Over-voltage protection for mini-computer
Suppose the relay interrupts the motor current, 20 amperes, in 20 nanoseconds.

Suppose there is a wire running from the relay to the motor controller, and that wire couples into the RPi regulated power; the wiring between regulator and RPI forms a loop with VDD and RTN.  &lt;- EDIT







simulate this circuit &ndash; Schematic created using CircuitLab

Suppose the distance from wire to loop is 1cm. Suppose the loop area is 1cm by 10cm (4"). What voltage is induced, on top of the regulated RPi VDD?

$$Vinduce = [MU0 * MUr * Area/(2 * pi * Distance)] / dI/dT$$

What is the induced voltage? anything higher than 1 volt transient is bad news.
Now insert value for MU0 of 4 * pi * 1e-7

Vinduce = 2e-7 * area/distance * dI/dT

Vinduce = 2e-7 * 1cm*10cm/1cm * 1amp/nanosecond

Vinduce = 2e-7 * 10cm * 10^+9 amp/second = 2e-7 * 0.1 * 1e+9

Vinduce = 2e+1 = 20 volts induced



To avoid this (e.g. reduce the induced voltage from 20 volts to 0.2 volts), 

(1) keep the MCU PCB a meter away from the high-current fast-off motor wires (hot and return wires).

(2) place a steel plate adjacent to the MCU PCB, insulated to prevent shorts; have the plate be larger than the MCU PCB

(3) insert a PI filter (C-L-C) at the MCU PCB, as shown







simulate this circuit

The one Ohm across inductor is for dampening. Rdampen = sqrt(L/C) =sqrt(1mh/0.1mF) = sqrt(100 = 3.16 ohms ideal. The 1 ohm is over dampening.