zener diode i-v confusion
I already mentioned the "dam analogy" in comments, so I'll avoid it here. I also won't re-use what's already been written here. I'll start fresh and be a little more direct (electronic-minded) about it.

There's a nice zener image I found at a page called the Working principle of Zener Diode:



It's jazzy-looking, but there is a lot of information contained in it, too. So let's look at it in pieces.

The upper right quadrant is the area where a zener is operating like a normal diode, where it is forward-biased. In that quadrant, you can see that the forward current, \$I_\text{F}\$, stays very low until the usual minimum *silicon diode voltage" of about \$600\:\text{mV}\$ is reached. Then the current shoots upward like a rocket. Since a zener diode isn't supposed to be used like a normal diode, let's ignore this quadrant. It's not terribly interesting, anyway.

It's the lower left-hand corner where all the action happens. In this particular chart, the author took some time to provide a variety of curves. This is because there are several different zener voltages and these are due to a couple of different effects: avalanche and zener. The technical folks can worry more about the differences there, but you don't really need to. If I want to say anything here about all those curves, I'd want you to notice just how "vertical" the line is for the \$6.8\:\text{V}\$ zener. It's almost exactly vertical.

What this means is that for this particular zener, when the voltage is oppositely arranged (reverse-biased, which is why we are on the LEFT side of the chart) and before it reaches about \$6.8\:\text{V}\$, there's very little "leakage current" through it. The curve stays very close to \$I_\text{F}=0\:\text{mA}\$.

But as you can see, once the voltage exceeds this magic value, the current magnitude gets very much larger very quickly. This is the nature of that vertical line portion of the curve. If the reverse-biased voltage is \$6.8\:\text{V}\$, the current might be growing beyond \$20\:\text{mA}\$ (just by glancing at that curve.) And if the reverse-biased voltage is \$7\:\text{V}\$? Well, that just reaches out to the end of the curve near about \$140\:\text{mA}\$! Just a few tenths of a volt makes that much difference!

Let's see what happens when we put a resistor in series here. Something like this:







simulate this circuit &ndash; Schematic created using CircuitLab

Usually, we know the value of \$V_\text{CC}\$ or, at least, a range of values for it. Let's say we only know that it will be at least \$8\:\text{V}\$ but might be as much as \$9.4\:\text{V}\$, since we'll be using a \$9\:\text{V}\$ alkaline battery for this purpose.

The rating for this zener diode is at \$37\:\text{mA}\$. So let's select a resistor by making the assumption that the zener voltage will just magically work somehow and also let's assume the lowest voltage we expect to work with (the worst case situation): \$R_1=\frac{8\:\text{V}-6.8\:\text{V}}{37\:\text{mA}}\approx 33\:\Omega\$.

So what actually happens? Well, we actually start with a fresh \$9\:\text{V}\$ battery. That's more than the planned figure, so it allows us to test out what happens when plans go awry. Let's use that higher starting value as the actual voltage to start. When the circuit is just connected up, with no current yet in the zener diode and therefore no current in \$R_1\$ and therefore no voltage drop across \$R_1\$, the entire \$9\:\text{V}\$ would seem to be applied to the zener. This would immediately suggest currents in the zener diode that are simply off the chart! But as the current in the zener diode rises (very rapidly) in magnitude (goes downward on that chart) there is also a growing voltage drop across \$R_1\$. This lessens the voltage at the zener.

Let's assume for a moment that there is actually the original, estimated \$37\:\text{mA}\$. Then the voltage drop across the resistor would be \$33\:\Omega\cdot 37\:\text{mA}=1.221\:\text{V}\$. So we'd predict that \$V_\text{Z}=9\:\text{V}-1.221\:\text{V}= 7.779\:\text{V}\$. But we can easily see that the zener diode's current would be so much higher, if that were true. So we know that the actual current in the zener will rise above this value.

Let's make another estimate. We came up with \$7.779\:\text{V}\$, which is \$979\:\text{mV}\$ more than we'd expected. So let's assume this added voltage creates an added current in \$R_1\$. So we get a new estimate of \$I=37\:\text{mA}+\frac{979\:\text{mV}}{33\:\Omega}\approx 67\:\text{mA}\$. This means \$V_\text{Z}=9\:\text{V}-33\:\Omega\cdot 67\:\text{mA}= 6.789\:\text{V}\$. That seems a lot closer, now.

However, we supposedly know that the datasheet tells is that it is \$6.8\:\text{V}\$ with \$37\:\text{mA}\$. Our current is a lot higher, so the voltage in the zener diode should (according to the curve) also be a little higher.

I think you can see that we could go back and forth for a while, trying to work this out. There are math equations we could try. But it's time for a new idea, I think.



At this point, it's time to introduce a new concept. This is called "adding a load line" to the curve. The "load line" is a little tricky to get at first. But once you understand it, it isn't hard to remember and apply. So let's give it a shot.

The resistor is a really simple device. The voltage drop across it is a very simple function of the current through it. It's just your basic Ohm's law. It's possible to "visualize" this resistor on a chart like the one above, by drawing a line that represents the current in the resistor for various voltages across the zener diode (which subtracts from the supply voltage.) So if the zener diode voltage is \$9\:\text{V}\$ then obviously there is no remaining voltage drop across the resistor, so the current in the resistor is \$0\:\text{mA}\$. And if the zener diode voltage is \$0\:\text{V}\$ (for some reason) then obviously all of the supply voltage appears across the resistor, so the current in the resistor is \$\frac{9\:\text{V}}{33\:\Omega}\approx 273\:\text{mA}\$. And in between these two points, the line is very linear. Resistors are like that. So let's draw \$R_1\$'s load line in green below:



Where the line intersects our \$6.8\:\text{V}\$ zener curve is where the two devices solve out, correctly. Looks like about \$63\:\text{mA}\$. So from this, we can figure that \$V_\text{Z}=9\:\text{V}-33\:\Omega\cdot 63\:\text{mA}= 6.921\:\text{V}\$. Which is likely.

See how much easier it is with the load line added?? We don't have to sit around with a piece of paper rolling numbers back and forth a lot.

So... what happens if the battery is \$8\:\text{V}\$, instead? Or \$9.4\:\text{V}\$? Well, we can work out the new load lines, too. Those would make the chart look like this:



Now. Notice how small the span is in voltage (it's almost invisibly small) for quite a range of current variation?? (The arrows point the way!)

So this means that the zener will do a pretty good job of holding close to its rated voltage. Even when the supply voltage is changing a lot.



Hopefully, this gets across a few useful ideas. The load line is one good idea. But another is just realizing that the zener diode "floods" rapidly when the voltage exceeds its rated value. And this dramatic flooding behavior is what keeps the voltage very tightly controlled even when there are huge differences in the current flowing through the zener diode.



There are other problems. Temperature is one of them. If there is too much current then the zener diode will warm up from the excess dissipation required and this will also affect the resulting zener voltage. The rating is based on the idea of dissipating about \$\frac{1}{4}\:\text{W}\$ and waiting until it stabilizes at the rated ambient temperature. As you can see, there could be quite a difference in the current and that means quite a difference in dissipation, too. So while it seems pretty nice already, there is a price hiding behind the scenes, too -- temperature rise due to varying dissipation with different applied voltage sources. So that's another concern. But for a later time, I think. Just a note to the wise.