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So I have been researching about protecting a MOSFET. I have used MOSFETs for basic on/off switches but never really thought about fully protecting them. I have used them for on/off switches to engage relay coils. The only thing I did was place a fly back diode across the the inductive load.

Then I came across Self Protected MOSFETs. Here is a link to a diagram: https://www.google.com/search?q=self+protected+mosfet&espv=2&biw=1920&bih=955&source=lnms&tbm=isch&sa=X&ved=0ahUKEwiY7M3j0MrKAhVIVT4KHWQvD_kQ_AUIBigB#imgrc=uUMrOVnQMAn6oM%3A

I am only really interested in the ESD and Over Voltage protection sections to implement in my own designs. If I were to make a robust design to drive, say, a DC motor, here are my thoughts:

Protect The Gate: I would place a series resistor of about 100 ohms to prevent too much charge the gate layer to puncture.

Ensure MOSFET Turns Off: I would place a 100k ohm resistor between the gate and source (assuming source is tied to circuit ground or battery negative) to ensure the MOSFETs gate capacitor like behavior discharges so that if the input floats, the MOSFET turns off.

Vgs Protection: I would place a TVS diode before the 100 ohm resistor so I could catch transients. Whether it is uni or bi directional I think depends on the gates tolerance for these voltages. I am going to assume that Vgs should only be positive and thus use a unidirectional TVS. And pick it so that the clamping voltage is not beyond the absolute maximum of Vgs for the MOSFET. Reverse voltage should clamp right away.

>>>Vds Protection:<<< This is where my confusion arises. Do I need this? Most MOSFETs have a body diode and should protect from voltages due to inductors being switched off quickly. I don't like taking risks so I would use a faster diode or maybe even a TVS across drain and source and place a flyback diode on the inductor. But in the diagram I linked to, the TVS is across drain and gate. Why?

If I assume a large voltage builds up on the drain, the TVS would conduct (before max Vds I assume of the transistor), go through the gate resistor, and go to both the postive rail driving the gate and through the 100k ohm bleed resistor to ground? But doesn't this also turn on the MOSFET? This is where things get murky and despite researching the answer on the internet, I couldn't find a clear answer.

Update: Here is a circuit diagram of what is in my head per request from user.enter image description here

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    \$\begingroup\$ In single MOSFET applications, the body diode points the wrong way to do any good in suppressing the voltage spike when switching off an inductive load. For example, for an N-channel low side driver with the inductor on the drain, the resulting spike when shutting off the MOSFET will be positive not negative and the body diode will be reverse biased. The positive spike can induce avalanche breakdown through the body diode which is sometimes used if the MOSFET is repetitive avalanche rated and can handle the energy. This is more like a high-voltage zener and is sometimes drawn that way. \$\endgroup\$ – Tut Jan 27 '16 at 19:42
  • \$\begingroup\$ This makes a lot of sense, thank you for this comment. So when designers use this body diode (or a diode in general), they intend to conduct reverse biased before max Vds. \$\endgroup\$ – xxgiuzeppexx Jan 27 '16 at 20:32
  • \$\begingroup\$ Datasheets may vary, but Vds MAX is the point of avalanche breakdown. I don't often trust using this for reliability, but should you wish to do so, you need to carefully look at the datasheet and not exceed either the rated avalanche current or energy (usually spec'd for both single pulse and repetitive). Extra margin is good. More margin is better. Some designers are careless and think that the body diode automatically protects the MOSFET. I usually use a fly-back diode across the load and sometimes add a TVS diode in series with that (anode to anode) if I need faster coil turn-off. \$\endgroup\$ – Tut Jan 27 '16 at 20:59
  • \$\begingroup\$ Adding a TVS in series with the flyback diode? Why not just use the TVS instead since it responds faster? I see the regular diode as being the slower one in the chain and not react fast enough? I am viewing this wrong? \$\endgroup\$ – xxgiuzeppexx Jan 27 '16 at 23:19
  • \$\begingroup\$ The regular diode is needed to block forward current between the power supply and the MOSFET. A Schottky diode is sometimes used for speed. You could use a bi-directional TVS if it is greater than the supply voltage, but it is normally done with a uni-directional TVS diode and regular diode as I described. \$\endgroup\$ – Tut Jan 27 '16 at 23:25
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By placing a zener diode (or similar) between the drain and gate, the gate voltage will be forced to rise if the drain exceeds the breakdown voltage of the diode. You are correct in thinking that this will turn the MOSFET on. This is how that protection method works. To see why you have to understand MOSFET avalanche. The body diode of the MOSFET will break down at some voltage like a zener diode (or avalanche diode to be strictly correct). A power MOSFET is made up of many cells in parallel with each other. Depending of the processing of the transistor, these cells may or may not share the avalanche energy well. Also the place within the cell where the avalanche energy is dissipated is not quite the same as where normal conduction dissipation occurs. For this reason some MOSFETs have a limited amount of energy that they can safely absorb in a single pulse of avalanche. This energy may be much less then they could take in a pulse of normal conduction. It may be zero. On the other hand, some MOSFETs have avalanche energy limited only by thermal considerations, in other words they can take any amount of avalanche energy as long as the die does not overheat.

By turning the MOSFET on before the drain-source voltage exceeds the avalanche point, the energy is dissipated by the normal conduction mechanism. This allows a MOSFET which cannot take high avalanche energy to safely dissipate a transient, for example from and unclamped inductive load. You still have to ensure that in your design the MOSFET cannot be exposed to a transient that is too large from a thermal point of view, the energy will heat the die up just the same and you must not exceed Tj max.

You have to be careful adding diodes to the gate terminal. In some circumstances diodes connected directly to the gate can result in very high frequency oscillation of the MOSFET. This results from parasitic inductance and capacitance in the circuit causing a feedback path at VHF. A diode can rectify this and pump up the DC level on the gate resulting in a stable oscillation maintaining the device in a partially on state when trying to turn it off. This is mitigated by adding a suitable gate resistor.

If your circuit topology ensures that the MOSFETs will never see excessive voltage, or if they can withstand any transients by avalanche then you may well not need to provide any protection components. This is a good solution because there can be more problems introduced by the protection devices (too big a subject to cover here). Decoupling the supply rails near to the output devices in power stages helps. Ensuring that devices in bridge configurations cannot cross-conduct is also very important.

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  • \$\begingroup\$ How you choose a TVS across drain and gate then? When the the FET is switched off, it will act as a fast voltage source + the voltage from the source (ex battery). So we have the node at drain rising which means the TVS has one end rising, what about the other end? If the designer does not use am ESD TVS or bleed of resistor between Gate and Source, wouldn't it be at the same potential and this never conduct? In the case of the bleed resistor, it is tied to the other end of the voltage spike which means it can conduct. But I am still not sure how one would pick a TVS? \$\endgroup\$ – xxgiuzeppexx Jan 27 '16 at 20:50
  • \$\begingroup\$ Could you post an example schematic, this is quite an involved subject and it is hard to talk in general terms. The first thing to decide is if you need to add a device at all. Over voltage can be prevented by choice of MOSFET, circuit configuration, decoupling the supply, adding a capacitor or snubber network across the drain-source, controlling the rate of turn off and many other ways. The solution will be specific to the situation. \$\endgroup\$ – user1582568 Jan 27 '16 at 20:57
  • \$\begingroup\$ I see. I guess since this is just for my sake of understanding, I assume the circuit I just updated my post with. I assume a situation where the motor will have enough L x (di/dt) to produce spikes upward of a couple hundred volts. I am just not sure how one would pick a TVS. I think you first determine the maximum reverse voltage, Vwm, and then your breakdown occurs typically within 10% of Vwm. And then clamping, Vcl, should be less than the max voltage you are protecting so Vcl < Vds max. \$\endgroup\$ – xxgiuzeppexx Jan 27 '16 at 23:08
  • \$\begingroup\$ I would not use any such device. I would use a diode or another MOSFET around the motor to prevent reverse potential from appearing across the motor when the MOSFET was turned off. \$\endgroup\$ – user1582568 Jan 27 '16 at 23:44
  • \$\begingroup\$ Then why do manufacturers have this appear in their line of Self Protected MOSFETs? It can't be just a marketing gimmick. \$\endgroup\$ – xxgiuzeppexx Jan 27 '16 at 23:58

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