# Why isn't fast decay mode on an h bridge damaging

I'm struggling to understand how fast decay mode works on an h-bridge used for stepper motors . Particularly I've been looking at fast decay mode in this data sheet: http://www.ti.com/lit/an/slva321/slva321.pdf

What I'm struggling to understand is why when BH and AL are turned on, this doesn't cause damage due to the voltage that the motor coil would wind up putting out as the magnetic field collapsed. As far as I can understand it if we take V(motor-coil) = L * (change in current) / (change in time) if we have a stepper motor coil that is holding at 1A and then we start a step therefore switching which transistors are one, and say we turn off our transistors in 10ns (I just grabbed that from a random transistor datasheet) then wouldn't the voltage across the motor coil wind up being:

V(mc) = L * (0-1) / .000000001

V(mc) = L * -1000000000

So even if the inductance is something really small like 3 mH (taken from random NEMA 17 pololou data sheet) that would still cause a few thousand volt difference across the coil no?

If I look at the diagrams from that data sheet linked above, wouldn't that voltage be applied directly to the input voltage source and ground?

How does this not cause arcing or damage to components that may be up or down stream of this h bridge?

• That is why we use clamp diodes on DC motor drivers and some very high powered audio amps. They absorb the strong back-EMF by dumping it into the supply rails. Their massive capacitors absorb the brief over voltage. – Sparky256 Mar 17 '19 at 3:05