fets

[hippiehunter]

im more then a little bit confused as to why we have so many fets in our brushless controllers. looking on digikey the spec for an IRFB4110 says 100V 180Amp cont. Why on earth do we need more then 3 of those, one for each phase? is it simply a matter of cooling? if so then why not mount a badass heatsink/cooler? sorry for my noobishness and thank you in advance for taking pity on my non electrically minded self.

 

[Anonymous]

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[Link]

Yeah, the package itself is rated for 70A. And when you factor in the current multiplication effect of PWM, you can't get even that through the system.

Plus, 180A is when they're completely on. When they're in a controller being PWM'd, they're only partially conductive (high resistance) for a bit of time when they're turning on and off. This heats them up WAY more than just full throttle does.

 

[dirty_d]

you NEED 2 for every phase for a 3 phase motor so an abosulute minimum of 6 fets, there are usually 12 because theyre paralleled up. i dunno for 35A i dont think you really need 12 fets. thats only like 6W resistance loss per fet. it would half the losses though.

 

[hippiehunter]

why do you need 2 for each phase? also I was looking around on mouser http://www.mouser.com/Search/ProductDetail.aspx?qs=Xjc0wECptJCAktBRj/IsyQ== From what i can tell that should be a much more effective FET. Its a much faster switch and its got lower resistance as far as i can tell. Is there something I am missing or would this be a good fet for controller experimentation.

 

[pwbset]

you NEED 2 for every phase for a 3 phase motor so an abosulute minimum of 6 fets, there are usually 12 because theyre paralleled up. i dunno for 35A i dont think you really need 12 fets. thats only like 6W resistance loss per fet. it would half the losses though.

My pedal first 72v20a Clyte has 6 4110s in it and since I filled the shunt with solder that controller vibrates/shudders like crazy when I floor it. That didn't happen before at the default 20A. I'm seeing 55-60A spikes on my CA now. I've already re-hot-glued the caps back onto the board. We'll see how long things last.

 

[Toorbough ULL-Zeveigh]

because max volt & current ratings are 'either/or' not 'and/both at the same time'.
for how much current the device can pass at a given voltage you have to consult the power dissipation curves to make sure you fall within the 'safe-operating-area'.
That depends entirely on the level of heat sinking such as if there's passive or active cooling.

going from memory here IIRC the 4110 is a 370 Watt device with little less than half the on resistance, compared to the IPP08CN10N you linked to can only handle 167W.
as a thumbnail calculation with an infinite heatsink the 4110 with 100V across it can handle 3.7A.
alternately if you need to switch 180 amps you can't exceed about 2 volts across the FET (2*180=360W).
If you need more amps at a given voltage than a single device can dissipate then you need to parallel up to divide the load.

 

[Jeremy Harris]

You need 6 FETs minimum as each phase needs one FET to pull it high and another FET to pull it low. At no time are both FETs on any phase on as that would short the supply.

In the main, it's not current or power ratings that dominate controller device selection, but on resistance. Those FETs you linked to are pretty poor, as they have an on resistance that's more than double that of IRFB4110s. 8.5mohms is not particularly wonderful if you want to reduce controller losses to a minimum.

IRFB4110s are pretty good at around 3.5 to 4mohms or thereabouts, but if you're using a motor with a low winding resistance then that's still a significant loss. For example, I've just bought a couple of motors that have a 32mohm phase winding resistance, so with a simple controller with just 6 FETs I could lose about 11% in the controller FETs at high power, zero speed (the "launch" condition). If I were to use 12 FETs then the loss comes down to a more reasonable 5.5%; using 18 FETs gets that loss down to under 4%.

Losses at speed and with partial throttle would be a lot lower, but this illustrates why paralleling FETs can be a good thing, apart from the package current and power limitations previously mentioned.

Jeremy

 

[hippiehunter]

The reason i posted that Fet wasnt because it had super awsome on resistance or great total power.... but because it has something like 1/10th of the transition time between on/off, off/on, isnt that where most of the power is lost?

Im Also a little bit unsure as to the whole losses thing, if 1 FET looses 11% of the power that goes through it(hypothetical) then 2 fets in parallel handling the same current should loose 11% of half the current, in each one.... for a total of 11%. Am I just running into a fundemntal misunderstanding of Ohms?

 

[Link]

Im Also a little bit unsure as to the whole losses thing, if 1 FET looses 11% of the power that goes through it(hypothetical) then 2 fets in parallel handling the same current should loose 11% of half the current, in each one.... for a total of 11%. Am I just running into a fundemntal misunderstanding of Ohms?

It's the ² part of I²R. Double the current equals four times the heat.

1A²1Ω=1W

VS

2A²1Ω=4W

 

[billvon]

The reason i posted that Fet wasnt because it had super awsome on resistance or great total power.... but because it has something like 1/10th of the transition time between on/off, off/on, isnt that where most of the power is lost?

That's where some of it goes, but for most designs IR losses dominate.

You have a few issues to deal with here:

1) The packages we usually use (TO220) are not very good, and aren't much good past 50-75 amps. So if you want to go near 50 amps you need two devices to deal with the package losses.

2) The larger the FET, the higher the gate charge and input capacitance. These slow down switching and increase switching losses. It is hard to drive large FETs well and most controllers do not do a very good job of it.

3) FETs carry flyback current in their body diodes as well, and this leads to heating. A simple example is coasting down a hill with the throttle at zero. At some point the FET body diodes will start conducting - and that's a lot of power to handle. (And you have the standard diode forward voltage losses.)

if 1 FET looses 11% of the power that goes through it(hypothetical) then 2 fets in parallel handling the same current should loose 11% of half the current, in each one.... for a total of 11%. Am I just running into a fundemntal misunderstanding of Ohms?

1 FET, 1 ohm, 1 amp = 1 volt loss. Total power lost is V *I or 1 watt.
2 FETs in parallel, so .5 ohms, so .5 volts lost at 1 amp. Total power lost is V*I or .5 watts, or .25 watts per device.