You can control a RGB common anode like this with only one transistor and 3.3V GPIO sources:
How can you do this with a common cathode LED array using only one transistor per LED and 3.3V GPIOs?
The forward voltage of the LEDs are 3.6V
Electrical – RGB LED common cathode with 3.3V GPIO +single transistor
analogcircuit-designled
Related Solutions
Since this is a constant-current sink IC, no resistors are needed in series with the LEDs.
The current is programmed using a reference resistor connected to one of the pins that sets the reference current used to drive the LEDs, and it is the driver chip's responsibility to make sure the correct voltage and current is applied to the LED. In practice, I have found this works well.
You could add a resistor in series, but this defeats the purpose of using this chip. It will increase your part count, increase your minimum VDD requirement, and if you need to increase VDD, then you will waste additional power.
Below is the figure used to determine your reference current for each resistor from the datasheet:
LEDs and the 74HC595 can go quick enough, no problem. But, note a few things:
You're multiplexing LEDs, that means that if a LED is constantly on it only gets 1/3 it's normal current. This is because it's only on for 1/3 of the time (2/3 for other LEDs).
Well, if you want full brightness per LED.. increase the current. You could, but then get the problem that you can burn out LEDs (if software gets stuck, during programming or flashing) and the hardware needs to support it. The 74HC595 can only sink 25mA per pin. The microcontrller probably a similar amount as well. This could already become an issue if you turn on Red Green and Blue at once all at 20mA. You may need to add a few MOSFETs or transistors that the shift register will drive, instead of straight on the shift register. Of course you don't always need to drive them that high, 5mA each LED is sufficient in most cases.
The only time I had trouble multiplexing a bunch of LED's was when I picked the wrong multiplexer. I had 13 7-segment displays setup. The segments would be driven via a 74hc595, buffered with some MOSFETs to switch the positive rail of the supply. The display was selected via 13 N-MOSFETs switching the GND of each display. I picked a 16-channel multiplexer for this, but forgot to look what it was. Turns out it was analog multiplexer.. Because I used N-MOSFETs they had to be 'forced' on/off. A analog multiplexer doesn't pull or push the output, it either connects it to the input or keeps to pin floating. With a floating pin the N-MOSFET wasn't turning off quick enough (tiny 'capacitor' inside the gate kept it on for 3 seconds).
I had to put a pull-down resistor on each gate to turn off faster, but had to limit the value to something the analog multiplexer could drive sufficiently. I was multiplexing at about 1kHz (which is quite slow for 13 segments) , but in the end had I programmed intermediate cycles to turn of the displays first, give it some turn-on time, re-set new display settings and turn it on. I can only actively use the display for like 70% duty cycle, 30% is wasted to turn on/off of the gates. It limited the amount of light from the displays, fortunately I only use the displays in a dark environment.
In your case you may see similar things if you leave the LEDs on while reprogramming the shift register, for example. This effect may be very small, because it's happening very fast.
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Best Answer
The temptation is to invert the circuit but this will fail.
Figure 1. The protection diodes on most logic chips creates a sneak-path to positive supply. This will keep the PNP transistor permanently turned on and may damage the chip. Source: High-side fail.
You need to add a second transistor.
Figure 2. To drive a high-side transistor from a GPIO pin we need a level translator. An NPN transistor does the job nicely.
There is a little more information in my article in the link.