On Mon, 10 Nov 1997 20:20:53 -0800 Ron Fial <ron@FIAL.COM> writes:
>>I am curious about one thing: What are the major differences between
>IGBTs
>>and MOSFETS? Lower Rds(on)?
>>
>>Sean
>
>Yes, very low Rds on - minimizes heat-sink requirements.
>Postive temp coefficient (turns OFF as it gets hotter, rather than the
>          other way around ala Bipolar, so avoids thermal runaway).

As I understand it, an IGBT consists of a MOSFET driving a bipolar in a
Darlington configuration:

            C
     ||-----|
 NFET||     |
 B---||-| | /c
        --|  NPN
        \ | \e
     Res/   |
        \---|
            E

Power FETs that can withstand higher Vds (off) voltage are hard to make
without increasing Rds (on) a lot.  Thus they don't have much current
capacity.  For high-voltage, high-current circuits, the IGBT was
designed.  The NPN acts as a current booster, so the FET only has to
handle a small proportion of the total current being switched.  The input
to an IGBT has the same large capacitance, zero DC current
characteristics of a FET (since it is a FET). The same driver circuit can
be used.  The configuration also prevents the bipolar from going into
full saturation, allowing it to turn off quickly.  I think a resistor is
integrated as I've drawn it to help the bipolar switch off.

But the IGBT has an unavoidable on-state saturation voltage of a couple
of volts (VBE of the bipolar plus the resistance loss in the FET).  This
voltage is relatively independent of the drain/ collector current.  In a
circuit using hundreds or thousands of volts, a couple of volts switching
loss isn't too bad.  But at low voltages, a regular FET will do better,
since the silicon can be designed for low on-resistance.  Conduction loss
in the FET circuit is easy to analyze by assuming the switched-on device
is a resistor.  The IGBT is more complicated and nonlinear.