Controlling a motor

Ian0

Joined Aug 7, 2020
13,132
The power that the FET is dissipating is governed by Ohm's law. The power form is simply:

P = I2R​

But what's R? That would be the FET's resistance between the drain and source or Rds(on), the datasheet lists it as 5.3 milliOhms.
somebody please tell Mr. McManis about switching losses.
 

CBTech_Joe

Joined Sep 9, 2020
39
I came here to discuss using an arduino output to a vfd set up in sensorless vector to get pretty close positional feedback, but it seems like we need more information from OP about what his intentions actually are.
 

Ian0

Joined Aug 7, 2020
13,132
I think he does mention it in one of his other pages. Can you give a better real world explanation than he does?
I could probably paraphrase IR’s application note AN-949 from the 1982 Hexfet data book, or the relevant section of Horowitz and Hill equally well if I wanted to.
 

Irving

Joined Jan 30, 2016
5,122
somebody please tell Mr. McManis about switching losses.
Its a valid point, but switching losses are a different issue from conduction losses. Generally, at least in my experience, for a motor controller at Fsw = 10kHz, they are usually < 10% of the conduction losses so I'll cut him some slack on that. They are a much bigger issue for high-frequency switching regulators with Fsw at 50kHz up to a few MHz where they can be significant.

For his design, using 8 off the specified MOSFET in parallel and assuming a 10kHz switching frequency, the gate driver and gate losses + the turn on & turn off switching losses are around 1.4W v 58W for the conduction losses. Total losses for the 8 MOSFET in both the high and low side switch are 2 x (58 + 1.4) = ~120W. This is where I do have an issue with his calculations. He works out the heatsink thermal resistance based on a single MOSFET (i.e. one heatsink per MOSFET), but actually there will be 16 MOSFET conducting at any one time - his assumption is that its a simple multiplier but in reality they will all be on the one heatsink and it rarely is a simple multiplier because of hotspots and airflow shadowing.
 

Ian0

Joined Aug 7, 2020
13,132
Ah yes. The trade off between iron losses and the customer complaining about the whistling.
Up at 100k I’ve come across a situation where the lowest Rdson MOSFET isn’t the coolest, due to switching losses being lower on the smaller device.
My method for multiple transistors on one heatsink is to assume that the heat sink is cut up in to individual sections, one per transistor, and calculate from the manufacturer’s data for the shorter heatsink.
Of course, the manufacturer would like to get the best figures and makes the measurements assuming that the short heatsink isn‘t right next to a similar and equally warm heatsink, but I think it’s a closer approximation.
 
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