Electronic – Art of Electronics – Zener Diode Example

diodeszener

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.

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!

Best Answer

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

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

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

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

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

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