Talk:MOSFET

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Splitup in progress[edit]

splitup of article Field effect transistor in progress, please see Talk:Field effect transistor Pjacobi 20:45, 19 Jul 2004 (UTC)

  • Done Pjacobi 21:19, 19 Jul 2004 (UTC)

Resistance Calculations....[edit]

The wiki says, "Conceptually, MOSFETs are like resistors in the on-state, and shorter resistors have less resistance." From the basic R=(pL)/A surely reducing the length would not change the resistance?

EG: for this example lets make rho=5; let the width of the channel=2 and we will vary the length. If we start with the length at 10, the resistance will be: (5*10)/(10*2)= 2.5 If we reduce the length to 5, the resistance will be: (5*5)/(5*2)= 2.5

(NB: this is my first wiki edit and I've probably broken loads of conventions. Could someone please edit/delete this, whatever is appropriate)

The A in your formula is the area formed by the other two directions, so to speak width and height of the channel, not length.
But the article chapter you are referring to is nevertheless a but fishy .
Your edit was just fine, you only missed to use the "auto signing" feature of the software, putting four tildes at the end of your discussion statement, would give user and date/time info.
Pjacobi 22:50, 5 Dec 2004 (UTC)

Graphical Representation of MOSFET operation in circuits[edit]

Would it not be usefull to include Vin/Vout graphs? i.e. what the 'resistance' across the MOSFET does as the gain-source voltage increases? Yossarian 10:19, 11 Jun 2005 (UTC)

Need some help on MOSFET and Square wave high amp inverting[edit]

Guys and gals, can you help me a little bit. I have this idea of buildng a square wave AC source from a constant current 300A DC. I can, hopefully, use an H bridge and MOSFETs. My question is, if I use (stepped down) AC to control the gate of the MOSFET, the resulting voltage would not be square wave, or would it be? Any thoughts? thanks. I will be watching this page.

Unlikely values?[edit]

The caption under the picture of the two power mosfets in D2PAK says that these devices are capable of dissipating 100 watts. This seems unlikely to me (by an order of magnitude). Is there a citation for this piece of data? — Preceding unsigned comment added by 68.65.89.98 (talk) 20:36, 9 May 2012 (UTC)

I can't read the markings on the device, but there is a similar MOSFET, based on the description, the Fairchild Semiconductor FQB32N12V2. The datasheet can be found here FQB32N12V2 from Digikey. It will breakdown at 120V minimum and can handle 32A continuous at a 25C case temperature. The junction is rated to 175C. The thermal resistance from the Junction-to-Case is 1 C/W. So for every WATT that the FET needs to dissipate, the junction will rise 1 degree Celsius from the temperature of the case, which for the D2 package is the large tab. So, if the tab is held at 25 C, then the FET can handle (175C - 25C)/(1C/W) = 150W. Holding the tab at 25C is difficult, but possible. The FET can actually dissipate 230W, (175C -(-55C))/(1C/W), if the tab is held at -55C, which is the lowest operating temperature of the junction. Dissipating a higher wattage requires that the tab be held below -55C, which would be outside the operating temperature of the device. This will require a special turn-on sequence to ensure the junction starts at -55C and never drops below it.

The Fairchild Semiconductor FDB024N06, has a Junction-to-Case thermal resistance of only 0.38 C/W for the D2-pak, which equates to 395W when the tab is held at 25C. It can dissipate 605W if the tab is held at -55C. Jeffrobins (talk) 07:25, 17 August 2013 (UTC)

Using the phrase "this devices dissipates" is just confusing to lay people. It should be removed, and the maximum voltage and current should be used instead. — Preceding unsigned comment added by 99.3.46.137 (talk) 23:38, 9 November 2013 (UTC)

QFET?[edit]

QFET redirects here, but neither that acronym nor the words Quantum Field Effect Transistor appear anywhere on the article... This was clumsy, and really should be corrected by someone who understands what these terms mean (unfortunately I am not one of those people). KDS4444Talk 06:25, 26 November 2013 (UTC)