Mitch Miller wrote:
>> MOSFETs are indeed specced interms of on resistance rather than saturation
>> voltage.
>
>Could someone explain saturation (as it applies to this thread) to me?
>

Saturation is a region of operation of BJTs [bipolar junction
transistors], where the base is being driven hard enough that the
collector-emitter voltage drops to ~0.2v, by virtual of almost
all the supply voltage dropping across the load. Most easily
viewed in terms of a stock NPN inverter ckt.

                  +Vcc
                    |      +
                  Rload  ~Vcc
                    |      -
                    |
                    C     +
Vin-----Rbase-----B     ~0.2v
                    E     -
                    |
                   gnd

More strictly speaking, for normal BJT operation, Icoll = beta*Ibase,
while for saturation operation, Ibase >= Icoll/beta

For driving digital type loads, ie on-off, saturation is a nice
area to run the BJT in, as little energy is being lost in it, and
little heat is being dissipated by it, and most of the energy goes
to the load.

With MOSFETs, you think more in terms of the channel resistance
and the amount of voltage being dropped across it. If you want
to keep the dissipation low in a MOSFET and also run a lot of
current through it, you need to use a device with a very low
channel Rds value.

A bad power MOSFET would have, eg, Rds = 1ohm, so if you run 5A
thru it, you get a 5v drop and 25W dissipation. Very inefficient
- you lose all your power in the MOSFET --> smoke city. Better
to have Rds = .05 ohm, for example, so Vds = 5A*.05ohm = .25v
and Pds = .25v*5A = 1.25W. [even Roman could live with that
- till something better comes along].

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