Your NPN transistor idea is exactly what I thought of... AFTER I sent that
message. I hooked up a pot to the gate to see what level the MOSFET turned on
and off... low and behold, I found out that my definition of 'above the
threshold' was backwards. Either way, I understand it now, thank you for re-
affirming that what I did was a valid solution!



> On Tue, Mar 11, 2003 at 10:09:29PM -0500, Jai Dhar wrote:
> > Ok, I'm really trying to understand these darned MOSFET's here, so I
> managed
> > to find some good articles regarding them. I will summarize what I know to
> see
> > if it's true (for pMOS)
> >
> > If the voltage drop from gate to source is greater than the threshold, the
> > MOSFET should turn on...
> >
> > I took my IRF9530 (just a basic pmos with a Vgs threshold of -4V) to test
> this
> > out. Connecting source to +12 and gate to GND, the drain showed a drop of
> > +12V. So far, that seems right.. since Vgs = -12V, and the threshold =
> -4V.
>
> How did you measure the voltage on the drain? The best way I've found to
> test
> this is to attach the drain to GND through a pulldown resistor. When the
> MOSFET is off, then the drain will measure 0V, whereas if it's on it'll
> measure
> somew voltage closer to the source.
>
> BTW in this configuration the MOSFET is on and will pull the drain to +12V.
>
> >
> > BUT...
> >
> > When I connected the gate to +5V, the drain still showed +12V. Why is
> this?
>
> Because the MOSFET is still on.
>
> > Vgs = 5 - 12 = -7V, which is still greater than the threshold, right?
>
> Right. Greater and less than can get you in trouble here. A threshold of
> -4V means that the gate must be at least 4V below the source for the MOSFET
> to turn on. Since it's 7V below, the MOSFET is on.
>
> I'm understanding now that this doesn't meet your expectation. You expected
> that the gate at 5V should turn the part off. It doesn't.
>
> To do it right the gate should be pulled all the way up to 12V to turn off
> the MOSFET giving a Vgs=0V which is not at least 4V below the Source.
>
> A single NPN transistor with the emitter grounded, collector tied to the
> gate, along with a pullup resistor between the source and gate/collector,
> and
> the base tied to the PIC I/O through a series resistor, will do the job
> fine.
>
> When the PIC I/O is 0V, the NPN is off, the gate moves to 12V, and the
> MOSFET
> is off.
>
> However when the PIC I/O is 5V, the NPN turns on, grounding the gate, And
> the MOSFET turns full on.
>
> Now a question: could a zener diode be used instead of the NPN? Consider
> if the gate is pulled up via a pullup resistor. Of course the MOSFET is off.
> Now connect a 5.7V zener between the gate and the PIC I/O pin. My though is
> that the voltage of the gate will be 5.7V above the voltage of the PIC I/O
> pin. So if the PIC pin is 0V, then the gate will be at 5.7V, and therefore
> the MOSFET is on. However if the PIC pin is 5V, then the gate will be at
> 10.7V
> and Vgs=-1.3V so the MOSFET will turn off.
>
> Thoughts?
> >
> > Another thing is the signs here... I think that's what may be messin me
> up.
> > Are we talking about algabraically greater?? Because -7 isn't greater than
> -
> > 4... nor is -12. So somewhere in my explanation (and understanding), I'm
> > mistaken. Can anyone clarify?
>
> I Hope that I did so above. I always read Vgs=-4 as "The gate must be at
> least
> 4V below the source for the MOSFET to turn on."
>
> Hope this helps,
>
> BAJ
>
> --
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>




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