I'm working through the Art of Electronics and I'm stumped by a zener diode example.

The text presents this circuit:

zener circuit

And then shows that the zener diode behaves like a voltage divider:

\$R_{dyn}\$ is the dynamic resistance of the zener diode.


$$ I = {V_{in} - V_{out} \over R} $$


$$ ΔI = {{ΔV_{in} - ΔV_{out}} \over R} $$


$$ ΔV_{out} = { R_{dyn}ΔI} = { R_{dyn} \over R } (ΔV_{in} - ΔV_{out}) $$


$$ ΔV_{out} = { {R_{dyn} \over { R + R_{dyn} }} ΔV_{in} } $$

Everything makes sense through #3, but I don't understand how we make the leap to #4. I'm probably missing something obvious, but if anyone can explain that last step I'd appreciate it, thanks!

  • \$\begingroup\$ #3 is the one that seems non-obvious to me. From #1, I is meant to be the current through R. #3 assumes the change in current through the zener is the same as the change in current through the resistor, with no explanation. There's a hidden assumption that the load has high impedance. \$\endgroup\$
    – The Photon
    Mar 13, 2018 at 20:53
  • \$\begingroup\$ Remember that a zener diode is a very non-linear device, while a (ideal) resistor is a very linear device. \$\endgroup\$
    – user105652
    Mar 13, 2018 at 23:25
  • \$\begingroup\$ The Photon, good point. I guess I assumed no load (since there wasn't one pictured in the diagram) and the open circuit would certainly have high impedance :) \$\endgroup\$
    – Carl Baron
    Mar 14, 2018 at 0:55
  • \$\begingroup\$ \$R_{dyn}\$ is a linerarization of the zener diode behaviour at the working point. You have to determine the working point to get this value. \$\endgroup\$
    – Janka
    Mar 14, 2018 at 9:09
  • \$\begingroup\$ Think of the Zener diode as of a resistor with resistance Rdyn. Then you will see a voltage divider consisting of two resistors - R (aka R1) and Rdyn (aka R2). I hope you know what is its ratio... Of course, there are some subtleties here: R is an ordinary static (ohmic) resistor while Rdyn is a "self-varying" (dynamic) resistor... and you have to imagine what this means... See also my comment here: electronics.stackexchange.com/a/467713/61398 \$\endgroup\$ Nov 18, 2019 at 11:07

3 Answers 3


$$ ΔV_{out} = { R_{dyn}ΔI} = \frac{R_{dyn}}{R} (ΔV_{in} - ΔV_{out}) $$

$$ ΔV_{out} + \frac{R_{dyn}}{R} ΔV_{out} = \frac{R_{dyn}}{R} ΔV_{in} $$

$$ ΔV_{out} \cdot (1 + \frac{R_{dyn}}{R}) = \frac{R_{dyn}}{R} ΔV_{in} $$

$$ ΔV_{out} = \frac{\frac{R_{dyn}}{R}}{1 + \frac{R_{dyn}}{R}} ΔV_{in} \| \cdot \frac{R}{R}$$

$$ ΔV_{out} = \frac{R_{dyn}}{R + R_{dyn}} ΔV_{in} $$

  • \$\begingroup\$ Like a boss with no words used. \$\endgroup\$
    – efox29
    Nov 19, 2019 at 20:46

I had this same issue and happened across this question - though the existing answers allowed me to follow the working, I found I still didn't understand what was going on.

Part of the problem is that steps presented are conceptually backward for a beginner, without intuition for what a zener diode is and how to interpret its V-I curve (which is shown in the book). I'm going to use the ol' fluid analogy to hopefully answer the intent of the question to understand what the steps really mean.


  1. A zener diode is effectively a one-way valve for current. If current is pushed through the back of the diode (the bottom of the triangle), it flows no problem. If current is pushed through the front of the diode (the point of the triangle with the line), the valve snaps shut and stops letting flow through.

  2. If you turn the pressure (voltage) up, current starts to leak through, and the more you turn it up, the more it leaks. The dynamic resistance describes how much a small change in the amount of pressure (voltage) will change how much leaks through (current). At a certain pressure, the valve gives way and lets it all through again.


We want to know how \$V_{out}\$ will change with respect to \$V_{in}\$. Keeping the analogy, \$V_{in}\$ is like a pump providing pressure (voltage), whereas \$V_{out}\$ is just a point where we're observing that pressure. \$V_{out}\$ in this is not an applied load.

Now, the amount of current flowing through the resistor is shown in equation 1, standard Ohm's Law stuff:

\$I =\frac{V_{in}-V{out}}{R}\$

HOWEVER, we also know that (ideally), our zener diode is blocking any flow, making \$I\$ approximately zero. This means that the voltage \$V_{out}\$ is forced to be approximately equal to \$V_{in}\$. In our analogy, this is the same as pressure building against the valve until flow stops entirely.

However, in reality, there will be a little bit of leakage, and this leakage is (nonlinearly) dependent on the pressure (voltage) \$V_{out}\$ at the valve. For very small changes in pressure, however, we can model it as linear and predict small changes in leakiness. This is what the dynamic resistance \$R_{dyn}\$ respresents - a value approximating how much a change in pressure will correspond to a change in leakiness for a sufficiently small change (in practice, this is the gradient of the I-V curve of the zener diode at a particular point).

This finally leads us to equations 2 and 3 - equation 2 is simply rephrasing equation 1 in terms of a small change so that our linear approximation will be reasonable;

\$\Delta I =\frac{\Delta V_{in}-\Delta V{out}}{R}\$

Since (in this case) any change in the current flowing through \$R\$ corresponds to current leaking through the valve (Zener diode), we can multiply \$\Delta I\$ by \$R_{dyn}\$ to figure out how much the pressure (voltage) at the valve (diode) will change (\$\Delta V_{out}\$) due to leakage increasing or decreasing.

\$ \Delta V_{out} = R_{dyn} \Delta I\$

Of course, we don't just want to know how much the pressure at the valve changed due to current flow changing, we want to know how it is changing compared to the overall pressure (voltage) we're applying with our pump in the first place - (\$\Delta V_{in}\$)

So we substitute in our earlier experssion for (\$\Delta I\$), and obtain equation 3; \$ \Delta V_{out} = R_{dyn} \Delta I = \frac{R_{dyn}}{R}(\Delta V_{in}-\Delta V{out})\$

Which can be rearranged as shown by Janka to obtain; \$ΔV_{out} = \frac{R_{dyn}}{R + R_{dyn}} ΔV_{in}\$

  • \$\begingroup\$ The one-way valve analogy is not suitable for explaining the Zener diode's behavior; it is more appropriate for explaining diode switch applications (e.g. diode rectifiers, limiters, logic gates, etc.). If you want to use the fluid analogy as an intuitive tool for understanding, consider how a pressure stabilizer works, for example in a hydrophore installation. But I don't think it needs to go down that low. The "electrical analogy" (dynamic resistor) is quite sufficient for an intuitive understanding of the Zener diode's behavior. \$\endgroup\$ Nov 18, 2019 at 11:29
  • \$\begingroup\$ @Circuitfantasist Forgive me, I'm very much still a beginner - do you mind giving me an indication of why that analogy is inappropriate? In this context, it appears to make sense and line up reasonably well with the mathematics given the caveats that I noted, so I'm not too sure where I should go hunting for the problem! As far as I can tell, the potential issues would be related to Zener diodes having having particular behaviour in the breakdown region, which I've already hinted at in the intuition section? \$\endgroup\$
    – sedgwick
    Nov 19, 2019 at 9:57
  • \$\begingroup\$ sedgwick, your reflections are interesting and original. I'm just trying to simplify them to clear your idea of the Zener diode as a voltage-stabilizing element. By the way, during today's lab work, one student accidentally made a "composed Zener diode" with VF = 0.7V and VZ = 0.7V. I encouraged him for his "invention" and suggested that he make more such "Zener diodes" with VZ = 1.4V, 2.1V, 2.8V, etc. Can you guess what they were? \$\endgroup\$ Nov 19, 2019 at 17:28
  • \$\begingroup\$ Sorry, I'm afraid I still don't understand what your objection to the valve analogy is? As I said, I am a beginner, so I really don't follow what you mean by a "composed zener diode". The question asked for understanding on the mathematical steps presented - it still seems like a pressure relief valve analogy is ideal for that purpose to me. \$\endgroup\$
    – sedgwick
    Nov 21, 2019 at 23:42
  • \$\begingroup\$ I agree with your explanations... The "composed Zener diode" consists of two diodes (or a diode and a diode string) connected in inverse parallel. This network is very attractive when LEDs are used. \$\endgroup\$ Nov 24, 2019 at 11:27

The best way to explain a circuit is to build it step by step. Here is a possible 6-step building scenario.

Step 1. We have a voltage source with voltage VIN but we need lower voltage VOUT. How do we reduce it?

Zener diode - Step 1

Step 2. A resistor would help us, we think. We connect the resistor R1 in series to the input voltage source... but surprisingly for us, the voltage VOUT after it is the same as the input voltage. We begin to think why...

Aha... A resistor resists the current... but there is no current... so the resistor has nothing to resist... it is not a resistor. There is no voltage drop across the resistor and it freely transfers the voltage to the output.

Zener diode - Step 2

Step 3. So we need to make some current flow through R1. For this purpose, we close the circuit with another resistor R2 to ground. A voltage drop VR1 = I.R1 appears across the upper resistor... it subtracts from the input voltage VIN and we take the resт voltage VR2 across the lower resistor as output voltage. Thus we have actually invented the famous voltage divider.

Zener diode - Step 3

Step 4. Only, if the input voltage varies, the output voltage will vary as well. But we need constant output voltage. What do we do?

Zener diode - Step 4

Step 5. We decide to use a clever trick - to vary the resistance R2 (RZ) when the input voltage varies. When VIN increases, we decrease RZ and v.v., when VIN decreases, we increase RZ so that to keep VOUT = VIN.RZ/(R1 + RZ) constant. In this way, we have made a dynamic voltage divider.

Zener diode - Step 5

Step 6. But we got bored of doing this routine. That's why we replace the variable resistor RZ with a real Zener diode... and go for something more enjoyable. It will continue to do our work obediently...


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