Electrical – Do Protected Power MOSFETs need a flyback diodes

You do need the protection diodes. When you activate a solenoid, it creates a magnetic field. This is what causes the mechanical action. This magnetic field also represents stored energy. How much? Well, a solenoid is an inductor. The energy \$E\$ stored in an inductance \$L\$ with current \$I\$ flowing through it is:

$$ E = \frac{1}{2}LI^2 $$

When you interrupt the current by opening the switch or turning off the MOSFET, the magnetic field goes away. But if it's stored energy, it can't just vanish. It must be converted into an equal amount of energy somewhere else.

Another property of inductors is that the current through them can't change instantly. It changes at a rate proportional to the voltage applied to them:

$$ \frac{dI}{dt} = \frac{V}{L} $$

If the inductor can't find a place to keep the current flowing, then it will make one, perhaps by making a spark across the switch or frying a transistor.

The diode is there so that the inductor only needs to make 0.6V to keep the current flowing. The current can then flow in a loop around the inductor and the diode until all the energy has been converted to heat by the resistance of the diode and wire in the inductor.

^{simulate this circuit – Schematic created using CircuitLab}

If you close SW1 for some time, a large current will be flowing in L1, limited only by its internal resistance. When you open the switch, that current can keep flowing through D1. The voltage across the inductor will be about 0.6V, because this is what it takes to forward-bias a silicon diode, and the current will die down at a rate given by the 2nd equation above.

You can not share one diode like this among several inductors. The point of the diode is to give each inductor a place to dump its stored energy that doesn't affect something else. If you are sharing diodes then you aren't doing this.

There seems to be another problem with your circuit: the gate of Q1 is floating when J1 is open. That is, it isn't connected to anything. This will make it very sensitive to stray electric fields (like, the really big one set up when S2 arcs across J2 because there's no diode). I bet you can also get it to turn on and off randomly by touching it with your finger. Add a resistor of maybe \$10k\Omega\$ so that it's either definitely on or off (it's hard to tell which you intended) when the switch is open.

First of all please note that there is no industry standard for doing avalanche rating, and that often avalanche rated FETs are tested under rather unrealistic conditions (e.g. single pulse only whereas using them in repetitive will cause self destruction). However characterizing and hardening it is sometimes not much harder, in fact there are some transistors that are not rated at all but known for having usable avalanche behaviour.

For the classic scenario of driving something inductive, I can imagine multiple reasons why you use such a transistor:

• A bit of extra safety for when the flywheel diode fails.
• Your chosen transistor has it anyways already and you can shave off costs by omitting the diode
• You are so space constrained that you can't put a diode there.
• You have other sources of voltage spikes that you have little to no control over (e.g. surrounding engine ignition systems) but want to be able to cope with.

As such it doesn't often seem to be a base of a good design, but there is a market for it, and so people cater this market.