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4th of July Fireworks

A project log for OpenBLDC

BLDC shield for arduino and stand alone controller

nerd.kingnerd.king 07/07/2014 at 13:351 Comment

So I hauled a bunch of my stuff to my parents for the holiday, much to my wife's dismay.  My goal was to see if I could get the motor spinning open loop control.  I brought a wall wart that output 14V@0.5A and I used three 1k resistors as the load.  I cycled through the commutation phases and could see some voltage on each resistor.  I changed the delay to 1 sec from 100msec and I could see 14V jumping from phase to phase.  I changed the wall wart for the massive 24v@20A supply and I verified that the 24V was moving correctly.  

I changed the timing back to 100msec and attached the motor.  The motor did not move so I measured the voltage between two of the leads.  As I did this one of the drivers popped, thus ending my testing.  I really wish I had a scope to verify the faster timing voltages.  I might try to see if I can use one at work, after I swap out the toasted chip.  I am at a bit of a loss by the driver fried,  I wouldn't think that the gate of the FET is pulling very much current.  I am also wondering if there is a problem with the motor.  I got it for free and I am thinking of purchasing a small motor just to test and develop the code.  One of the secondary goals is to create a motor profile that can be loaded based on the type of motor being used.

Discussions

K.C. Lee wrote 07/15/2014 at 20:42 point
> I wouldn't think that the gate of the FET is pulling very much current.

You'll be quite surprised at how much gate current a beefy MOSFET actually needs.

Datasheet shows that STB60NF10 has input capacitance (Ciss) of 4270pF (huge!) I would normally use a driver rated 3-6A for driving that size.

see Microchip Appnote: AN799 "Matching MOSFET Drivers to MOSFETs" for sizing your gate drivers.
http://www.microchip.com/stellent/idcplg?IdcService=SS_GET_PAGE&nodeId=1824&appnote=en020629

It is just a guideline. The current is highly dependent on the rise/fall time to charge/discharge the gate capacitance. Q=CV -> I*dt = C * dV

Hope this points you to the right direction.

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