Mfcttrf

In literature we commonly see the use of a freewheel diode to protect the circuit against inductive kick when the switch goes from closed to open (Fig 1). However what would be the pros/cons of discharging the energy to ground through say a Zener diode (Fig 2)? I have never seen this in practice and I wonder why.

Compared to (1) where the diode conducts in the forward direction, the zener diode in (2) breaks down in reverse to present a larger voltage drop to the inductor which it act as a power source when it kicks. This drains energy from the magnetic field more rapidly and therefore collapses it more quickly.

My understanding is this helps solenoids reset faster (because it kills the magnetic field holding the solenoid faster) but seems like a needless waste of energy in PWM scenarios since you're just dumping all the energy stored in the magnetic field only to immediately build it up again. Like rushing up to a red light and slamming on the brakes only to accelerate again from zero rather than coasting to the red light so you still have some momentum when the light turns green

It's frequently used in applications such as actuation of fuel injection solenoids where it's important to collapse the magnetic field as fast as possible.

Edit: See, for example, the LM1949 fuel injector driver IC which shows an external 33V 5W zener diode. Typical 'on' timing is 2.5 to 3.5ms, and a typical inductance is in the ~2mH range.

It's also possible to use an avalanche-rated MOSFET or add a zener from MOSFET drain to gate in order to cause clamping at a predetermined voltage. See, for example, TI's TPIC6C595 power shift register. One big advantage in this particular case is that it saves a pin on the IC.

Since more of the energy is dissipated in an active device, this is harder on the switching circuit. When you use a diode, most of the energy is lost in the coil resistance and only a small amount is lost in the diode.

You can also simply add a resistor in series with the normal flyback diode. Since the peak current is the same as the current prior to switching off, the peak voltage will be the supply voltage plus a diode drop plus the current times the series resistance. The diode is not necessary for the purpose of limiting the voltage, but it prevents the resistor from dissipating power while the coil is energized.

Edit: (below is a simulation of 1mH inductance with 1 ohm series resistance and 10A rectifier, shortly after 10A current is switched off)

Pink trace is diode current, cyan trace is power in the coil resistance, red trace is power in the diode. Force of a solenoid will be more-or-less proportional to the coil (and thus diode) current (pink trace).

Integrated energy for the rectifier: 7.7681mJ
Integrated energy for the coil: 41.844mJ

As a check, the total energy stored in the magnetic field is obviously 50mJ, and the two add up to 49.6mJ, pretty close. I've used a real MOSFET model for the switch

By comparison if you remove the diode and add a 36V zener across the transistor, the current drops to zero in about 350us rather than 2.5ms, a 7:1 improvement, but the zener sees a peak power of 360W and absorbs most of the stored energy.

It depends on what the coil is for and what you want it to do. For relays, a 100 ohm resistor is often placed across the coil to dampen the energy yet permits the relay to open/close quickly. If the coil is part of an actuator and the coil is pulse width modulated, then a diode is placed across it like you have shown. If you connect the zener to ground the inductive voltage will rise until the zener clamps it. Try modeling the circuit in LT Spice (it's free) and see what happens.