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I have had a look through all the information available about flyback or snubber diodes in a relay system and was hoping to confirm if one is necessary in my circuit or not. All the info about why a flyback is necessary is due to the use of sensitive chips. My circuit is driving a pneumatic solenoid valve and has no chip on it; using two SPDT switchs to set a reset a latching relay.

My Circuit Diagram

Do I need the flyback diode that is currently included? To my mind the answer is no since I have no sensitive electronic parts.

Thanks!

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    \$\begingroup\$ Given the very low cost of a diode, just do it. Good habit to get into. \$\endgroup\$ – tcrosley Jun 9 '15 at 6:55
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    \$\begingroup\$ On the page you cite the relay specs are 1A, 12V Here BUT on the specificaation page it says "DC Switching Current (Opening) 11.4mA" and elsewhere on the same page for "voltage source" mentions 12V, 11.4A. As Icoil = Icontact (claimed) = 11.4 mA and is 11.4A/1000 it means that the data is an absolute utter complete and irretrievable mangled mess. BUT the relay is NOT rated at vast switching currents and a diode is the least you can do for it. \$\endgroup\$ – Russell McMahon Jun 9 '15 at 7:02
  • \$\begingroup\$ See datasheet here - 12V 1050 Ohm coil, BESTAR reed relay. Note that various coils all use the same contacts. \$\endgroup\$ – Russell McMahon Jun 9 '15 at 7:18
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Arcing can still cause physical damage to the switch contacts. And if the contacts happen to be close enough while arcing, they can actually fuse together from the heat generated.

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  • \$\begingroup\$ What sort of voltage would I be looking at for this relay when running of a 12v source? The microswitches I have on hand are 30 amp 250 volt switches. Would they be significantly damaged? jaycar.com.au/p/SY4038 \$\endgroup\$ – Rhino_Aus Jun 9 '15 at 6:24
  • \$\begingroup\$ This arcing also causes RF (Radio Frequency) signals to be transmitted (this is how Mr Marconi generated RF signal in the early days of radio communication !). So it might disrupt some sensitive circuits in the neighborhood. \$\endgroup\$ – Bimpelrekkie Jun 9 '15 at 6:24
  • \$\begingroup\$ @Rhino_Aus: Impossible to say without knowing the internal physical construction. The switch rating isn't enough to tell. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 9 '15 at 6:29
  • \$\begingroup\$ @Rhino_Aus The question how high the "arcing" voltage will go can't be answered by knowing the supply voltage (12 V in your case). You store energy in the inductance of the coil of the valve or relay, and this energy will want to go somewhere once you open the supplying contacat. In doing so, a voltage spike will be created, theoretically up to "infinity volts", and something will clamp it (usually the opening contact, across the source (battery). The amount of energy in the coil will define how "hot" (better: warm) the arc can get. But arc, it will! \$\endgroup\$ – zebonaut Jun 9 '15 at 6:43
  • \$\begingroup\$ The problem with this (correct) answer is that the biggest source of arcing is the solenoid, not the relatively low power relay! Therefore a second diode is recommended, to protect the relay contacts too. \$\endgroup\$ – Brian Drummond Jun 9 '15 at 19:39
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A diode is essentially essential in that application.
Why? - see "Why use a diode" below.

The Jaycar specifications have been badly scrambled.
See **Jaycar's mangled specs:"" at the end of this answer.

The relay sold by Jaycar is a BESTAR reed relay with 1050 Ohm coil for 12V operation.

Data sheet here
There are a series of relays with the same contact ratings but different coil resistances allowing 3, 5, 8, 9, 12 or 24V operation depending on coil.

Maximum DC ratings for the contacts are 0.5A at 24 VDCmax carrying current
but according to the data sheet, when switching V x I should be <= 10.
eg notionally 24V at 417 mA max, or 0.5A max at <= 20VDC

Your cited Jaycar relay has a 12V rated 1050 Ohm, 11.4 mA rating. The Jaycar website specification has been transmogrified.
I wrte comments on this before but they are now more of interest (perhaps) than value so I've moved them to the end. They may serve as an example of how to spot scrambled specifications.

BUT the relay is NOT rated at vast switching currents and a diode is the least you can do for it.


Seeed Studio sell the same relay (5V version) on a PCB with transistor driver. here

They place a reverse diode and also a resistor across the coil. In this case they are protecting their transistor driver, but the same general principles apply.
See "Why use a diode" below

enter image description here


Why use a diode?:

All the info about why a flyback is necessary is due to the use of sensitive chips

No. Not so.
Relay switching specs are in part based on such things.

For an inductor the relevant equation is

V = L.di/dt

ie voltage is proportional to rate of change of current.
Fast change = great voltage.
Infinite rate (opening a switch) = infinite voltage.
This is usually a bad idea [tm].

ie When current is flowing in an inductor it CANNOT be instantaneously stopped (except by allowing an infinite voltage to occur).

While all inductors are non ideal, an ideal one is lurking inside along with all the non idealities, and it acts in the theoretical manner, and the results are modified by the other parts. In an inductor there is some parallel capacitance from the windings and some series resistance and other usually less relevant components. The capacitance saves you, to some extent.

When an ideal inductor with a parallel capacitance across it has a current 'i' interrupted then the voltage will rise until the energy stored into the inductor is transferred to the capacitor. This occurs when

i^2.L = 0.5 x C x V^2 or V = sqrt( 2 x i^2 x L / C)
Bigger L = bigger V (sqrt law applies) Bigger C = smaller V. (sqrt law applies)
bigger i = bigger V (linearly related)

That was for an ideal inductor.
In a real world situation, as the i tries to find new path, which it ALWAYS will, any R in the path it finds dissipates energy.
Voltage will rise across the resistor.
Big R = big V but large dissipation
Small R = low V but low dissipation.
Place a diode across the coil and V will be diode drop or 0.5V - 2V range typically plus any IR drop.

Omit the diode, do not add a shunt resistor or parallel capacitor and rely on stray capacitance to contain the energy and, depending on current and inductance, and leakage resistance, the voltage may rise to 100V, or 1000V, or more. Usually hundred of volts starts to get secondary effects happening, but if necessary Murphy and Mother-Nature will generate the voltage necessary to turn your contacts into a spark gap to dissipate energy in.

Look at the ratings for power relays designed to switch substantial domestic loads.
A relay that is rated at a say 16A, 250 VAC MAY be rated at 16A, 30VDC. Maybe less. Almost never more.
And, that's resistive. They assume SOME inductive component is unavoidable, but they are not expecting to deal with large reactive energies.
Relays very seldom have inductive DC ratings - they expect you to deal with the reactive energy some other way.

A diode is what's usually used.


Jaycar's mangled specs:.

Note that the relay specs on that page are scrambled beyond recovery.

On the page you cite the relay specs are 1A, 12V Here
BUT on the specification page it says "DC Switching Current (Opening) 11.4mA"
and coil current = 11.4 mA
and elsewhere on the same page mentions "voltage source" spec of 12V, 11.4A.
As Icoil = Icontact (claimed) = 11.4 mA
and as Icoil = 11.4A/1000 it shows that the data is an absolute utter complete and irretrievable mangled mess.

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