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Mosfet vs BJT

Discussion in 'General Electronics Chat' started by icetea, May 16, 2008.

  1. smanches

    smanches New Member

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    It's a 40% increase in resistance, which means a 40% increase in power dissipation. Might seem small, but it's all about percentages here.

    Correct. Drain-source current does have an affect on the channel resistance. Although I don't know the exact details as to why. Yet. :)
     
  2. BrownOut

    BrownOut Banned

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    Ok, by "small" I mean in comparison to the load. If the supplies were, say 30V, and Id was 5A, then the total circuit resistance is 6 Ohms. So going from 6 Ohms to 5.99 Ohms is pretty insignificant to the circuit operation, and 175mW vs 125Mw isn't usually going to be much of a concern, unless you're on a really limited energy source.

    But anyway, thanks! I think I'm getting there.

    Edit: "
    Thanks!
     
    Last edited: Jun 18, 2009
  3. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi there,


    " 'Fully On' is in the meter of the beholder" :)

    What is more important is the answer to this question:
    "Does the device turn on enough for my application or not?".

    If the answer is 'yes', you are good to go. If not, better look for
    another device or a better way of turning it on.
    Keep in mind that some devices turn on with roughly 4 ohms min,
    while others turn on with say 0.040 ohms min. What works for
    your application works, and what doesnt, well, doesnt.

    1. Review the data sheet
    2. Review the application design
    3. Make the go/no go decision
    4. You're done.
     
  4. dave

    Dave New Member

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  5. OutToLunch

    OutToLunch New Member

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    The MOSFET will be "on" whenever Vgs exceeds Vt. Once this occurs, the channel between the source and the drain is created. As Vgs is increased, the channel becomes "wider" and thus the impedance between the drain and the source decreases.

    Now keep Vgs constant and start increasing Vds. The channel near the drain starts getting narrower and narrower until Vds = Vgs - Vt. At this point, the channel shape will remain the same which limits the current regardless of Vds. So the enhancement channel is affected by Vds which can limit the amount of current through the FET. So if Vds keeps increasing but the current remains the same, that, in effect, translates to an increasing impedance.

    In a switching application you really don't want to put the FET in the saturation region. You want the Rds(on) to be predictable and you want the losses to be low. That means a low voltage across the drain to source and as low an Rds(on) as you can get. Basically a switchable resistor.

    If you are operating the FET in the saturation region, you'll really need to watch the power dissipation - unless the duration is short or you have some really good cooling to keep the junction temp down, the FET will likely burn up.
     

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