Integrated Circuit – Using 7447 to Drive Common-Cathode 7 Segment Display

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.

My first assumption from the symptom is that you have a single cathode resistor, which would explain everything. However, you say this is not the case and that you have individual 100 Ω resistors on each anode line.

The answer then must be that whatever is driving the cathode can't handle the combined current. Apparently you are driving the cathode with just a microcontroller output, which has insufficient current sink capability. Do the math. You want 10 mA thru each of 7 segments. That means whatever is holding the cathodes low has to be able to handle 70 mA. A microcontroller pin is very very unlikely to do that. See the datasheet.

The solution is to add a transistor that can sink the current from each of the cathode connections:

70 mA is easy for any discrete transistor you will be able to get. Let's say the transistor has a minimum guaranteed gain of 50. That means the base current needs to be 70mA / 50 = 1.4 mA. Figure the B-E junction drops 700 mV. With 3.3 V drive from the micro, that leaves 3.3 V - 700 mV = 2.6 V accross the base resistor. 2.6 V / 1.4 mA = 1.9 kΩ, which is the absolute maximum allowed base resistor. Given that, I'd use 1 kΩ between the digital output and the base.

Note that this will invert the common cathode logic from what you had. The microcontroller output now has to go high to enable a digit.