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ignorant MOSFET question

Discussion in 'Electronic Components' started by Walter Harley, Jan 19, 2004.

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  1. Please help me as I try to better understand MOSFETs.

    I think that the physical construction of power MOSFETs is such that the
    drain and source are approximately symmetric: so, electrically speaking
    there's not a whole lot of difference between the drain and the source.

    But, I see that Vgs(max) << Vds(max). For instance, a 500V MOSFET might
    have Vgs(max) of 30V.

    Now, suppose I take a MOSFET and I put a 400V supply on its drain, a 1k load
    to ground on its source, and I ground the gate. The MOSFET will be "off"
    and no current will flow. Vgs = 0V; Vds = 400V. But, the voltage from gate
    to drain is -400V.

    So how come it can stand 400V from gate to drain, when a tenth of that from
    gate to source would have fried it? In at least that regard, the drain
    seems very different from the source.

    Is one of my assumptions wrong, or am I just not understanding properly?
  2. They are not "symmetric" in the sense you are suggesting. The
    drain and source are definitely not interchangeable. Mosfets only
    block current in 1 direction. N-channel mosfets can only block
    current flowing from drain to source. P-channel mosfets can only
    block current from flowing from source to drain.
    This is a fairly typical value.
    The load connections are wrong here and would probably fry the
    part if it ever turned on. With an n-channel mosfet, the load is
    connected between the power supply and the drain (ground connected to
    the source). With a p-channel, it is connected between the drain and
    ground (power supply connected to source).
    They are very different.
  3. Okay, suppose we connect the 400V supply to the source, and we put a 1k load
    from drain to ground. Connect the gate to the source, so Vgs = 0V. No
    current flows. Now, there's 400V from drain to gate.

    I guess my question could be rephrased:

    I think the construction of a MOSFET is basically a channel of one type of
    material, in a substrate of another type, with an insulated gate on top of
    the channel. One end of the channel is the source, the other is the drain.
    How come the gate insulator can stand 400V from gate to one end of the
    channel (drain), but a mere 40V from gate to the other end (source) is
    enough to fry it?
  4. A helpful person responded to me offline. Paraphrasing his response:

    It's not because of asymmetry, it's because of whether the channel is
    conducting or not. When the channel is conducting, all that's between the
    gate and the channel (S or D, either way) is the metal oxide insulator,
    which is fragile. So, Vgs(max) applies when the device is turned on. Note
    that if Vgs is large-ish then the device is fully on, meaning S and D are at
    about the same potential, so Vgs ~= Vgd.

    When the device is turned off, then the channel isn't conducting, because
    there's a carrier-depleted region around the gate. But in that case, the
    depleted region serves as an insulator along with the metal oxide; so it's
    substantially stronger than the metal oxide alone would be. So in that
    case, Vds(max) is what applies.
  5. Tim Wescott

    Tim Wescott Guest

    Some power MOSFETs have protection diodes that will conduct if the drain and
    source are reversed. Also, most commercial parts aren't 100% symmetrical,
    for heat dissipation on the gate if for no other reason, so you can't just
    willy-nilly turn them upside down.

    You _can_ however, take great advantage of this property in designing 2- or
    4-quadrant switching drives if you mind your P's and Q's.
  6. If you're thinking of what I'm thinking of, I think that's not a protection
    diode per se; it's a consequence of the physics of a MOSFET, not something
    the designers put in on purpose. I think it's because the substrate is
    bonded to the source, rather than bringing it out as a separate lead.

    You're right, that is an asymmetry. When I said that S and D were
    approximately symmetric, I was intentionally being vague. I believe there
    are also other asymmetries, in power MOSFETs, such as different capacitance
    from G to D than to S, as well. (Some small-signal MOSFETs are apparently
    truly symmetric, but they have a separate lead for the substrate.) The
    point is that that asymmetry is not the explanation for the effect I was
    asking about, as was later explained to me.
  7. Jim Adney

    Jim Adney Guest

    They're really not symmetrical at all.

    If you can find a copy of the IR (International Rectifier) databook on
    these, it has a very good explanation of all the characteristics of
    MOSFETS, including the intrinsic body diode which is an inherent part
    of the design.

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