ignorant MOSFET question

[Walter Harley]

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?

 

[Gary Reichlinger]

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.

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.

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

This is a fairly typical value.

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.

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).

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.

They are very different.

 

[Walter Harley]

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).

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?

 

[Walter Harley]

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?

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.

 

[Tim Wescott]

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.

 

[Walter Harley]

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.

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.

 

[Jim Adney]

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.

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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