dc motordiodesflybackmosfet
I'm trying to determine a suitable flyback diode for a DC motor controlled by a N-channel MOSFET. The specs the motor read:
Rated current: 2 A
200 mA no-load current draw @ 12 V
12 A stall current @ 12 V
No load speed of 5,500 RPM
The MOSTFET I have is IRF520.
Should the flyback diode have a "Maximum average forward rectified current" above 2A, 200mA or 12A?
(EDIT: The motor will only be spinning in one direction, so there is no H-bridge.)
In your first example, before the switch opens, there is a current flowing through L. When the switch opens the current wants to continue flowing in the same direction so this has to create a positive voltage on the right-hand side of L relative to the left hand side (which is fixed at +Vs).
This positive voltage causes current to continue to flow and this current finds the path of least resistance through diode D. It continues to flow until all the magnetic energy in the inductor is gone then current = 0 and the voltage on the right hand side of L is the same as the left hand side i.e. +Vs.
In the 2nd example with the H bridge, all the diodes are needed because it has to cater for scenarios when the motor is deactivated i.e. pin 3 and pin 6 are open circuit. This is the simplest way to explain it I think. The motor current may have previously been upwards or downwards so all four diodes are needed to catch the back emf when the motor is switched off.
Also, current flow back into the H-Bridge circuit does not appear very practical.
That's how they work.
It's not unusual to find that under near-full-power-out conditions the motor will be about 50% efficient. This is a big motor but it still applies: -
Next, you estimate how much mechanical power you need (2 pi n T) where n is revs per second and T is torque in newton.metres. If your dc motor's data sheet doesn't provide this information (like the graph above) then get one that does.
If you don't know how to calculate your mechanical power requirements you need to learn and start with: -
Force = mass x acceleration courtesy of Isaac Newton
Mass is what you want to move and acceleration is final speed divided by time to get to that final speed (i.e. what you want it to do).
Force x distance = work done and work done per second = power.
Don't forget friction as a loss and if it moves fast then there will be other losses due to moving air around.
Best Answer
At turnoff the flyback diode will momentarily carry the same current as your motor is consuming. This current will then reduce quickly depending on the inductance and resistance of the motor.
You will find a rating in the diode data sheets for non repetitive surge current normally about a 10ms rating. The inductive spike at turnoff should not last anywhere near this long so that so as long as the surge rating exceeds your maximum motor current (12A) you will be good to go.
The maximum average current of the diode is unimportant unless you are repeatedly turning off the motor rapidly say for PWM. In this case use the average current the motor is taking.
The IRF520 is not rated at 12A which is the current you will get at startup or if you stall the motor. You should find a mosfet rated at least at 15A.