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How is an active output clamp implemented on a high-side MOSFET switch?

If there are several possibilities, then I am focused on how it's implemented on Freescale's MC15XS3400.

I can wrap my head around how it's done on a low-side switch, but the high-side has me stumped. With reference to the schematic below (of a low-side switch), when the MOSFET is turned off, the source voltage overcomes the TVS and drives current through \$R_1\$ leading to \$V_g > 0 \$ and opening the MOSFET's channel a bit to allow the inductor to demagnetize (or, at least, that's how I understand it).

The reason I included this explanation/schematic is to highlight my expectation that there be some sort of resistor near the gate perhaps which will pull-up the voltage when current flows through the TVS, wherever it happens to go on the high-side version.

schematic

simulate this circuit – Schematic created using CircuitLab

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Simplest way is to put something like a zener diode structure from the drain to the gate. That way when the zener diode voltage is exceeded by a bit the MOSFET turns on and clamps the voltage. Something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

If I read the datasheet correctly, the part you mention clamps at about -22 to -16V with a supply voltage 6~20V, so there's probably something a bit different going on- the zener is changing a logic level relative to ground (before a level shifter).

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  • \$\begingroup\$ When the MOSFET turns L1 off, Vs should see a large negative voltage spike which causes current to flow down through the diode, but shouldn't the current that's coming through R1 down to the diode drop the voltage at the gate to a negative value such that the gate doesn't open? (My apologies, I think I may have accidentally added rough-draft comments.) \$\endgroup\$ – Gerhard Oct 29 '14 at 16:22
  • \$\begingroup\$ @Gerhard R1 would be picked so that the current is low compared to the load current. If the current through R1 keeps the gate from turning on, then it is doing the clamping, but typically that wouldn't be enough and you want the MOSFET to turn on and absorb the energy. \$\endgroup\$ – Spehro Pefhany Oct 29 '14 at 17:41
  • \$\begingroup\$ I hate to be a blockhead, but I don't understand how a positive gate voltage is generated at the gate when source voltage drops very low. \$\endgroup\$ – Gerhard Oct 30 '14 at 14:25
  • \$\begingroup\$ @Gerhard The p-channel MOSFET turns on when the gate is pulled below the source by more than a few volts. \$\endgroup\$ – Spehro Pefhany Oct 30 '14 at 16:03
  • \$\begingroup\$ I checked the specification, and on pg. 3, they show an enhancement-type N-channel MOSFET, which they are explicit about on pg. 25. That's too bad because I was hoping we had the problem figured out. \$\endgroup\$ – Gerhard Oct 30 '14 at 18:13
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My confusion comes as a result of not appreciating that the MOSFET gate voltage is relative to the source.

Therefore, I believe the correct way to clamp an inductive load on a high-side switch is to put a Zener from the gate to ground, such that the gate-driver output can't drag it's voltage down with the source in order to keep \$V_{GS}\leqslant V_{GS(TH)}\$ and the channel closed.

In @Sphero's schematic, I would first change the MOSFET to an n-channel type, then disconnect the TVS from the load and set it to ground.

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