Looking at your project's netlist, it appears that terminal block J1 is connected to the output (pin 'OUT') of the L7805 voltage regulator. Given that L7805 is a 5V voltage regulator, J1:P1 should indeed be providing a 5V output.

As for the current, the L7805 typically can provide a maximum output current of 1.5A. However, the actual output current will depend on multiple factors such as the load connected to the terminal block J1:P1, the input voltage to U1, the temperature, etc.

You should refer to the L7805ACD2T-TR data sheet for a more specific value for your exact conditions and make sure that your actual load on the terminal block J1 doesn't exceed the maximum current spec for the voltage regulator. Be also aware that increasing the output load will increase the heat dissipated by the 7805 chip, so cooling considerations may be needed.

what is the voltage output at the terminal block now? is it still 5v?

The output voltage at the terminal block J1 is determined by the voltage regulator, U1. Since U1 is an L7805ACD2T-TR, a 5V voltage regulator, the output voltage at J1:P1 should be 5V. This assumes that the conditions necessary for U1 to regulate voltage correctly are met (i.e., correct input voltage range, also taking into account temperature, loading conditions, and other environmental conditions as stated in the datasheet for the L7805ACD2T-TR).

Please note, these conditions include a minimum load current for the regulator to maintain regulation. From the component information provided for the L78 series, the minimum load current for regulation is 5 mA. If the load connected doesn't draw at least this amount of current, the output voltage may rise above the expected 5V. Do validate this with actual measurements during your prototype testing.

if you look at teh net, it has R2 which has a value of 10k. should not it reduce as there is a voltage drop?

The L7805 voltage regulator provides a constant output voltage of 5V, unchecked with any load changes within its power limits. And that's why it gets connected directly to J1:P1 through Net 3.

Now coming back to R2, it is connected to J1:P1 through Net 2. Essentially, it forms a voltage divider with the load that will be connected to the terminal block. The voltage at J1:P1 will still be 5V.

However, you are correct that there will be a voltage drop across R2. So if you were to measure the voltage across R2, you'll find that it's less than 5V due to the voltage divider formed by R2 and the load. This does not affect the voltage at J1:P1 since R2 is not in series with the output of U1. It's in series with the load that will be connected to J1.

You can calculate the voltage drop across R2 using Ohm's Law, $V=IR$, where I is the current passing through it. The current can be calculated as $I = V/R$, where V is the voltage across the entire circuit (5V) and R is the combined resistance of R2 and the load in J1.

Please note that the load at J1 should be chosen carefully to ensure the minimum load current of 5mA for regulation is maintained for U1, and not exceeded to the point of overheating the voltage regulator.

The value for the capacitor C2 isn't specified in the information provided. The correct capacitor value can vary based on the specific requirements of your circuit.

However, in a typical 7805 voltage regulator circuit, a capacitor between the output (pin OUT on U1) and ground is used to improve transient response. Common values are 0.1uF or 1uF, but refer to the datasheet for your specific regulator (L7805ACD2T-TR in this case) to select the optimal capacitor value. Always consider your specific requirements regarding stability, noise, and transient response.

The L78 series regulator recommends an input bypass capacitor for stable operation under all load conditions. The datasheet suggests a $0.33 \mu F$ or larger capacitor with low internal impedance at high frequencies.

So, for C1, which is connected to the input (IN) pin of the L7805ACD2T-TR (U1), the preferred capacitance value would be $0.33 \mu F$. So it's better to go with $0.33 \mu F$ over $0.22 \mu F$. This results in good high-frequency characteristics ensuring stable operation of the regulator under all load conditions.

Please ensure to mount this bypass capacitor with the shortest possible leads directly across the regulator's input terminals to minimize ground loops and lead resistance drops.

@copilot what happen if i put a 47uF capacitor?

Using a 47uF capacitor as an input bypass capacitor for the L7805ACD2T-TR regulator (U1) is definitely possible and it will provide even better filtering of the input voltage. However, keep in mind that a larger capacitance value means a larger physical size of the capacitor, so ensure that you have enough space on your PCB.

The main purpose of this input bypass capacitor is to filter any high-frequency noise present on the input supply line and prevent it from disturbing the operation of the regulator. The datasheet recommends a capacitor with a value of $0.33 \mu F$ or more, so the 47uF value is well within this specification.

The value of 47 uF is typical for electrolytic decoupling capacitors in many power circuits. It will provide more "bulk" decoupling, improving the regulator's response to rapid changes in load current, but might be more than you need. The higher capacitance, in general, provides better smoothing or noise reduction.

Please remember that if you use an electrolytic or tantalum capacitor, take care to observe the correct polarity when you connect it, as these capacitor types are polarized.

The resistor, R1, is connected in series to the LED (D1) in your voltage regulator circuit. Its main purpose is to control the current that flows through the LED. The value of this resistor can be calculated by using Ohm's law:

R = (V_input - V_LED) / I_LED,

where:

Typically, a red LED has a forward voltage drop of about 2V and a desired current of 20mA (0.02A) for normal brightness. Given your input voltage of 12V, the value for the resistor would be:

R = (12V - 2V) / 0.02A = 500 Ohms.

Here, a 500Ω resistor is already in use, so no changes are necessary for a 12V input.

It's important to know the specific forward voltage and desired forward current of your LED to get the best results. These parameters can typically be found in the datasheet of the specific LED in use.