except that the parts for individual LEDs will be replicated 16 times instead of just 4 as shown.
Since you have 16 LEDs each drawing 20 mA, the total current when all are on at the same time is 320 mA. R1 will cause about 10 mA base current in Q1 when the processor drives the common enable line low. That means the transistor needs to have a gain of at least 32, which this transistor can do. The other transistors are only handling the current for individual LEDs, which is 20 mA, so don't need as much base drive. That is why R2-R5 can be larger than R1.
R6-R9 are sized so that after a little voltage drop in the anode and cathode transistors, about 20 mA should flow thru the LEDs if they are normal green ones. Red LEDs have a little lower voltage drop, so you may need a little higher series resistor.
Best Answer
Yes you can do like Schematic 2. The LED will light till the battery wears down.