Art of Electronics - Zener Diode Example

Question

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

The text presents this circuit:

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

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

1.

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

2.

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

3.

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

4.

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

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!

Answer 1

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

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

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

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

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

Answer 2

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.

INTUITION FOR ZENER DIODE:

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.

INTUITION FOR THE SCHEMATIC:

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}\$

Answer 3

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?

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.

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.

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?

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.

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...