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In a common relay circuit a zener diode, or other diode, is used to shunt the spiked that the relay could cause.

Now, why is the zener's cathode connected to the battery (or other source)'s positive terminal, and the anode to negative?

Should it not be the other way around?

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    \$\begingroup\$ If not, the battery or other sources would see a short if the relay coil is switched on. Please, draw a schematic. It helps explain things. \$\endgroup\$
    – Long Pham
    Commented Sep 9, 2018 at 17:10

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If it is connected the other way around it behaves like a regular diode and will short out the relay supply to ground.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The two options.

  • Figure 1a shows your suggested configuration. It should be clear that D1 shorts out the supply. The relay won't pick and either the diode will burn out or the supply voltage will collapse to about 0.7 V.
  • Figure 1b shows the correct orientation. When SW2 opens a circulating current flows as shown by the green arrows.

The easiest way to think of this is that inductors try to maintain current flow.

  • If the current is zero and you connect a supply the inductor resists the change somewhat and it takes some time for the current to rise to the steady-state value which is determined by the inductor resistance.
  • If the current is non-zero and you interrupt it - open SW2, for example - the inductor voltage increases dramatically to a large negative value if the diode is not there. This creates a spark on the contacts or plays havoc with switching transistors. By adding D2 we can allow the current to decay safely and eliminate sparking at SW2 or presenting a high voltage to the switching capacitor.

Should it not be the other way around?

Nope!

There is a downside to adding the diode. Since the current is maintained the relay releases more slowly causing a short time delay. Adding a Zener in reverse series with the diode causes the inductor to dissipate its energy much more quickly. The reason is that the power dissipated in the diode is given by P = VI and the forward voltage, Vf for a silicon diode is 0.7 V whereas it can be whatever you want with a Zener and the higher it is the more quickly the relay will lose its energy.

The diode + Zener gives the protection of the diode but with less of a release delay.

See "A zener with series diode"? for an expanded answer on this topic.

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  • \$\begingroup\$ No. If you want to use a zener in the manner you're suggesting, you need to put it in series with the regular diode, but pointed the other way. \$\endgroup\$
    – Dave Tweed
    Commented Sep 9, 2018 at 18:17
  • \$\begingroup\$ Thanks, Dave. Brain lapse. I've already written about that in detail in this answer. My only excuse is that I had to peel the potatoes somewhere in the middle of writing that. I have no idea what I was thinking this time. We mashed the potatoes though, added some fried onions and had them with gravy. That part went right. \$\endgroup\$
    – Transistor
    Commented Sep 9, 2018 at 18:21
  • \$\begingroup\$ Thanx for the input and helpful illustrations ;) \$\endgroup\$
    – SilverNodashi
    Commented Sep 19, 2018 at 7:35

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