saeed_pk-
Your analysis is completly incorrect. The LEDs would not be in parallel, they would be in series - the total current is therefore 150ma as serpa4 has stated.
serpa4 -
The triangle symbol in the schematic denotes an "Operational Amplifier" which is a basic building block for many different types of electronic circuits. If you take a look at a data sheet for the 14pin device, you will see the same schematic symbol used in a top view of the physical chip (see
https://www.onsemi.com/pub/Collateral/LM324-D.PDF ). The pluses and minuses are clearly marked.
Note that there are four of the amplifiers in the same package - your circuit only needs to have two of the four connected. You can pick any two that are the most convenient for you to wire. Pin 4 and pin 11 (VCC and VEE) are the power supplies for the chip that are shown in the schematic as straight lines coming out of the top and bottom of the triangle symbol. VCC is the positive supply and VEE goes to ground (the negative supply).
For the IRF9520 field effect transistor (FET), G, D, and S stand for GATE (G), DRAIN (D), and SOURCE (S). If you look at the data sheet from the Harris Semiconductor databook (see **broken link removed** ), you will see a diagram for the FET that uses the same schematic symbol as your reference schematic. The source (S) is the top line, the gate (G) is the middle line that connects to R4, and the drain (D) is the bottom line that connects to R5.
Observe that Bob's project photos show a large aluminum heat sink as part of the construction. That is because this circuit will get hot while it's in use. Without some way to remove the heat from the immediate location of the chips, the heat would destroy the circuit within a short time.
Be sure to read Dr Bob's article carefully. The output voltage is controlled by the ratio of R2 To R3, and his example is set for a 6volt output from the first stage. That 6 volts is then amplified to 12 volts in the following stage, so the regulator output is 12 volts with 15 volts input. If your input falls below 12 volts, there will be no regulation. If you want a constant 11 volt output, you'll need to adjust the values of R2 and R3 slightly to change the output of the first stage to 5.5 volts instead of 6. This could be accomplished by replacing the fixed resistors with an 8K potentiometer - that would make the regulator adjustable.
The 25uf capacitor in the parts list will work in the circuit. "WV" stands for working voltage. A 50WV capacitor can tolerate a higher working voltage than the minimum 16WV shown in the schematic. Any capacitor with a working voltage rating higher than 16WV will work in the circuit shown - you could use 20WV, 25WV, 35WV, etc. Note that the higher the working voltage, the larger the physical size of the capacitor will be.
Since you are being brave and venturing into a project with very little knowledge, you may have some failures. Use a fuse in series with the supply to your little circuit so any mistakes are less likely to wipe out all your parts. A blown fuse is a minor annoyance while you are learning, and it beats having to replace a bunch of burned out semiconductor parts.