The PNPs will drive from 74HC595.You mean if I change it to a NPN it will drop only its saturation voltage?But if I use a PNP it will add the drop of Base Emitter drop + 74H595 drop?
Yes, the PNP emitter-follower has a base-emitter voltage loss and the 74HC595 adds its output saturation voltage loss.
The 2N2904 is obsolete and is in an old metal case. Its max base-emitter voltage is 2.6V when its collector current is 500mA.
How much will be its maximum current?
Wait a minute. The 2N2904 was an old PNP, not an NPN.
The datasheet of the 2N3904 NPN transistor shows a max allowed current of 200mA and poor performance above about 50mA. Its max collector to emitter saturation voltage is 0.3V when its collector current is 50mA and its base current is 5mA.
Again, how much will be its maximum current?
I changed my PCB layout by replacing ULN2803 with 2N3904 transistors.I got some improvement in LED brightness.I think I could lower down the value of the resistor further....!!!
The 2N3904 has about 5mA of base current which should be a little more to get a little more collector current with the same saturation voltage loss.
The 2N4401 or BC337 will be a little better.
The output of the 74HCxxx is high at +5V only with no load. It is about +4.2V with a 10mA load.
The typical saturated base voltage of the 2N3904 is shown on a graph at 0.9V. Then the base current is only (4.2V-0.9V)/330 ohms= 10mA but less if the 74HCxxx and/or transistor are not typical.
The 2N3904 has a max allowed current of only 200mA and its typical saturation voltage loss is shown on a graph at 0.13V at only 100mA.
The 2N4401 has a max allowed current of 600mA and its graph shows a typical saturation voltage loss of 0.10V at 100mA.