Electronic – How to use Darlington arrays to source current in a 7-segment display driver

I do not believe you can reasonably run your display (or any multi-digit common anode display) using this chip. You may be able to finagle it to drive a single digit display through some creativity by hard-wiring the common anode to VCC and not connecting any of the DIG pins of the chip.

The way the chip works is that the SEG* pins "source current to the display" and the DIG* pins "sink current from the display common cathode". The way it's meant to be wired is that the common cathode of each digit on the 7SD is wired to a corresponding DIG pin and the SEG pins are wired to the corresponding anode segment pins of the 7SD. You serially load up "what should be displayed" into the memory of the chip and some configuration settings and it takes it from there.

The chip "scans" the 7-segment display for you and does this by, for each digit N:

• switching the DIG N pin to GND (and all other DIG pins to high-impedance).
• setting all the SEG pins to what is stored in its memory (optionally decoding the stored value as BCD first)
• then moving on to digit N+1 modulo the number of digits its scanning...

There's just no way you are going to be able to take advantage of the chip's multiplexing algorithm because it relies on the DIG pins driving the common cathode to GND to "enable" each digit in turn.

You could insert an inverting buffer between all the SEG and DIG pins of the chip and the display and then you would communicate with the chip and wire it up "normally" as though it was a common cathode display. Not worth it if you ask me...

Here is a possible mechanism for the fault. You say it works (dimly) without the transistors, but doesn't work with them. So it sounds like the fault is in the transistors. Is there some reason that the transistors might be letting some current through when they shouldn't?

Yes. You're using a PNP transistor. As you know, these transistors are on when the base voltage is lower than the emitter voltage. They are off when the base voltage is higher than or equal to the emitter voltage.

The problem with the shift register chip is that the outputs are always lower than the emitter voltage. I couldn't quite make out the part number of the chip you're using, but according to the datasheet for the 74HC595 (page 6), the outputs don't quite reach Vcc. If there is a tiny voltage difference, then you could find that a small amount of current is leaking out of the base of the PNP transistor. With a gain of about 100, you could find that there's enough CE current to give noticeable light output on the LEDs.

Something to try: Add a schottky diode between Vcc and the emitter. This should drop the collector voltage by a fraction of a volt, just enough to allow the shift register to fully turn off the transistor.