If using mosfets like IRLB8721PbF , they already have a protection diode built in.
What device is actually in use?

yes it has a built-in diode (IRLB4132PBF). I don't understand what the body diode is for and never did. A freewheel diode goes across the load, not the fet. So what do it do?

 

The theory is that the inverse connected diode suppresses the inductive spike produced as the current is instantly cut off. The concept is thus that the diode protects connected devices. But if the current is not instantly cut off then the voltage spike is not so very large, and thus not a problem. And in a mosfet, it is difficult to develop much voltage spike across a forward biased diode.

 

A freewheel diode is connected across a motor (/ inductive load) to allow the current to decay, when the 'supply' is switched off.
A motor was classically supplied by thyristors from an AC voltage supply, providing a variable voltage - hence speed.
When the thyristor(s) stopped conducting, the flywheel diode provided a path for the decaying motor current.
There was no path back through the thyristor into the mains AC supply for that current.

With a battery supply, (ie. DC), the fet(s) are always connected to the motor.
Of course, the nature of PWM, is devices switching on and off rapidly, so the in built diode has a purpose.
If the fet is switched off, there is a path via the inbuilt diode for the decaying current.
There is likely a capacitor across the supply prior to the fet, so if the battery is actually disconnected by a switch, the current still has a path through the inbuild diode to the capaitor.

Have a closer look at the hand tool trigger switch.
Does it switch the main current path to the motor or just inhibit the electronics.
It would be a much cheaper switch if it only has to switch a signal to the controller.
The controller will then look after how the voltage is reduced to zero.

 

Any tool (drill driver for instance) that is reversible cannot have a freewheeling diode.

 

Any tool (drill driver for instance) that is reversible cannot have a freewheeling diode.

That's true, as a separate component.
But an H fet bridge, all having built in diodes would provide the freewheeling function in a reversing circuit.
But I think most tools use a gearbox selection to do this usually, don't they?

 

if the current is not instantly cut off then the voltage spike is not so very large, and thus not a problem. And in a mosfet, it is difficult to develop much voltage spike across a forward biased diode.

that's what i was wondering about. If the pwm control has low frequency, like 1kHz, then maybe it could have a gate resistor to slow the slew and suppress the spike. If you're a product engineer, you know that a diode is at least 10x the price of a resistor.

 

If the fet is switched off, there is a path via the inbuilt diode for the decaying current.

that's the part i don't understand. The voltage spike from the motor is in the same direction as the driving voltage. So how is the built-in diode gonna pass that current. It's wired in the opposite direction.

 

That's true, as a separate component.
But an H fet bridge, all having built in diodes would provide the freewheeling function in a reversing circuit.
But I think most tools use a gearbox selection to do this usually, don't they?

no the gear selection on a drill is for high n low speed. They all have bridges for reverse direction.

 

There's virtually no cost for a diode if it's included on the fet silicon chip.

The voltage spike is in the opposite direction to the running direction. Check out how the diode is connected.

 

The voltage spike is in the opposite direction to the running direction. Check out how the diode is connected.

ok this is frustrating. According to https://www.allaboutcircuits.com/textbook/direct-current/chpt-15/inductors-and-calculus/, "a rapid [decrease] in current will induce a high voltage across the inductor, oriented with negative on the left and positive on the right, to oppose this decrease in current"
So far so good. An induced voltage oriented from the left (end of the inductor) to the right is a voltage that's the same orientation as the driving voltage, it has to be if it is going to oppose the decrease in current. But then the article says that v = L di/dt, and if di/dt is negative (decrease in current), then the induced voltage must have the opposite orientation as the driving voltage. But then if the the induced voltage is opposing the decrease in current, that would mean that v = - L di/dt. WTF?

 

That's correct isn't it.
The voltage is -ve.
That's why the flywheel diode conducts and allows the current to decay to zero. (When the flux has collapsed to zero).

 

Another scheme is to put a capacitor across the coil. The cap becomes charged and then discharges thru the resistance of the coil.

 

That's correct isn't it.
The voltage is -ve.

ok that makes sense, there must have been a typo

 

Another scheme is to put a capacitor across the coil. The cap becomes charged and then discharges thru the resistance of the coil.

yeah there's many ways to skin a cat. But if the body diode provides protection for the mosfet then it looks like that's all i need. THANX GUYS