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I am having trouble choosing a flyback diode for my 5v relay. The switching current of the relay for 83Ohm is ~60mA. Goind through the parameter list of the RS catalog I selected:

  • Maximum Continous forward current > 100mA
  • Peak reverse voltage >30V
  • Power dissipation >500mW
  • SMD

One of the first entries were 1N4148WSF and BAT54WS. The BTA54WS is a Schottky diode, but still in the parameter range.

Does it matter which diode I use as long they are in my parameter range? Do I need more parameters to choose the correct diode?

EDIT: follow up question

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    \$\begingroup\$ No it does not matter, 1N4148, BAT54, 1N4001, 1N4002, 1N4003, 1N4004, 1N4005, 1N4006, 1N4007, BAS21, LS4448, they're all OK. Just use the cheapest. You can't go wrong with the 1N4148. There just needs to be a diode. Is it a diode and can it handle ~ 100 mA or more? Then it is fine. You're overthinking this. \$\endgroup\$
    – Bimpelrekkie
    Feb 4, 2020 at 14:36
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    \$\begingroup\$ Isn't that voltage spike in the forward direction to the diode (and the whole point of the diode), or am I misunderstanding? \$\endgroup\$
    – evildemonic
    Feb 4, 2020 at 15:21
  • \$\begingroup\$ @evildemonic I sure hope so, because the power supply is trying to push current in the opposite direction! \$\endgroup\$
    – DKNguyen
    Feb 4, 2020 at 17:10
  • \$\begingroup\$ Well, the guy I was responding to deleted his comment so now mine makes less sense. He was saying since the coil is inductive the diode needs to have a much higher breakdown voltage to support the voltage spike. \$\endgroup\$
    – evildemonic
    Feb 4, 2020 at 17:14
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    \$\begingroup\$ @evildemonic you're right. A few minutes later I posted that comment I realized that I totally blathered. I meant the transistor, not the flyback diode. Since I couldnt edit the comment due to the time limit, I deleted the comment. \$\endgroup\$
    – Rohat Kılıç
    Feb 4, 2020 at 17:16

2 Answers 2

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The BTA54WS is a Schottky diode, but still in the parameter range.

Schottky diodes are the fastest and good as well. It will be helpful in rapidly switching load but rather costlier than the cheap 1N400x series.

Chosen breakdown voltage of the diode shall be 8 to 10 times the supply. The peak current handling capability of 1N400x series are perfectly fine.

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    \$\begingroup\$ The rate that the coil current decreases is proportional to the voltage opposing the current. So using a Schottky will mean the inductor discharges more slowly, not more quickly. \$\endgroup\$
    – The Photon
    Feb 4, 2020 at 17:15
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    \$\begingroup\$ The higher you allow the voltage peak to be, the faster the inductor will discharge. If you really want to discharge quickly, use a zener in series with the PN or Schottky diode, chosen so the peak voltage is near the limit that the rest of your circuit can handle. \$\endgroup\$
    – The Photon
    Feb 4, 2020 at 17:18
  • \$\begingroup\$ both in the same direction? \$\endgroup\$
    – v3xX
    Feb 4, 2020 at 17:38
  • \$\begingroup\$ @ThePhoton isn't the point of a flyback diode to protect the rest of your circuitry from transient voltage spikes? A lower forward drop on the Schottky might mean that the current drops more slowly, but the difference in time is likely on the order of microseconds for small inductors. If your schottky is properly rated to dissipate the power you'd expect from inductive kick, why does this matter? Wouldn't a schottky be better in the general sense for low voltage applications because of the fast reverse recovery time? \$\endgroup\$
    – Ocanath
    Feb 4, 2020 at 17:58
  • \$\begingroup\$ @Ocanath, but if your circuit can survive a 3 V pulse, then limiting the pulse to 1 or 2 V will discharge the inductor more quickly than limiting the pulse to 0.25 V. My point is that even if Schottkys are "faster" than PN diodes, they won't make OP's circuit (switching a relay) operate faster. \$\endgroup\$
    – The Photon
    Feb 4, 2020 at 19:40
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Standard 1N4148 and compatible should work just fine, as it is a relatively fast small signal diode. 1N400x should also work but it is mainly meant for rectifying of mains voltage frequencies so it is not so fast, and otherwise overkill specs-wise. Since the current via relay is max 60mA, initial surge current via diode will be max 60mA and fall rapidly towards zero. And since voltage over the diode will be max 1V at 60mA with worst case estimation, it will dissipate only max 60mW peaks during relay release. So you really don't need 500mW continuous dissipation for the diode. But as The Photon said, it will be better to add a Zener diode like relay application notes suggest so it will release faster and contacts will last longer.

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