So I've been looking into bldc and motor drivers lately (ESCs). Generally larger the BLDC motor the larger the FETs need to be in order to suppluly needed current. One can do this either by using a higher powered FET or stacking them up (quite literally). Tesla cars use motors that are quite large. How large do the FETs need to be to run them?

# Electronic – how big are the FETs used to drive Tesla cars

motor

#### Related Solutions

You probably want a stored energy system to reduce required motor power.

Mean power can be low - if you lower the firing rate the mean power drops.

Easiest way to get consistent results is probably to have a rotating wheel with substantial mass that spins at the required circumferential velocity. If this is used to accelerate and "sling" the puck, and if Mwheel >> Mpuck then it will not slow much as the puck is accelerated to exit speed. Wheel is spun up by a motor of whatever power is desired. Larger motor = quicker recovery time. Note that if the wheel stores say 10 x the puck max energy you have a potentially lethal flywheel which would not be too hard to build safely, but care is required.

Wheel balancing machines spin the tires & wheels up with an electric motor. Some have quite a small Wattage motor and a slow run up time. Others use larger motors and faster runup. I've seen older ones with a 24 VDC motor and maybe a few 10's of Watts max power.

Spin up a wheel on such a machine to desired speed, place a sprung "floor" under the wheel, not touching but less than puck height separation, feed in a puck [from the BACK :-) ! ] and watch it vanish as it touches the tire. An old deadish wheel balancer may be available at minimal cost. Probably larger than what you had in mind :-).

**Energy & Power:**

- Liberties: The following takes some liberties with power and energy profiles when accelerating a puck. If you accelerate at constant power then velocity change is not linear. If you accelerate at constant acceleration then power input is not linear. Time to accelerate depends on power input profile and is not the simplistic figure derived below. But, the following should give a good feel for power end energy levels involved.

**Energy / Power / Acceleration / time in 'launcher' / ... :**

Given puck with mass = 100g = 0.1 kg and

velocity_max = 60 mph = 27 m/s -> say 30 m/s then

Puck energy \$(E_K) = \frac{1}{2} \times m \times v^2 = 0.5 \times 0.1 \times 30^2 = 45 Joule = 45 Watt seconds. \$

(\$ E_K \$*=kinetic energy, m= mass of the object in kg, v= speed of object in m/s).*

**Linear motor**: If you accelerated it at constant power over a say 1 metre linear track them

Vmean = (30-0)/2 = 15 m/s so time to accelerate = 1/15s = 67 mS.

As puck energy = 45 W.s and you are delivering this in 1/15s the power during acceleration is ~= 45 Ws / (1/15s) = 675 Watts.

That's more Watts than you'd like to need in a linear launcher if you are driving it directly electrically.

**"Crossbow":** Instead you can wind up a "crossbow" type mechanism where a spring or torsion bar or whatever is wound up to store the desired amount of energy and then "tripped".

For a potential firing rate of 1/second you need 45 Watt of energy to storage transfer rate - say about double this to get electrical motor input or about 100W.

For a 5 second repeat rate you are down to 20W and 10W for 10 second cycle time.

**"Putter:"** A 10 Watt motor winding up a spring via a reduction box or screw thread or ...? is easy to build and play with. It would (probably) be easy to pull back a weight on a pivot (rigid shaft pendulum) and let it go so it swings and strikes the puck, golf putter style. Potential energy in a mass is mgh

or about 10 x kg x height.

You want 45 W.s max.
10 kg x 0.5 metre x 10 = 50 W.s

Height is the vertical height above rest level.

Velocities are now wrong and you are going to need to deal with impulse energy transfer, but it should be in the order of right.

**Rocket:** A potentially workable system would be to use compressed air and a pressure reservoir.
A completely DIY system could use a "water rocket" type launcher as built by amateurs world wide. Liable to be somewhat noisy [tm] on release.

**Treadmill motor** / Direct drive rotary flinger: A potentially excellent excellent motor for a flinger is a treadmill motor. These are typically rate in the 0.5 to "several" HP range and 200 VDC permanent magnet units powered from 230 VAC rectified mains (and no doubt the 110 VAC based versions in the US) are common. Speed control by PWM or other voltage variation allowing wide range speed control. An arbitrarily large disk could be attached directly to suit the motors preferred RPM maximum. eg an 1800 RPM motor = 30 RPS will need a 1 metre circumference disk to deliver 30 m/S = a diameter of 318 mm or about 1 foot.

FET gates MUST NOT float.

Nothing can be guaranteed in that state.

Miller capacitance will happily couple large drive signals onto the gate from drain transients. A gate driven above its Vgsmax value will often enough puncture the gate oxide and any combination of hard shorts between GDS can result. I have seen DS short with G open, GS short with D open, GDS all short and perhaps GD short with S open but I'd not be 100% sure of that.

For ANY power FET with an inductive load I add a GS zener mounted as close to the FET as possible, with a voltage rating above Vgs_drive_max and comfortably below VGS_abs_max. This transforms circuits which fail in minutes to hours into circuits which fail never.

## Best Answer

A little googling suggests these are the parts IXGX72N60