> From: Murray McGregor <murray.mcgregor@stonebow.otago.ac.nz>
> Steve Hardy wrote:
>
> >Another untested option is to use a charge pump to charge a capacitance
> >connected between the driver and the gate.  This requires an initial
> >'priming' of the charge at power up, and the occasional transition to
> >maintain the charge:
> >[cut]
>
>  I particularly like the ingenuity of this approach - but, with due
> deference to Steve, I suspect the above circuit will be slow in turning the
> motor OFF and that discharge resistors (R1,R2, below) will be needed, plus
> more regular pulsing to maintain the ON voltage. A capacitor from gate to
> source *might* also be necessary.
> [cut]

At the risk of getting a bit off-topic, I should have made it clearer
that the OFF state of the FET is achieved at Drvr(low)+boost voltage
(2V in the example).  The FET will switch just as fast as the driver
can drive its gate capacitance, so long as C1 >> C(fet).  It does not
depend on leakage through a resistor to switch off, especially since
any FET worth its silicon will retain its gate charge for at least a
few minutes if open circuit.  Most VMOS N-ch fets will be off if their
gate voltage is <= 2-3V with respect to source.

A cap from G-S would worsen a bad situation, since the C looking into
the gate is already multiplied by the Miller effect in common source
configuration.

By the way, a string of 3 silicon diodes could be replaced with one
LED for the purpose of voltage drop.

Regards,
SJH
Canberra, Australia