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Warning: this may be an extremely naive question (if so, please enlighten me).

Many applications of relays require a flyback diode to protect against inductive voltage. I'm unable to find any relay that incorporates a flyback diode.

Since it's such a common need, why don't relays include a flyback diode inside the relay package? Are there just too many factors to consider, making it hard to guess the circuit's need?

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MOSFETs often come with such a diode in them, so there is built in protection on the switching side of the coil, rather than in the relay housing. This makes more sense for many reasons given in answers below and also for the fact that it's easy to fab a diode into a MOSFET and somewhat less economical to solder one that you may or may not need into a relay. –  Big Endian Feb 19 at 5:23
The diode in a MOSFET is not constructed additional to the MOSFET. It is an artifact of the structure of the MOSFET itself. –  Ignacio Vazquez-Abrams Feb 19 at 5:26
When a driver chip like the ULN2003 or ULN2803 is used the flyback diodes are provided in the chip, so an extra diode in the relay would be a waste. –  Wouter van Ooijen Feb 19 at 7:35
@BigEndian When a single MOSFET is used to switch a relay coil, the body diode does not help as it conducts in the opposite direction from what is needed. For example: an N-channel MOSFET used as a low-side switch, the voltage spike from the relay coil (when it is switched off) will be positive and the MOSFET body diode does not conduct (except possibly avalanche breakdown which can be destructive). An exception to this is a MOSFET that is "Repetitive Avalanche Rated" where the body diode acts like a high-voltage zener diode, for example IRFD220 –  Tut Feb 19 at 14:30
Proof that naive questions, worded well, can be quite good indeed. –  JYelton Feb 19 at 16:55

5 Answers 5

up vote 13 down vote accepted

There is simple answer to this question - there are many flyback schematics and the reverse diode is the simplest one. Although it has one big disadvantage - it makes the relay to switch off very slow.

This way, sometimes other schematics are to be used. There are several examples:


simulate this circuit – Schematic created using CircuitLab

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+1, Nice example –  shuckc Feb 19 at 9:12
isn't the design for #2 just Ohm's law? Why is that hard to design? –  markrages Feb 19 at 15:38
@markrages It would be "just Ohm's law" if D4 was shorted. –  hoosierEE Feb 19 at 15:51
e.g., 100 mA relay, Vcc=12V, 0.7 Vf on D4, Q2 60 Vceo. Say we want 40 V max on Q2. Then max voltage on R is (40 - 12 - 0.7) or 27.3. R = V/I so 27.3 / 100 mA or 270 ohms. –  markrages Feb 19 at 15:54
@markrages - the problem with the second schematic is that the current through the relay is never known. The resistance of the relay is very sensitive to the temperature and the current may vary significantly. Then you need to design for the worst case. In all other schematics, the voltage is relatively constant and depends only on the zener voltage. –  johnfound Feb 19 at 16:48

There are quite a few relays (also contactors/breakers) which are powered from AC. Putting a diode inside renders them useless for AC applications: -

enter image description here

There are several types of relay that are of the latching type and these also require a reversal of voltage - a diode would render this type of relay useless in that application: -

enter image description here

See also this answer for high speed switching of a relay.

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There are actually are (or were) a few which have had internal coil suppression. For example this Teleldyne one. They've not been very successful commercially in the general market. Maybe in the automotive market.

It probably has more to do with stuff like history and second-sourcing than anything technical, though I fail to see much advantage in putting a diode internally. It forces a polarity on the pinout and results in a sub-optimal electrical life for all users, for a very small saving.

There's no PCB generally inside a relay so it would have to be welded or crimped or soldered into place.

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There is also a reliability problem. Most relays are sealed or, at least, not made to be disassembled by the user. If an internal diode fails, the relay becomes useless. Since the diode cost is much less than the value of the relay, it makes more sense to add the diode externally.

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There are relays that include a fly-back diode, but percentage wise, not very many.

If you go to this Digi-Key page which lists signal relays (less than 2 amps) and scroll to the right, you will see under the column Features that one of the attributes you can pick are Diodes.

Here, only a little over 5% of the relays listed incorporate a diode. For power relays, the number with diodes is a little over 3% of the total.

So they do exist. But why so few? Obviously, it keeps the cost of the relay down, even if the user has to add their own diode. This also lets the user select a diode that exactly meets their needs. And it is cheaper (and easier) for a user to add a diode to a PCB (which is an automated process) than the manufacturer add it across a relay coil (which may have to be manually done).

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Why would the load being inductive matter? –  Nick T Feb 19 at 0:48
I mistyped. The relay itself is the inductive load in this case (relative to its driving circuit), whether or not it's contacts are driving an inductive load has no bearing on it needing a fly-back diode. Answer edited. However if the contacts are driving an inductive load, then that load needs to have a fly-back diode. –  tcrosley Feb 19 at 5:19

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