Electronic – How much current can a 74HC00 supply

Since the TPIC6B595N registers can only sink current, they are connected to PNP transistors on the positive side to source the current.

That's fine, but it's a bit of a waste. TPIC6B595N is notable in that it's a power shift register. A quick read of the datasheet suggests that each output can sink 150mA continuously. This is way more than the ubiquitous 74HC595, which is way cheaper. I'd go ahead and use TPIC6B595N on the bottom, and use the much cheaper 74HC595 driving some power MOSFETs on the top.

This setup is explained in more detail: Explain the use of NPN and pMOSFET in this 8x8x8 LED cube. You will need to select a P-channel MOSFET that can handle at least all the current required when all the LEDs in that section are on. Your conclusion of this being \$7.6A\$ appears sound.

the datasheet suggests 150 uA for each register so the current needed for the ICs may be insignificant.

That's correct.

I don't want the LEDs to burn if only a few of them are on so how do I prevent them from drawing more than 20mA of current?

You need a current-limiting resistor in series with each LED, or some other current-limiting component. Keep in mind you can't share resistors between parallel LEDs.

How would I go about distributing the power over various components if they have different voltage and current ratings?

Your power supply will pump whatever current is required to maintain 5V, so your life is simplest if everything can operate with 5V. Shift registers for 5V operation are probably the most common kind. Select your LED resistors for a 5V supply and you are all set. I don't think you have any other devices which do not operate from 5V.

One final consideration: parts of this circuit will be carrying some significant currents. Be sure to check out a trace width of wire gauge calculator (Google has hundreds) against the high-current traces. If your traces or wire aren't fat enough for the high-current paths, you will lose a significant fraction of the supply voltage and energy in the wire, and in the worst case, start a fire.

I'd guess 2 possibilities. The first (and one you should be dealing with anyways) is that you don't have a decoupling capacitor on your flip-flop. You need a 0.1 uF ceramic between V+ and ground, as close to the chip as you can get. Putting it on the breakout board is best, assuming you can without damaging the chip.

The second possibility has to do with your connection to the signal generator. How long is the wire, and how fast are the transitions? If you're not properly terminated and the leads are long, the flip-flop may be responding to ringing on the clock line.

Your connection does not require worrying about timing from input to output. A delay will not make a difference.

Neither of your chips are differential, so I really can't advise you on this issue.

You are handling set and reset lines exactly right.

And while it's a little late, you might consider a different flip-flop. Since you're working with protoboards and are clearly new to the subject, I'd suggest that you get a few 74HC74s in DIP packages. These will simply plug into your protoboard without the need for adapters. For learning how logic chips work, this is a whole lot easier on the eyes and pocketbook that trying to start with little surface mount devices.