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This guy had a bit of problem installing the flyback diode to his relays due to physical obstruction. Now if that is really an issue will this be a good alternative way, for perfboards and breadboards?

schematic

simulate this circuit – Schematic created using CircuitLab

When the MOSFET is turned on, C1 is essentially shorted out and completely discharged.

When the MOSFET is turned off L1's flyback current charges capacitor C1 to a voltage.

Capacitor C1 is chosen to be able to hold all the energy L1 can hold without reaching a voltage so high that it can damage the MOSFET.

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  • \$\begingroup\$ What happens to the charge in the capacitor when you switch on the MOSFET? It might be damaged by the pulse of current... \$\endgroup\$ – tomnexus Jan 4 '15 at 20:12
  • \$\begingroup\$ @tomnexus I really doubt it. Switching power supplies subject capacitors to similar conditions many thousands of times per second. It also happens to every power supply decoupling capacitor every time the power to a device is connected. \$\endgroup\$ – Phil Frost Jan 4 '15 at 23:53
  • \$\begingroup\$ Ah, I meant the MOSFET, sorry, not the C. Depends on the MOSFET Ron which isn't all that low anyway. \$\endgroup\$ – tomnexus Jan 5 '15 at 2:49
  • \$\begingroup\$ That would usually be referred to as a 'snubber' and would typically include a resistor and for some configurations a diode too. \$\endgroup\$ – brhans Jan 5 '15 at 6:08
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The capacitor as shown has a couple of problems. First of all, the peak current when the relay is switched on will be much greater than without the capacitor. Secondly, the combination of the capacitor and inductor may cause "ringing". Adding a resistor in series with the capacitor will mitigate those problems.

Another approach would be to use a capacitor with a resistor in parallel with it and a diode so that current can flow from the coil to the cap but not vice versa. This approach will avoid having the cap waste power when turning the relay on; if the relay is switched on and off repeatedly before the cap significantly discharges, the cap voltage will increase but this in turn will help it dissipate energy faster. A caveat with this approach is that the relay will be release more quickly when the cap has a higher voltage than when it does not.

A final suggestion would be to use both high and low side switches along with a couple of diodes; using that approach would cause the energy stored in the relay to feed back into the supply. This would cause the relay to release more quickly than it would with many other flyback approaches, and if the relays are switched on and off frequently this approach would improve energy efficiency. One caveat is that one must ensure the supply has adequate filter caps to accept the energy supplied by the relays.

schematic

simulate this circuit – Schematic created using CircuitLab

As shown, the circuit will switch the "relay" (1 1mH coil) "on" and "off" cleanly. If one closes the switch (right-click, select "properties", and then set the state to "closed") and reruns the simulation, the relay's current will dissipate much less quickly (which will in turn mean that the relay will respond more sluggishly).

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Yes that's a perfectly good way of doing it but bear in mind there is very little (other than the coil's series resistance) to burn off the energy - the resistance of the coil is likely enough to rapidly dissipate the energy and thus ensure rapid turn off of the relay (should it be needed).

You can also use a resistor in parallel with the coil or wired as you've shown the capacitor - you just have to make sure that the peak current through the relay and the resistance (product) don't generate a voltage that could fail the mosfet.

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  • \$\begingroup\$ relays are built with solid cores so when the curcuit is opened some energy will go into eddy currents in the core and the rest into the capacitor. so long as the mosfet and capacitor can handle the peak voltage driving the mostfet slower (eh: 10K in series with gate) will also reduce the peak voltage, but cause heating of the mosfet during turn-off and turn on. \$\endgroup\$ – Jasen Jan 5 '15 at 0:22
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If you need snubbing at the transistor then you should use a reverse-biased zener or avalanche diode with a voltage lower than the transistor's breakdown voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I have used NE2 neon bulbs before in place of the zener. Diodes are mostly cheaper solutions nowadays. \$\endgroup\$ – steverino Jan 5 '15 at 1:05
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Other methods are to use an avalanche-rated MOSFET (no additional parts), or put a zener diode as so:

schematic

simulate this circuit – Schematic created using CircuitLab

These methods dissipate the energy stored in the inductance partly or mostly in the MOSFET rather than mostly in the coil as happens with a diode across the coil.

They will maximize the relay life because the relay can open quickly. The rapid dv/dt could cause EMI problems and in fact a small capacitor (eg. 10nF) across the MOSFET may help with that.

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