Electrical – Common Collector Biasing

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.

The voltage from V1 is divided between R1 and the B-E junction of the transistor. Since that junction is a forward-biased diode, the voltage drop across it is only about 0.65V; the rest of the voltage appears across R1, setting the magnitude of the base current.

The voltage from V2 is divided among R2, the C-B junction of the transistor, and the B-E junction. Since we've already established that the last one of those is 0.65V, we can say that V2 - 0.65V is divided across R2 qand the C-B junction. As long as this quantity is positive (i.e., V2 is greater than 0.65V), the C-B junction is reversed-biased.