The MPGuino uses a 5.1v zener diode to read an input from the switching of an injector. MPGuino circuit diagram According to my very basic understanding of how zener diodes work I used the formula:

$$ I_{diode} = \frac{U_{in} - U_{out}}{R} = \frac{14.51V - 5.1V}{50k\Omega} = 0.178 mA$$

which seemed far to low given that a 1N4733A is \$5.1V\$ at \$49mA\$. Would the circuit still work at these low levels, or would a smaller resistor be needed? And would the output from this be safe to feed in to an input pin on an Arduino? I hope this all makes sense and I haven't misunderstood how zener diodes work.


1 Answer 1


Those zener diodes are clamps.

The circuit will still work at those currents, but the zener may only have 4.8V across it. (see the dynamic impedance numbers in the datasheet).

  • \$\begingroup\$ So if I understand right, the zener is just there as a sort of backup/fail-safe to make sure the voltage is never higher than 5.1v? \$\endgroup\$
    – Session
    Jul 28, 2013 at 8:19

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