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In the schematic below, I would like to know the best way to dissipate the FreeWheel Current in the very shortest time possible.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ What is the inductance of L1? Have you considered the use of a flyback diode rated for the necessary transient current? \$\endgroup\$ – nanofarad May 18 '16 at 22:33
  • \$\begingroup\$ Are you protecting M1 with a clamp diode as well? \$\endgroup\$ – user105652 May 18 '16 at 22:35
  • \$\begingroup\$ I do not have that data. However, although L1 is not a motor, it should approximate that of a motor that would draw that amount of current. \$\endgroup\$ – Dave May 18 '16 at 22:42
  • \$\begingroup\$ The main idea is to reduce the flyback current in the fastset possible time. \$\endgroup\$ – Dave May 18 '16 at 22:44
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    \$\begingroup\$ Use another MOSFET that gets slapped open when the first gets turned off. \$\endgroup\$ – Ignacio Vazquez-Abrams May 18 '16 at 22:49
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The fastest current decay will be achieve when you allow the coil voltage to rise to the maximum possible voltage. That is also the hardest on the transistor.

Possible approaches are a resistor and series diode or a Zener clamp or even an avalanche- rated MOSFET.

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  • \$\begingroup\$ Spehro Pefhany, thank you very much for taking the time to answer my question. \$\endgroup\$ – Dave May 23 '16 at 2:37
  • \$\begingroup\$ Spehro, if 20 amps was flowing in the coil when the MOSFET was turned off, what wattage the zener Diode needs to be? \$\endgroup\$ – Dave May 23 '16 at 18:32
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The only way to stop a current in an inductor is to apply a reverse voltage. When you open-circuit an inductor that has current flowing through it, the inductance itself provides that reverse voltage.

So, stopping the current in the minumum time really boils down to just how much reverse voltage the rest of the circuit can tolerate. For example, you could use a 200-V MOSFET and put a 150-V zener diode across the coil, which will cause the current to ramp down 6× faster than it would if you used a 60-V MOSFET and a 25-V zener.

schematic

simulate this circuit – Schematic created using CircuitLab

The regular diode in series with the zener is required to prevent the zener from passing current in the forward direction when the MOSFET is on. Note also that the MOSFET experiences a voltage pulse that is equal to the supply voltage PLUS the zener voltage. You can also replace the zener with a power resistor, assuming you have a pretty good idea what the worst-case inductor current is.

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  • \$\begingroup\$ Dave, thank you very much for taking the time to provide this detailed answer. \$\endgroup\$ – Dave May 23 '16 at 2:37
  • \$\begingroup\$ Dave, if say, 20 amps was flowing in the coil when the MOSFET was turned off, what wattage the zener Diode needs to be? \$\endgroup\$ – Dave May 23 '16 at 18:30
  • \$\begingroup\$ It's impossible to say without knowing the zener voltage or the value of the inductance. \$\endgroup\$ – Dave Tweed May 23 '16 at 19:20
  • \$\begingroup\$ I was told that I could use a 175 volt Zener; I was thinking of using that value. \$\endgroup\$ – Dave May 23 '16 at 20:35
  • \$\begingroup\$ OK, since you still haven't said what the inductance is, we'll treat that as a variable, \$L\$. The current ramps down from 30A at a rate of \$\frac{175 V}{L}\$ amperes per second, taking \$\frac{30 A\cdot L}{175 V}\$ seconds to get to zero. The instantaneous power at the start is \$175 V \cdot 30 A = 5250 W\$, and of course, at the end it's zero. So the average power during the pulse is 2625 W and the total energy is the power times the time: \$2625 W \frac{30 A\cdot L}{175 V} = 450\cdot L \$joules. The long-term average power depends on how often you switch the MOSFET. \$\endgroup\$ – Dave Tweed May 23 '16 at 21:18

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