Thank you, CrossRoads!
1-2 days ago we used a different calculation to obtain the value of the current limiting resistors. You wrote back then:
"Resistors = inexpensive resistors, 1/8W, carbon composition. Value will depend on the LED color & the current you want to put thru them.
If have a 5V source, and the anode transistor has 0.45V across it, and the cathode shif register has 0.25V across it, that leaves the remaining voltage across the LED and the resistor.
For and LED with Vf of 3.2V way, and using 20mA as the current, then:
(5V - .45 - .2v - 3.2)/.02 = 55 ohm. 56 ohm is a standard value."
If I understand correctly, now instead of the .45 voltage drop across the anode transistor we have the (at least) 1V drop across the IRF9540. In your earlier explanation you wrote that the cathode shift register drops 0.25V, but I think your current calculation is the correct one and it drops only about 0.07V. I use standard blue LEDs with a voltage drop of 3.3V. So, basically, if your old calculation was wrong and the new one is right, it means that the remaining voltage that needs to be handled is 5V - 1V - 0.07V - 3.3V = 0.63V. Our target current is 20 mA, so R = V/I = 1.18/0.02 = 31.5. In other words 33 Ohm resistors are fine.
Is my above calculation correct? If yes, I'll go for the IRF9540 MOSFETS. That's great news! 