Buildint series voltage regulator using transistor and zener diode
Overview
The following is both a specific diagram but then also put into a more behavioral block diagram, as well:



simulate this circuit &ndash; Schematic created using CircuitLab
Combined, current source \$R_1\$ and voltage reference \$Z_1\$ are supposed to form a resulting combination voltage reference that produces a reference voltage value that holds its value well.
The meaning of well usually includes things like:

relatively immune to variations in temperature over some given range
relatively immune to variations over time as it ages
relatively immune to load current variations
relatively immune to device variations from a single manufacturer
relatively immune to device variations across different manufacturers
relatively immune to the position of the moon, etc, etc, etc.

Of course, manufacturers cannot anticipate everything. So they just tell you how better to operate their device and let you worry about the rest. So the zener datasheet is vital to you.
Zener Datasheet
The zener is a device that is supposed to be operated at a specific current in order for it to yield the manufacturers' intended reference voltage. That's how they work. So get that right.
For now, let's assume we don't yet know what zener to select. (We don't, so it's not really much of an assumption at all. It's a fact.) I'll refer to Vishay's datasheet. Let's assume for now that you'll need \$6.2\;\text{V}\$. That's the 1N4735A. Note that this voltage is nominal. The datasheet actually says that the tolerance is \$\pm 5\%\$. Combined, this is usually taken to mean that the initial accuracy of the device can be expected to be within 5% of the listed nominal value.
Whether or not this kind of variation is acceptable to you, is a matter for you to decide -- probably by carefully studying up on the load you intend to operate. So that's your first design decision to make. If you can't accept that variance, then you either need to find a better zener or else go looking for some entirely different approach. (Later, we'll develop equations that will magnify this error, so the \$\pm 5\%\$ figure is going to get worse as we start to analyze the circuit.)
Note that the test current is \$I_\text{ZT1}=41\:\text{mA}\$ and \$I_\text{ZT2}=1\:\text{mA}\$ and that the resulting dynamic resistance is typically \$Z_\text{ZT}=2\:\Omega\$ at \$I_\text{ZT1}\$ and is worst case \$Z_\text{ZK}=700\:\Omega\$ at \$I_\text{ZT2}\$. This is already a strong indicator that you want to operate the zener at \$I_\text{ZT1}\$. But it's not proof of anything.
The only argument you might actually have with Vishay is that if you operate their zener with a current between \$I_\text{ZT1}\$ and \$I_\text{ZT2}\$ and if the zener isn't dissipating at all (short pulses to keep it that way) and is sitting squarely at \$25^\circ\text{C}\$ then you can expect better than a dynamic impedance of \$700\:\Omega\$. Which isn't much to hang a hat on, to be honest.
The good thing is that they say you should get nearer \$2\:\Omega\$. But then you'd be operating it at \$41\:\text{mA}\$ and dissipating about \$\frac13\:\text{W}\$ and operating about \$35^\circ\text{C}\$ above ambient if you ensure that the leads at \$4\:\text{mm}\$ from its body are held at ambient. Of course, the zener die is then no longer at \$25^\circ\text{C}\$ unless ambient is \$35^\circ\text{C}\$ below that, or \$-10^\circ\text{C}\$.
In short, ... yeah, maybe, kind-of. If you haven't already started wondering about zeners, you should be. There was a time when that's what we had to live with. So, usually, you'd have to include methods for periodic calibration. (Things like potentiometers and more complicated circuitry around them and tools for calibration like &quot;standard references.&quot;) In practice, modern zeners are acceptably good for some purposes. The important thing to keep in mind is that it is only by providence and some good engineering that they even make it to &quot;acceptably good enough for some things.&quot;
One last note on zeners. I want to stop digging into the above details for a moment and have you just look at the more behavioral viewpoint illustrated above. It's important.
In a perfect world, you would have something you could insert in series with the zener to guarantee that it always is operating with a fixed current because that's the better way to make sure that the zener voltage remains closer to its nominal value. But you can't. Any such circuit will have variability in the voltage it adds to the zener and that would make the zener voltage worse, not better.
An alternative would be to replace \$R_1\$ with a current source. This isn't a bad idea, in fact, because the supply voltage above might be noisy or have ripple on it. It may not be perfectly solid. A current source would cope with those variations and hold the current relatively constant, regardless. But a resistor can't do that. (See next section for an elaboration.) The current through it will vary linearly with the voltage across it. So power supply ripple, for example, will turn immediately into current ripple. And given how zeners work, this translates into voltage reference ripple.
Resistor as Current Source
Using a resistor is a trade-off. It's cheap and with enough voltage headroom (you have about \$6\:\text{V}\$ of it) they do a credible job of keeping the current relatively constant. It's just that better can be had, if you are willing to replace the resistor with an active circuit. But that's for another day.
Today, you wanted mathematics. So today you will get that. The resistor is supposed to be holding a steady current so that the zener is supplied with a steady current. (We are temporarily leaving out the base current requirements of the BJT [system.]) But it is faced with its own tolerance (resistors don't have perfect values), variations in the supply voltage (supply voltages aren't perfect, either), and variations in the zener voltage itself (as already noted zeners also aren't perfect, even assuming they are supplied with a perfect current.)
Infinitesimals can be used to work out the details. A tiny % variation in current is \$\% I=\frac{\text{d}I}{I}\$ (from a calculus point of view.) Let's start out by applying the derivative operator:
$$\begin{align*}D\left[\: I_{R_1}\:\right]&amp;=D\left[\:\frac{V_\text{SUPPLY}-V_\text{ZENER}}{R_1}\:\right]\\\\\text{d}\,I_{R_1}&amp;=\frac{1}{R_1}\,\text{d}\,V_\text{SUPPLY}-\frac{1}{R_1}\,\text{d}\,V_\text{ZENER}-\frac{V_\text{SUPPLY}-V_\text{ZENER}}{R_1}\,\frac{\text{d}\,R_1}{R_1}\end{align*}$$
(Note the factor \$\frac{\text{d}R_1}{R_1}=\% R_1\$ and is just the resistor tolerance value.)
If we choose to look at the partials (holding the other variations as constant for the purpose), then we find the following three approximations:
$$\begin{align*}
\frac{\%\,I_{R_1}}{\%\,V_\text{SUPPLY}}=\frac{\frac{\text{d}\,I_{R_1}}{I_{R_1}}}{\frac{\text{d}\,V_\text{SUPPLY}}{V_\text{SUPPLY}}}\quad \quad&amp;=\quad\left[\frac{1}{1-\frac{V_\text{ZENER}}{V_\text{SUPPLY}}}\right]\tag{1}\\\\
\frac{\%\,I_{R_1}}{\%\,V_\text{ZENER}}=\frac{\frac{\text{d}\,I_{R_1}}{I_{R_1}}}{\frac{\text{d}\,V_\text{ZENER}}{V_\text{ZENER}}}\quad \quad&amp;=\quad\left[\frac{-1}{\frac{V_\text{SUPPLY}}{V_\text{ZENER}}-1}\right]\tag{2}\\\\
\frac{\%\,I_{R_1}}{\%\,R_1}=\frac{\frac{\text{d}\,I_{R_1}}{I_{R_1}}}{\frac{\text{d}\,R_1}{R_1}}\quad \quad&amp;=\quad \bigg[\quad-1\quad\quad\bigg]\tag{3}
\end{align*}$$
Those are mathematical descriptions of how well \$R_1\$ regulates its current. You can reach the following conclusions, now:

Equation 1 says that regulation vs changes in \$V_\text{SUPPLY}\$ is better when \$V_\text{SUPPLY}\gg V_\text{ZENER}\$ and that increases in \$V_\text{SUPPLY}\$ will lead to increases in \$I_{R_1}\$.
Equation 2 says that regulation vs changes in \$V_\text{ZENER}\$ is better when (again) \$V_\text{SUPPLY}\gg V_\text{ZENER}\$, but that increases in \$V_\text{ZENER}\$ will lead to decreases in \$I_{R_1}\$.
Equation 3 says that regulation vs changes in \$R_1\$ is fixed at 1:1 (but with opposite sign.) So a +1% change in the resistor value will correspond to a -1% change in the current.

Let's take your case and apply it to the zener and resistor tolerances, which are easily had. Let's say your resistor has a \$\pm 2\,\%\$ tolerance and we already know the zener as \$\pm 5\,\%\$. Let's just assume a power supply rail that is \$\pm 5\,\%\$, just because. From equation 2 above, we find that we can expect \$\approx\mp 5.4\,\%\$ current variation. From equation 3 above, we find another \$\mp 2\,\%\$ current variation for the resistor. From equation 1 above, we find \$\approx\pm 10.3\,\%\$ current variation for the power supply. Things add up fast.
These current variations need to be turned into voltage variations. For that, see the next section.
Zener Dynamic Impedance
So far, I've not addressed what any of the above means to the generated reference voltage. That's only getting at the variation in the current into the zener. So, finally, we now get to understand the purpose of \$Z_\text{ZT}\$ and \$Z_\text{ZK}\$!!
We need to multiply the above sensitivity equations, which provide us the %-variation of the zener current with respect to certain other variations, by the dynamic resistance of the zener and the nominal operating zener current, in order to get the magnitude variation in the zener reference voltage.
All of the above has been leading up to this moment.
You can now work out the variation in the reference voltage you get, versus any %-variation of some other factor!! I think that's cool! Yes?
But I'll leave this work for you. ;)
Note that it is now that we really begin to feel the weight of \$Z_\text{ZT}=2\:\Omega\$ and its worst case scenario of \$Z_\text{ZK}=700\:\Omega\$. The smaller this value can be made, the better, since it multiplies our current variation to get the voltage variation. From this, we know that we do NOT necessarily want to go cheap on the zener's operating current. We may prefer to stay closer towards its \$I_\text{ZT1}=41\:\text{mA}\$, instead. But don't forget that operating at a higher current means that the same %-variation yields larger magnitudes of absolute current variation. And it is the absolute current variation that is multiplied by the zener's dynamic resistance. So using lower currents might be just fine, accepting the higher dynamic impedance, but realizing a more optimal final result. And another consideration is that the datasheet only specifies a worst case, \$Z_\text{ZK}\$, for the \$I_Z=1\:\text{mA}\$ operating point. We really don't know the worst case when operating elsewhere. (We just know it's probably better at higher currents and probably worse at lower currents.) So a conservative design might instead choose to operate the zener where an absolute worst case number is specified and live with the implications.
In short, don't jump too fast to conclusions. You need to &quot;do the numbers.&quot;

Part of the reason why your question stimulated this response from me
is exactly because of where we are, here. There are no bright lines.
You must always think for yourself. And what initially, with a
simplistic view, may look &quot;one way&quot; will actually be &quot;something else&quot;
when you take into account more within your perspective.
Comprehensiveness is always the watchword. It's not enough to just
look at one thing and stop. This post is intended to illustrate this
point more than any other. (And in saying so, I may fall as easy prey
to others with still more complete or less incorrect perspectives than I possess.)

Let's compare the typical \$Z_\text{ZT}\$ case with the conservative \$Z_\text{ZK}\$ case. And without getting into details, let's say that we've now concluded that our power supply, resistor, and zener variations lead us to expect \$\pm 15\,\%\$ variation in the zener current (without considering variations due to the current boost section.) A quick &quot;look-see&quot; at the numbers tells us that there is a \$41\times\$ change in operating current (good) that is countered by a \$350\times\$ change in dynamic impedance. So we'd expect the conservative design to be about \$10\times\$ worse. (But it would then be relying squarely upon a datasheet guarantee instead of a &quot;typical&quot; from the manufacturer. And also, then, the zener dissipation will be negligibly small, which is also nice.)
Compute \$\pm 15\,\%\cdot2\:\Omega\cdot 41\:\text{mA}=\pm 12.3\:\text{mV}\$ and \$\pm 15\,\%\cdot 700\:\Omega\cdot 1\:\text{mA}=\pm 105\:\text{mV}\$. Which confirms the rough, earlier impression.
But also remember that this doesn't include anything related to the current boost circuit or any variations of its needed current supply from the left side of the schematic.
Current Boost
Above, we've assumed no disruptions caused by the current-boost block. (A BJT?) But BJTs require recombination current to operate. And this varies with the load requirements. Luckily, BJTs do have an approximate value of \$\beta\$ that is relatively constant (though unknown) over a several orders of magnitude of emitter current (load current.)
So you have yet another calculation to make. You now need to know \$\frac{\%\,V_{Z}}{\%\,I_\text{BASE}}\$. This will require a re-analysis of the earlier equation. But it will provide some interesting results.
When finished, you will be able to work out %-variation for \$V_\text{CC}\$ with respect to a %-variation in any of: resistor tolerance, power supply ripple, zener tolerance, and now finally also base current!!
I think you can see now how to proceed to work that out, as well. Just note that you can assume some average value for \$I_\text{BASE}\$ (because you will just add that value to the required zener current to work out a value for \$R_1\$) but that you want to see what \$\%\,I_\text{BASE}\$ do to your reference voltage.
Again, I'll leave this as an exercise.
Shockley Equation
Then, and no it just never ends really, there is yet another variation you need to account for. Whether you use one BJT or a Darlington or a Darlington with yet another added BJT for your current-boost block, there will be variability of the base-to-\$V_\text{CC}\$ nodes. The reason for this is because of the Shockley equation. For a single BJT, this works out to about \$60\:\text{mV}\$ change for a factor of 10 change in collector current at room temperature. But that's only for a small-signal BJT. In your case, you will be seeking out BJTs that can handle more than just a small signal. And these include additional factors (like the emission coefficient) that may worsen this issue. And with a Darlington, made of two BJTs, it's twice as bad. Etc.
This means one final new equation to re-analyze. And you'll need to include device variations there, as well. But when that's done you can finally figure out the %-variation for \$V_\text{CC}\$ with respect to a %-variation in any of: resistor tolerance, power supply ripple, zener tolerance, and now finally also load current!!
Temperature
Did we forget about ambient temperature? Yup. So, you now need to go back and insert temperature into the above development. This readily impacts forward-biased PN junctions, so it will certainly impact the zener as well as the current boost circuit. It directly impacts the Shockley equation itself and also affects the saturation current used in that equation, too. (In fact, the impact of the saturation current is greater than the impact of the thermal voltage. And this is a significant element to consider. It's not small.)
All of these details must all be included into your final design plan. You will then need to work backwards from your output requirement specifications and see if you can make this consistent with the above %-variation analysis I offered you above.
And by the way, this is just your basic zener-diode-regulator-with-current-boost circuit. It's more fun with more complicated circuits.