I am not 100% sure how the current flows through flyback diodes when the motor turns off. Path seems weird because the current is flowing from ground to Vcc. From low potential to high potential. But i don't see any other path for the current. Did I draw the current path right?

 

Whichever direction the current is flowing in the inductor (in this case the motor's winding), that's the direction the current will flow through the inductor when then circuit opens.

 

Whichever direction the current is flowing in the inductor (in this case the motor's winding), that's the direction the current will flow through the inductor when then circuit opens.

- I know that. Current path just seems weird to me. Current is basically flowing from ground to Vcc through diodes. From low potential to high potential.
- But I guess the current path is right since you didn't comment on it?

 

When the current through the motor inductance is interrupted, the voltage on the motor terminals rises until the diodes conduct.
In this case 5V plus 2 diode drops, or about 6.2 Volts.

If the motor is spun externally, the diodes also conduct to direct the back EMF to the power rails.

 

- I know that. Current path just seems weird to me. Current is basically flowing from ground to Vcc through diodes. From low potential to high potential.
- But I guess the current path is right since you didn't comment on it?

I have named the diodes in the circuit above, starting with the grounded anode and going clockwise: D11, D1, D5, D7 (numerals on a clock face).

If while running, the connection of D11 and D1 is positive, then connection of D5 and D7 will be at ground. When the power is disconnected, the cathode diode D11 goes to one diode drop above the positive power supply rail while D5's cathode goes to one diode drop below ground.

 

I think I understand now. Motor becomes a "power supply" when you cut off the power. Thanks for all the answers.