> Can any of you figure this one out?  I have a PIC16F628-based circuit using 2
> multiplexed switches like this...

>   .__________.
>   | 16F628   |
>   |          |
>   |      RB0 |---------o-------------| MOSFET 1 gate
>   |          |         |
>   |      RB1 |---------|-------o-----| MOSFET 2 gate
>   |          |         |       |
>   | RA5      |         |       |
>   '----------'         |       |
>      |               |/        o |
>      |  A>--'\/\/\,--|  NPN      |] Pushbutton N.O.
>      |         R1    |\>       o |
>      |                 |       |
>      |                 |       |
>      |                (D)      |
>      |                 |       |
>      |                 |       |
>      `-----------------o--|<|--'
>                        |   1N4148
>            ,--'\/\/\,--'
>            |     R2
>           GNDb

> Port B is all outputs, and RA5 is an input.  The point at (D) is just a direct
> connection for now.  R2 is a 10K pull-down resistor.  Only one output on port
> B will be High at any time, with the rest Low.  When either RB0 or RB1 is
> high, the input value at RA5 is read to determine the state of the
> corresponding switch.  Well, it works properly.

> Oddly, I have another circuit, which is identical with respect to this portion
> of the circuit, but it does NOT work properly.  When point A is connected to
> +5V, the transistor switches on, and is properly detected, but the pushbutton
> cannot be detected until the transistor is switched off.  However, the
> circuit will function properly with another diode added at point D (banded
> end away from the transistor).

> In the second circuit, I would think that the emitter serves/acts as a diode
> when the transistor is on, so that another diode at the emitter is not
> required.  I do realize that the proper diode behaviour is achieved in the
> B-E junction of the transistor, but thought that the C-E junction is somewhat
> unidirectional as well.

> I've re-checked the initialization and usage code, but cannot see any
> difference in this portion of the circuit.  Any idea why this additional
> diode is required in the second circuit?

> Thanks,
> -Neil.

Whenever I have driven the emitter of a transistor, I have always had to
watch out for breakdown of the base-emitter junction. This can happen
with a reverse BE voltage as low as 3 volts in some transistors. The
actual voltage varies from one transistor to another, which may be why
one of your circuits works and the other doesn't. Putting a diode in
series with the emitter blocks the reverse voltage causing the breakdown.

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   PMB 387                                        ELECTRICAL ENGINEER
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