How to drive a 12V common cathode 7-segment display with 5V signals and BJT transistors?

Question

I'm having trouble designing a driver for a large (5") 12V common cathode 7 segments display using a 74HC595 shift register and BJT transistors. Below are the schematics for the component (display) I want to drive.

_{(source: soldafria.com.br)}

I have a working driver for a similar 12V common anode display using the shift register and 8 NPN transistors (BC547) (see schematics below). I've even seen folks replacing the transistors by darlington arrays sych as ULN2003 and ULN2803.

But I can't seen to find the equivalent circuit for common cathode displays.

My first (failed) attempt at it is shown below.

The problem seem to be that the 5V signal is not enough to turn off the PNP transistor. I would have to raise the signal to 12V to turn the transistor off, but I don't have that voltage coming from the microcontroller or the shift register.

So, how can I drive common cathode displays? Can anyone point me in the right direction, please?

Answer 1

You need to put a NPN low-side drive in front of your PNP high-side drive.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

I'd redo your power supply scheme to have the positive of the 12 V supply tied to +5 V and therefore the negative end of the 12 V would be at -7 V.

In effect you'd have +5 V, 0 V and -7 V. Then I'd use PNP transistors feeding the anodes of the LED displays (from 5 V) with resistors between collectors and anodes.