>Looks like 5 to 8 mA is all that it takes to drive each segment >reasonably. I will take your advice to drive the segments thru an >unregulated supply so that I can use a 78L05 regulator which further >reduces cost. The HC595 is fine, particularly if price is a concern. Hi there Bill. Two things to keep in mind: you can daisy chain the hc595 chips so that you need only 1 data line (plus the clock snd strobe). You can further reduce the pin count by 1 line if you use a small diode/resistor/capacitor network as shown in by the sketch below - the downside of this is that the time delay for the latch must be substantially longer than the clock period. This may or may not be a concern. One final note: when I mention driving the anodes of the LEDs directly from the battery supply, you MUST ensure that the unregulated supply is no more than about 1.5 Vdc higher than the 5V supply. This implies 2 possible situations: use a 6V supply (4x 1.5v) with a low dropout regulator *or* use a lower voltage battery with a boost convertor. I have used both techniques but my preferred is 3x 1.5v for an unregulated supply that varies between 2.5v to 4.8v; with a LT1300 boost regulator to give me the 5V for the circuit. Works well and reasonably low cost (at least as far as industrial users are concerned). If I were designing for consumers, I would probably opt for the 6V and LDO choice: cheaper to produce even though you don't get all the energy possible out of the batteries. Here is the simple 2 wire interface for hc595 shift registers: ------------------------ data ------------------------ clock | | | | R --- cathode R /\ in4148 R -- anode | | | | -------------------- latch | | + --- chose RC time constant to be about 5X the clock period --- | | GND Shift data out as normal. On the last rising clock edge, hold clock HI for about 10X the clock period. Then bring clock LO and start over again. dwayne Dwayne Reid Trinity Electronics Systems Ltd Edmonton, AB, CANADA (403) 489-3199 voice (403) 487-6397 fax