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I understand the use of the Zener but what I don't understand is one topology uses two components, but why can't the other topology be used ?

schematic

simulate this circuit – Schematic created using CircuitLab

The circuit on the left uses two components (D1 and D2), whereas the circuit on the right only uses one (D3).

I can't really see a difference between the two, other than that the left might decay a bit faster because of of the higher clamp voltage, but aside from that, I would imagine they operate the same.

So why do most schematics I've seen with regards to coil suppression use a series diode zener pair ? There must be something I am not seeing or understanding.

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  • \$\begingroup\$ This article addresses this question quite nicely. The author shows that the circuit with the zener permits a faster collapse of the magnetic field and thus a shorter transport delay between the microcontroller deactivating the darlington and the relay actually changing state. \$\endgroup\$ – Paul Stiverson Jun 21 '15 at 4:45
  • \$\begingroup\$ @PaulStiverson sorry, but it doesn't address the question. The question is about topology and not about what the zener is doing to speed up the relay operation. \$\endgroup\$ – efox29 Jun 21 '15 at 5:00
  • \$\begingroup\$ this may be relevant (not nearly a duplicate, though). \$\endgroup\$ – Lorenzo Donati supports Monica Jun 21 '15 at 12:38
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Not a lot of difference if the zener diode is specified as higher voltage by the supply voltage in the second case. The current loops are different so there might be some difference in the EMI but probably not significant in most cases.

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  • \$\begingroup\$ While operation might be the same, would one be preferred over the other ? I suppose the left topology can work for either high side or low side switching whereas the right topology can only be used for low side. \$\endgroup\$ – efox29 Jun 21 '15 at 3:52
  • \$\begingroup\$ If there is only access to the relay coil connections.. but other than that I can see no reason to not prefer the right topology. Of course a single bipolar TVS could be used in either case, but that would typically be more expensive than a zener diode. \$\endgroup\$ – Spehro Pefhany Jun 21 '15 at 4:05
  • \$\begingroup\$ What am I missing - why are either left of right preferable to a single reverse-biased diode across the coil? \$\endgroup\$ – Techydude Jun 21 '15 at 4:43
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    \$\begingroup\$ @Techydude That's been covered here in detail- bottom line is that the diode slows the relay opening and thus shortens the contact lifetime. The datasheet life specs are typically without a diode. Using a zener is harder on the transistor but better for relay life. \$\endgroup\$ – Spehro Pefhany Jun 21 '15 at 4:49
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In the second case, the current loop goes through the voltage source. That means in the second case when the zener conducts during the flyback period, it not only dissipates power stored in the coil, but also additional wasted power supplied by the voltage source.

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  • \$\begingroup\$ Not sure I follow the last part. When the zener conducts, it will only conduct when the field is collapsing. The zener breakdown voltage would be much higher than the power supply, so there shouldn't be any contribution from the source - only from the coil. \$\endgroup\$ – efox29 Jun 21 '15 at 6:41
  • \$\begingroup\$ If you draw the current loop when D3 conducts, you can see the current passes through the V1. Power (V1 x current) is supplied and wasted. \$\endgroup\$ – rioraxe Jun 21 '15 at 20:24
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Time constant of inductive circuit=L/Rl, for faster collapse of mag. field R should be as large as possible. Zener dynamic resistance is current dependent so it helps in fast relay operation t=L/(Rl+Rz).further protecting the transistor from excessive voltage spike. Diode is used as a blocking diode for zener. vtingole

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In the first design, you are limiting the induced voltage of the inductive relay to whatever the diode and zener diode are rated for. This allows the current in the inductor to dissipate at a stable rate through itself.

In the second design, you are allowing the inductive relay to create a very large voltage spike when the switch is turned off. And then you are shorting all that stored energy through the zener diode. Since a zener diode is designed to maintain its voltage by converting the extra voltage to current, this creates a very large current through the zener diode in order to maintain its stable voltage across the transistor. Not only that, but from your diagram, you'd be putting a larger voltage drop across the transistor compared to the first design.

The problem is that with larger currents, the zener diode in the second design would burn out and fail, causing most likely a short across the transistor. Whereas in the first design you limit the relay voltage and subsequent current through it to a much lower amount.

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