I would caution against a newbie trying multiple parallel FETs. There are subtleties with current sharing among parallel FETs as they switch on and off. First of all, a significant amount of heat is dissipated during the turn-on and turn-off transients. If you don't turn all FETs on and off at exactly the same time then one will take all of the turn-on/turn-off transient power each PWM cycle. Also, as the FETs heat up, their characteristics change and you can get positive-feedback effects where they start out drawing equal current at the same time but as one gets hotter, it turns on more quickly and begins to take all of the turn on/turn off power. How much current do you actually expect to put through the motor? At what duty cycle? If you get first-class modern FET (about 3-4milliohms ON resistance in a TO-220 or D2PAK) and heatsink it well, you should be able to do true 50A continuous. It will get hot, but not more than the rated max temp. I am using the IRFS4310 but you should be able to do better for low RDS on as this is not the absolute latest design and it is rated 100V and you don't need that high voltage. Bear in mind that you cannot get very close to the FET max Vds voltage. If you are working with a 24V bus voltage, I'd go for at least 50V FETs. This is because there are spikes much higher than the bus voltage. There should also probably be considerable bulk capacitance on your power bus near the motor/FET. This will help to smooth out the ripple current so it doesn't travel all the way back to the batteries and cause excessive heating. As has been pointed out, there NEEDS to be a fast current limiting scheme. Probably the easiest way would be to use a low-value (perhaps 0.01 ohm) resistor in the source-lead of the FET, with a comparator and a way of cutting off gate drive when the voltage drop across the resistor exceeds a certain value. The reason why people are telling you that you need a proper FET driver is that you actually need significant peak current to drive the FET gate. This is to turn the FET on and off very fast (to keep the turn-on/turn-off losses low) and also to prevent voltage swings at the drain of the FET from affecting the gate-source voltage through the gate-drain capacitance (look up Miller effect). Sean On Wed, Aug 5, 2009 at 8:42 AM, M.L. wrote: > On Wed, Aug 5, 2009 at 7:17 AM, Chris Loper wrote: >> >> I know that driving the gate at only 5V is not good, >> and I plan to replace the IR540 with a logic level FET. >> I did a search on Digikey and found this one: >> =A0 DIGIKEY IPP0165N03L G =A0 30V =A0 50A =A0 Rdson=3D.0065 =A0 $1.03 >> =A0 http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=3DI= PP065N03LGIN-ND >> Datasheet: >> http://www.infineon.com/dgdl/IPP065N03L_rev1.02.pdf?folderId=3Ddb3a30431= 441fb5d01148c401f250e27&fileId=3Ddb3a30431441fb5d011492371ebc0fe2 >> One US dollar! =A0It seems almost too good to be true. >> Are there other parameters that I need to check? > > > You can't put 50A through a TO-220 part for more than a few > microseconds even at low duty cycle. Don't count on more than 20A > continuous. You won't know until you figure out how much heat you can > dissipate via a large heatsink. Since you won't be able to figure this > out easily you need to use many FETs in parallel and figure 15A or > less for each one (regardless of how great the datasheet says it is) > > You also need to use a real FET driver or totem-pole PNP+NPN driver > running to 12v or higher. You aren't going to get the results you want > by trying to use logic level devices. A real, if somewhat outdated, > motor controller looks like this: > > > I would recommend you study everything to the right of the output of > U5C for starters. > > Also you absolutely need fast current sensing + limitation or the > first time you step on the pedal too hard the thing WILL pop like a > firecracker and you'll have just blown up your expensive components. > This current limitation has to act before the maximum peak pulse > current occurs due to a short circuit. What I'm trying to say is that > time =3D (L*I)/V where L is the inductance of the motor wires, I is the > peak current your FETs can handle, and V is the system voltage. Time > is short. > > -- > Martin K. > > -- > http://www.piclist.com PIC/SX FAQ & list archive > View/change your membership options at > http://mailman.mit.edu/mailman/listinfo/piclist > -- = http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist