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I want to open a solenoid electromagnetic lock using an IRFR024NPbF MOSFET.

Datasheet: https://datasheet.lcsc.com/lcsc/1809062114_Infineon-Technologies-IRFR024NTRPBF_C2583.pdf

Generally speaking, I do understand why flyback diodes are needed when controlling an inductive load with a relay or MOSFET. I also know how I should connect the flyback diode.

Let's take a look at the symbol in the datasheet:

enter image description here

and assume that I want to use this as a low side switch. Now, if that diode would be a regular diode, then it would be of no use regarding the flyback effect. The current flows from drain to source during normal operation, and when I "open" the switch, the diode does not help, the MOSFET still dies.

But, on the symbol we can see, that it is not a regular diode, but a zener diode. A zener diode can conduct current in both directions without actually burning up. And I have read somewhere, that if it is a zener diode, it would help and protect the MOSFET.

So, if I have something like this:

enter image description here

where the MOSFET has the zener body diode, there is no flywheel diode, and the lamp is actually some inductive load, would that work? (Vdd = 24V, In=0.16A)

(the specific MOSFET here is an overkill, but important is the concept rather than the actual application)

My questions are the following:

  • is the body diode a zener diode? Because in the datasheet it is not specified, only the symbol represents a zener diode
  • if that is a zener diode, do I still need an additional flyback diode, or does this protect my MOSFET on its own?
  • how does it work? In the datasheet, I can see a drain-to-source breakdown voltage. Is that the voltage where the zener diode begins to conduct in reverse direction? Then it conducts the current force by the inductive load, the MOSFET is safe and sound?
  • if so, then I suppose I need to also take into consideration the inductive load's voltage tolerance, because if e.g. the breakdown voltage is 55V, then the inductive load should have no problem handling that voltage (in case of a solenoid lock, the winding isolation should be able to handle that). Is that correct?
  • I have also read that the diode is actually a byproduct during production. Do all MOSFETs have a body diode? If so, are all of them zener diodes?
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    \$\begingroup\$ All diodes will breakdown with an avalanche at some reverse voltage. Most discrete MOSFETs do have a specification for avalanche energy that can be dissipated in the reverse breakdown. Therefore, yes the MOSFET will conduct from Drain to Source when off but the voltage needed is very high (higher than the rated blocking voltage). The power dissipation is also very high, so the MOSFET will die immediately or faster than usual if you operate it in this way. \$\endgroup\$
    – tobalt
    Commented Oct 18, 2021 at 10:49

1 Answer 1

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is the body diode a zener diode? Because in the datasheet it is not specified, only the symbol represents a zener diode

No. It is an intrinsic diode due to manufacturing process, so it's not a zener, you can find the forward voltage and current characteristic in any MOSFET datasheet.

I have also read that the diode is actually a byproduct during production. Do all MOSFETs have a body diode? If so, are all of them zener diodes?

Exactly, none of them are zener.

enter image description here

The datasheet describes the body diode, more exactly it says integral p-n junction diode. It's clear that isn't zener.

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  • \$\begingroup\$ Sad... But thank you :) \$\endgroup\$
    – Marcell Juhász
    Commented Oct 18, 2021 at 12:53

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