# zener diode i-v confusion

So zener diodes in reverse bias apparently only work within a specific current, Is min and Is max which generally have basically the same X value. On the zener diode graph usually lets say Is min is at like 5.6v and Is max is at like 5.61v or something really close. So zener diodes only work within that voltage? If I apply 6v it will break? If not, then I don't understand the relationship between I-V on the graph

How do zener diodes create a stable voltage output even then? How do diodes have a constant voltage drop? I don't understand how to input a specific current into the zener diode without having to input a super specific voltage since voltage is just the different charge density that drives current

I'm just entirely confused

• The voltage will be clamped to 5.6V, and the wire (and insides of the PSU) will have a voltage drop of 0.4V across it. – Ignacio Vazquez-Abrams May 5 '18 at 21:48
• @Ignacio But how would that not make the current go over Is max and thus risk breaking the zener diode? – nunch May 5 '18 at 21:48
• It wouldn't, which is why you need to use another device to limit the current. – Ignacio Vazquez-Abrams May 5 '18 at 21:50
• Think of the zener like a dam. The height of the dam is equivalent to the voltage rating on the zener. If the water behind the dam rises up enough, then it will flow over the spillways at the top (current will flow.) But there is no way for the water behind the dam to rise up beyond the height of the spillway, no matter how much flow arrives behind the dam. Because all that happens is that more water flows out the spillways if more is arriving. But the height stays the same. The zener is like that. If the voltage exceeds its rating, then it "spills current" until the voltage doesn't exceed it. – jonk May 5 '18 at 22:07
• @nunch Before the height reaches the rated voltage, pretty much no current flows through the zener. Once the height is reached, there is no theoretical limit to the current. But there are practical limits, of course. As current spills through the zener, this causes an increased voltage drop across the resistor that is supplying that current, which of course lowers the voltage at the zener. The resistor itself is what limits the current. So do not hook up a voltage source directly to a zener, without a resistor present. You will destroy it very quickly. – jonk May 5 '18 at 22:28

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 – 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.

A Zener diode is intended to be a current-driven device. It has a low 'slope resistance'.

In the case of your 5.6v device, let's say it draws 1mA at 5.6v, and 11mA at 5.61v. The slope resistance is dV/dI = 10mV/10mA = 1 ohm.

When designing a circuit using a zener diode, ensure there is enough resistance in series with the diode so that over the whole range of operating conditions (and transient/fault conditions) the circuit will see, the zener diode is not presented with a continuous current of more than Is max.

In its typical shunt regulator voltage reference mode, a zener diode will be supplied with a current of nominally Iz (the current that the diode has been tested with) which tends to be in the 5mA to 10mA range. Its voltage drop is then its nominal zener voltage. Where low accuracy is adequate, this current is supplied through a large resistor connected to a much higher input voltage.

If the current varies, the zener voltage will vary, but not by much. The 'better' the zener, the lower the slope resistance. With our 1 ohm slope device above, a change between 5mA and 10mA would cause a change in zener voltage of 5mV.

Where higher stability is required, a common trick is to use the circuit's regulated output voltage to help define a constant current through the diode. This improves the effect of slope resistance by several orders of magnitude, rendering it a negligible error compared to the thermal variation of zener voltage. Depending on the specific part used for OA2, this circuit may or may not start up. It will for most types. If it doesn't, an additional biassing resistor or startup capacitor can be used to ensure it starts.

It's worth pointing out the single transistor 'amplified zener'. Q1 maintains 0.7v across R5, which acts as a constant current source for the zener, in this case 5mA. The tempco of Q1 VBE is nominally matched by the tempco of a 6.2v zener, so the overall combination has a reasonable tempco.

simulate this circuit – Schematic created using CircuitLab

• Sorry what's precisely meant by slope resistance (like I know that it's the slope between those two points but I don't understand what it actually means)? And what is meant by "use the circuit's regulated output voltage to help define a constant current through the diode"? Like what's physically being done? thank you – nunch May 6 '18 at 5:19
• @nunch change the current by dI, observe the voltage change by dV, slope resistance is dV/dI. I'll add circuit diagrams to my answer. – Neil_UK May 6 '18 at 6:04
• As I said I understand that it means the slope between two points; rather, again, I don't know what that even means... the resistance of the zener diode at some point? If so, at what point? Those two points were chosen arbitrarily, if I chose two slightly different points I would get a slightly different answer. Are you just trying to say there that a small change in voltage means a large change in current? – nunch May 6 '18 at 6:17
• @nunch I don't know what you mean by 'means'. It's called a resistance because it has units of volts/amps. It's called 'slope' (also dynamic, also incremental) because it's measured by looking at the slope of the graph, or with changing or increments in current. It's given a name because it's an important parameter in the accuracy of the reference voltage. It's reasonably constant over a range of currents, which is why it's worth naming. You'll get much the same answer whether you pick currents of Iz_nom +/- 10% or +/- 50%, similar, same ballpark, but not the same to 3 decimal places. – Neil_UK May 6 '18 at 6:25

Consider 1k resistor in series with Zener diode. Add its characteristic to Zener I-V curve. For powering both from 6V source the ch-c for R would be the line through points (0mA,6V and 6mA,0V). For 10V it would be the line through points (0mA,10V, 10mA,0V). Where the line croses with diode I-V curve you get the voltage at Zener diode. Compare the difference betwean 6V and 10V powering and the difference at Zener diode in both cases. It is the stabilising effect of Zener diode.

• I don't know what ch-c means and why do the two resistor equations have negative slope? spent awhile drawing it out trying to make sense of it – nunch May 5 '18 at 23:45
• ch-c was short from characteristic. Resistor lines has negative slope because driven not in its (resistor) U/I coordinate but in Zener diode (or enything else powered by resistor) U/I . – Piotr May 6 '18 at 17:34
• For 10V the resistor ch-c has I the same as Zener diode, and its U with 0V at point 10V of Zener diode U/I coordintes and increase to the left - then it has positive slope. – Piotr May 6 '18 at 17:43
• Consider 10V source powering resistor 1k and anything else in series with resistor. When at that thing there is 9V then current has to be 1mA (because of resistor), when voltage is 8V current has to be 2mA and so on. – Piotr May 6 '18 at 17:45
• I used driven instead of draw (may be drawn). Sorry for my English. – Piotr May 6 '18 at 17:51