A 7805 might be used, but it also may require an additional load resistor depending upon how much the logic circuit draws. A 7805 requires a minimum load of 5mA for guaranteed regulation; that would be a 1K Ohm resistor.
Another thing; an NE555's output decreases about 1.7V with a 100mA load, due to the Darlington voltage follower arrangement in it's output circuitry, and may decrease as much as 2.5v with a 200mA load. If you are considering using a 9v "transistor" battery, these frequently have an 8.6v output when new when tested under light load (25mA, or about 350 Ohms). So, 8.6v - 1.7v = 6.9v remaining; 8.6v-2.5v=6.1v remaining. There are some "industrial" type 9v "transistor" batteries available that output around 10v; they have seven internal cells instead of six.
A 7805 regulator has a specified dropout of 2v. If you are supplying it 6.9v, the most it can output is 4.9v; if you are supplying it 6.1v, the most it can output is 4.1v. This may not be enough to power the logic.
Earlier, eblc1388 mentioned (basically) that the blue LEDs would not light if more than two were placed in series due to their higher Vf, but in the subsequent schematic, three are shown in series. That won't work if you're only going to have 6.9v left to work with; blue LEDs may have a Vf of 3.4v to 4v.
Another thing; an NE555's output decreases about 1.7V with a 100mA load, due to the Darlington voltage follower arrangement in it's output circuitry, and may decrease as much as 2.5v with a 200mA load. If you are considering using a 9v "transistor" battery, these frequently have an 8.6v output when new when tested under light load (25mA, or about 350 Ohms). So, 8.6v - 1.7v = 6.9v remaining; 8.6v-2.5v=6.1v remaining. There are some "industrial" type 9v "transistor" batteries available that output around 10v; they have seven internal cells instead of six.
A 7805 regulator has a specified dropout of 2v. If you are supplying it 6.9v, the most it can output is 4.9v; if you are supplying it 6.1v, the most it can output is 4.1v. This may not be enough to power the logic.
Earlier, eblc1388 mentioned (basically) that the blue LEDs would not light if more than two were placed in series due to their higher Vf, but in the subsequent schematic, three are shown in series. That won't work if you're only going to have 6.9v left to work with; blue LEDs may have a Vf of 3.4v to 4v.