosity: What if VCE = VBE? C-B pn junction won't
be reverse-biased then, so it won't attract electrons in base region.
Thus, IC will be zero, and IE will be equal to IB?
But the C-B diode is not forward biased. This is an application where the BJT is used as a diode and no "classical" amplification is possible (transition region between saturation and amplifying region).
IC and IE are controlled and only controlled by VBE; IB is just a side
product; Once VCE is greater than VBE, its specific value does not
matter, because E-B junction is reverse-biased. Am I right?
It does not matter too much - on the other hand: Look at the Ic=f(VCE) curves. Ic slowly rises with VCE because of the Early-effect.
Given VBE, IE is fixed, and as a result, the sum of IB and IC is
fixed. When VCE < VBE, what IB and IC are depend on VCE. The greater
VCE is, the greater IC/IB is. However, the value of IC/IB is capped by
"beta", which is reached when VCE = VBE. " Is this right?
In this case (VCE < VBE) the C-B diode is open and there is a small current Ic which has a direction opposite to the "normal" Ic direction. Example: For VCE=0 we have a current Ic which is negative (The Ic=f(VCE) curves do NOT cross the origin!).
Best Answer
As far as the mentioned Early effect is concerned, there is no difference between CB and CE configuration. The effect is as follows (modulation of the base width Wb) :
As the reverse bias across collector-base junction increases (due to a rising collector-emitter voltage) the width of the collector-base depletion layer increases - and the base width Wb decreases correspondingly. This effect has - in principle - the following consequences:
1) When VBE is held constant, the emitter current IE=f(VBE) remains constant but the collector current increases and the base current is reduced correspondingly (due to IE=IB+IC=const).
2) When IB=const, both IE and IC increase at the same time (B value is larger).
In reality, neither of the cases 1) or 2) will be exactly met - instead, we will have a mixture of both cases.
Remark: As can be derived from the above (and as shown in the corresponding data sheets for constant IB and VBE, respectively) the slopes of the curves Ic=f(VCE) are not equal in both cases. Hence, the derived values for the EARLY voltage VA are different.
Thus, do we have two EARLY voltages? Surprisingly, this question has not been answered in textbooks (as far as I know). Some sources define the EARLY voltage for VBE=const, some others for IB=const. and some make no statement at all.