Now in this case how does one get a Vbe? If we can only drive to 5V then the
maximum supply voltage that can be turned in is 5V + Vbe = 5.7V



Suggestion
if the current is low then:- connect something like a MMBT2907 (PNP) as a
high side switch, with a pull up resistor on the base to the rail that you
are switching, then use a MMBT2222 (NPN) to pull the base of the PNP to
ground (Via a resistor). This will let you drive the NPN with +.7V or better
to turn on (Current is required as suggested). Then the PNP will switch up
to 75V (Watch the current and dissapation). Simple and it does not turn on
when the power is on either and yes it also passes the relevant GM standards
(And Fords too)
Bit this all assumes low current, how much do you need to switch?

Dennis





> -----Original Message-----
> From: Dan Michaels [SMTP:oricom@LYNX.SNI.NET]
> Sent: Tuesday, 29 August 2000 13:58
> To:   PICLIST@MITVMA.MIT.EDU
> Subject:      Re: [EE]: How to switch 8 volts with a PIC
>
> John Hansen wrote:
> >Here's a dumb question (which if I'd done the right thing in college and
> >studied EE I could probably answer... but I didn't). ...
> >
> >Aside from using a relay, is there a simple way to use a I/O pin from a
> PIC
> >(running at 5 volts) to switch a higher voltage (say, 8 volts).  The
> >current is negligible.   I'm assuming a bipolar transistor switch is not
> >going to work here because I'm attempting to switch a voltage that is
> >higher than that which is applied to the base.
> >
>
> NPN inverter with collector pulled up to whatever voltage you like,
> base driven by PIC thru 10-50K resistor.
>
> --
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