At 12:52 AM 6/13/01 -0400, you wrote:

>When you pull up the base of a pnp wired like this, you are
>actually turning off the transistor, so the base current goes to
>zero. So, source current is not a problem.

Yes, and modern CMOS outputs as on the PIC pull cleanly to the
+ve rail, so the base resistor effectively is connected across
the emitter-base junction when the output is "high".

>Actually I would have thought you might have a problem with sink
>current instead. Transistor current gain decreases for higher
>collector currents, and may not be 100 for this case. You may not
>be able to pull a full 90 mA of collector current with only
>~ 4v/4.7K = .8mA of base current. May need a smaller Rbase,
>maybe 2.2K, in this case.

I agree. I try to keep forced beta to ~20 for low voltage transistors
in deep saturation (high voltage ones tend to have even worse beta),
so if the fan is 90mA, the base current should be around 4.5mA, so
about a 1K resistor (back of the napkin calculation).

I suspect the transistor was getting way hot and thus not turning
off completely (at 200'C Tj).

                +5
                | E
              |/
PIC --[1K]----|     PNP
              |\
                x----|<|---x
                |          |
             [FAN]         |
                |          |
                x----------x
               0V

BTW, I'd use a slightly beefier TO-92 transistor for this, such as
a 2N4403 (500mA Ic(max)). Price difference is pretty much negligible.
You can run into problems with SOA (safe operating area) with
wimpy transistors switching inductive loads because at the turn off
you have high voltage and current simultaneously present at the
collector.. result can be 2nd breakdown. There should be curves on
this on the full data sheet, but a lot of data sheets are truncated
to only a few pages on these "mature" devices.

Best regards,

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