1
\$\begingroup\$

I have a bank of x6 MOSFETs (SUG90090E) connected in parallel, which I am using to switch a current of maximum 30A through a coil (resistance 1R and inductance 1mH), and am looking for some general advice with regards to the "flyback" section of the circuit. The current will not be switched at high frequency - only once per second or so. My schematic at present looks as below (each MOSFET has its own dedicated driver chip): enter image description here

My goal is to have the current reduce to zero in several tens of microseconds. Based on pieces of designs I have seen elsewhere, I have tried to create an arrangement of components which I believe may do the job. The idea is that when the MOSFETs are switched off, the large induced voltage that appears at the bottom of the load is allowed to rise until it reaches the clamping voltage of the TVS diode D1. When this happens, the TVS allows a small amount of current to flow to the base of Q7, switching it on, and allowing the large circulating current to dissipate away. As far as I understand, the fact that a large voltage is allowed to build up means that the rate of decrese of current can be greater, due to V = -L*dI/dt, when compared to the case of using a simple single "flyback diode" (which would allow it to build up only to a diode drop or so).

I have tried to choose the components suitably - for example, the clamping voltage of the TVS diode 1.5KE100A is maximum 178V: enter image description here and so this should be enough to protect the SUG90090E MOSFETs, which have a VDS(max) of 200V. I have also chosen the transistor Q7 BUF420AW and diode D2 IDW100E60 such that their forward currents are at least greater than the steady state current through the load when switched on (their max rated values are 30A and 150A, respectively). I have added a small capacitor C1, but am sure if it will make any difference in practice?

My question is whether this is generally a reasonable approach to switching off the current fast in this type of situation, and am looking for reassurance that it's wise to continue down this route? Are the components chosen suitably enough, or is there something that I have not accounted for? Any advice or tips would be helpful.

With thanks.


EDIT

At the suggestion of @Bimpelrekkie , I have tried to make a simplifed LTSpice model of this situation. The simulation also compares the case when the transistor Q7 is removed, leaving only a TVS diode across the coil (as suggested by @Andyaka).

The simulated schematic is the following: enter image description here

The the resulting currents and voltage spikes just after switch-off: enter image description here

There is little difference between the two cases - the current is reduced from 35A down to zero in ~250us, and the back EMF spikes are capped below safely below the 200V rating of the SUG90090E MOSFET. The difference is that in the case with the BJT, it is this transistor which absorbs the power of P = I*V = (170V - 35V)*35A ~ 4.7kW (peak value, when current is maximum). With only a single TVS, then it is the TVS which takes the 4.4kW peak.

enter image description here

I feel now that @Andyaka suggestion is reasonable, and as long as I choose a TVS which is able to handle this power then it could be the only snubber element necessary.

\$\endgroup\$
  • 3
    \$\begingroup\$ For me the next step would be to put this design in a circuit simulator, for example LTSpice and check that it behaves as you think it should. You can't break anything in a simulator but for sure something can break when you build this due to the high currents and amount of energy in the coil. Don't forget about the parasitic resistances of wires, PCB (if any) and components! The more problems you can solve before building, the better. \$\endgroup\$ – Bimpelrekkie Feb 20 at 13:27
  • \$\begingroup\$ Why not just use a TVS across the coil and forget about the transistor? \$\endgroup\$ – Andy aka Feb 20 at 15:58
  • \$\begingroup\$ @Andyaka, the idea was that it might manage the power dissipated during the flyback pulse better. The way I've drawn it, the BJT takes the power, rather than the little zener. Do you know of a disadvantage to doing it this way (aside from there being an extra component)? \$\endgroup\$ – teeeeee Feb 20 at 16:57
  • \$\begingroup\$ @Bimpelrekkie, thanks - I already verified that the basic idea works in LTSpice using the built-in components (I struggled to get the exact manufacturer's models working correctly though, due to my inexperience with SPICE). I was more worried about something that I missed that could kill one of the components, which I might not know how to spot or what to look out for in the simulation, due to lack of intuition for this sort of thing. \$\endgroup\$ – teeeeee Feb 20 at 17:05
  • 1
    \$\begingroup\$ Power is power and turns to heat one way or the other. Please justify why you think a single TVS diode connected from drain to source would give problems. \$\endgroup\$ – Andy aka Feb 21 at 8:18
1
\$\begingroup\$

I would just place a TVS diode from drain to source of the mosfets, that makes it easier to match to the maximum ratings of the FETs and removes the need of a series diode.

If power dissipation is an issue, either add the transistor or use multiple TVS diodes in series with resistors (see below) in parallel.

Something that you might not have considered is power sharing among MOSFEts. Depending on where you are in their gm/temperature curve you could enter a positive-feedback regime in which the hottest one takes an increasing portion of the current. This quickly leads to a cascade of failures that destroys all of them.

A small added source resistance can remove this possibility. The same is true for multiple TVSs, the same resistance can work for both.

schematic

simulate this circuit – Schematic created using CircuitLab

But. As you are talking of tens of microseconds you would need to let the voltage build to ~1kV. This would require separating the protection of the MOSFEts from the voltage build-up on the inductor, which is not easy to do. It is much more practical to simply increase the maximum drain-gate voltage of the existing Mosfets (>800V Mosfets are not that uncommon nor terribly expensive)

An example of how to isolate the drain of the mosfets (which requires quite a few extra components and an auxiliary winding in the solenoid to simplify their gate drive), is shown in the following schematic. Note that this is not a practical circuit, as (1) it will clearly cause breakdown on the gate-source junction of the PFETs, and (2) it more than doubles the number of required components, as it places them in series with the main current.

schematic

simulate this circuit

A more practical circuit, but still greatly increasing the number of components is below. Note that the gate drive resistors are taking a considerable portion of the power dissipation:

schematic

simulate this circuit

\$\endgroup\$
  • \$\begingroup\$ Thanks for the comments - would you mind including a few simple sketches to illustrate the various arrangements of components, so I can understand better what you are suggesting? \$\endgroup\$ – teeeeee Feb 21 at 23:18
-1
\$\begingroup\$

I would have put a Schottky diode reverse biased across the coil. The very fast switching time and the low Vf are both advantageous. The low cycling rate limits the amount of heat that can build up. So all you need to do is make sure that the max pulse voltage does not exceed the diode’s rating.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.