Todd and everyone who replied

Thanks for all your help. I really need to swot up on my analog stuff.  I
DID read Horowitz & Hill Art of Electronics (transistor section)  before I
posted the question, but I find it heavy going.  I need to find a book which
explains things in more basic terms - rather like the responses I have
received from my posts to the PICLIST on this subject.

So, if I understand your response Todd et al, that means I can drive the
base of a PNP BJT with the collector of a NPN , the emitter of which goes to
ground, and the base of which will connect to my PIC pin - rather like the
circuit that Rick C posted on this thread, and Matt Pobursky pointed out?

Appreciate your time
Les

PS So my circuit has +9v at the collector, +4.4V at the emitter...have I
invented a logic controlled 4.6v heater? :o)

> -----Original Message-----
> From: pic microcontroller discussion list
> [mailto:PICLIST@MITVMA.MIT.EDU]On Behalf Of T.C. Phelps
> Sent: 11 August 2002 16:53
> To: PICLIST@MITVMA.MIT.EDU
> Subject: [PIC]: Membrane latching power switch for PIC circuit
>
>
> Hi Les,
>
> In response to your question (below), you're right in
> that applying +5V to the base of your NPN transistor
> will make it act like a switch; but the transistor is
> basically a controlled _current_ source. So you design
> the circuit to pass the desired current into the
> collector and you get almost the same current out the
> emitter (depending on your beta -- for your purposes,
> you can probably assume it's the same) when the
> transistor is on. BUT, the voltage difference between
> the base and the emitter is 0.6-0.7 volts, like a
> diode (there is a pn junction between the base and
> emitter pins after all). So, if you apply +5 volts
> from a PIC pin to the base then your emitter will be
> at 4.4 volts.
>
> Regards,
> Todd.
>
> > Thanks for the help..but I don't understand why I
> > will only get 4.4v. I thought that if I supply 5v to
>
> > the NPN base from a logic high PIC output pin, the
> > transistor would act as a simple switch and pass the
>
> > +9v across the C-E junction.  Why only 4.4v?
> >
> > Thanks
> > Les
>
>

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