Toggle transistors with an shiftregister to light up an LED
Assuming your RGB LEDs are common anode, you will need 5 PNP transistors for level control and 75 NPN transistors for LED control.

For PNP it has to handle maximum 75 LEDs and for NPN it has to handle maximum 5 LEDs.
For PNP I recommend using Darlington transistors.

To saturate PNP, \$V_B&lt;V_E\$ and \$V_B&lt;V_C\$. And to saturate NPN, \$V_B&gt;V_E\$ and \$V_B&gt;V_C\$.







simulate this circuit &ndash; Schematic created using CircuitLab

In saturation region, transistor amplifies base current, so there is no need to pass high current to base-emitter. Thus, you can use base resistor. In fact, you should, because of current limit of shift register and transistor.

For minimum base resistor value:

$$R_{R2}=\frac{V_{CC}-V_{Q1_{be}}}{(V_{CC}-V_{Q1_{ce}}-V_{D1}-V_{Q2_{ce}})/R_{R1}/h_{fe}*2*75}$$
$$R_{R3}=\frac{V_{CC}-V_{Q2_{be}}}{(V_{CC}-V_{Q1_{ce}}-V_{D1}-V_{Q2_{ce}})/R_{R1}/h_{fe}*2*5}$$

Refer to transistor's datasheet for \$V_{be}\$(Base−Emitter Saturation Voltage), \$V_{ce}\$(Collector−Emitter Saturation Voltage), \$h_{fe}\$(DC Current Gain).
(There may be voltage drop in shift register, but it's not significant, so I omitted it from calculation.)

To be safe, use minimum \$h_{fe}\$ and maximum \$V_{be}~and~V_{ce}\$.

Also, there is high-power shift register. You might want to consider it.