V G wrote:
>> FETs look more
>> like a resistor when on, and bipolars more like a voltage source.
>
> I've seen you say this before, but I don't understand this. Could you
> please explain this? Aren't they both like base/gate controlled
> variable resistors?

No, that's the point.  A saturated bipolar will have a roughly constant
voltage accross C-E.  Of course that's not perfect, and under high current
it goes up some.  As a result, there are some operating points where a
bipolar can result in less voltage drop than a FET.  This is more likely
when the transistor has to withstand higher voltage (this forces a tradeoff
with Rdson in FETs) and at moderate to high currents.

>> A FET with
>> really low Rdson can often have a lower voltage drop accross it, and
>> therefore lower power dissipation, at the design current.  This
>> becomes
>> less
>> of a advantage as the current goes up.
>
> Can you explain that last sentence? Of course, as the current goes
> up, the
> FET will dissipate more power, but will it dissipate more than a
> competitive BJT (is that what you're saying when you say "less of an
> advantage")?

The FET when on looks roughly like a fixed resistor to first approximation.
That means the dissipation is proportional to the square of the current.  I=
f
a BJT has a constant C-E voltage, then the dissipation is proportional to
the current.  As the current goes up, the FET's relative advantage
diminishes, until in some cases the BJT can have lower dissipation.

> What effect will the capacitance have?

It takes large current to switch quickly.  For example, let's see what it
takes to change a 1nF capacitor by 15V in 50ns.

  15V * 1nF / 50ns =3D 300mA

Take a look at the effective gate capacitance of some large power FETs and
you will see why high end FET drivers are rated for 1A or more switching
current.


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