Leo van Loon wrote:

> For simplicity I must use a 9V block battery, but the PIC does not
> like 9V.

  Dead right there, recently mentioned on this list.

> How can I reduce 9V to 5V with a minimum of components?
> Does only a serial 4,7 V zener work?

  As I see it, you go either of two ways.  For robustness, you use a
78L05 (as previously advised), because this allows you to maintain
optimum performance over a wide battery voltage.  Preferably use a low-
dropout regulator version.  And these are almost as cheap as
transistors.

  Alternately, consider the range of workable voltage of the PIC (about
3 to 6V is it not?) as a resource.  Connect the two LEDs which must
operate together (the diagonal ones) in series and use them to feed the
PIC and simultaneously drop 3V from the battery.  To drop another 0.6V
just to get it within ratings, use a series diode.

  These LEDs may barely light (but not in sleep mode) especially if you
use high- efficiency ones (but you wouldn«t, would you?).  To truly
light them, switch a resistor across the PIC using one of its outputs.

  Similarly the other two pairs are in series plus a silicon diode, from
the +9V supply to a PIC output each but *via* a resistor of the same
size.  The single LED must needs be driven separately, either by a
transistor or using a series "ballast" LED and diode as the others.

  The big concern with electronic dice (you meant you were producing an
electronic "die") is that switching the display LED current can warp the
randomisation algorithm.  With a PIC you have the opportunity to
actually run the randomisation very quickly (timing the button press in
µS, mod 6) then knowing the desired result, implement a "run-down"
display which eventually stops on the already-determined value!

  Actually, on reflection, a single silicon power diode in series with
all three LED pairs will both reduce the voltage to bring it into the 6V
rating and provide reverse-battery protection.

  Cheers,
        Paul B.