Hello,
I am building my first PIC circuit.. for the power supply I am using a cheap wall-wart (rated 500mA) and setting the input to be 12V.
This feed a 7805 but when I check the output with no load I get ~1.45V.. So I have the following questions now:

Have I managed to fry the 7805? (I might have connected it wrong way first.. hence the doubt)

Based on the circuit attached.. how can I calculate the current required, The reason I am asking is, I want to calculate what kind of battery life I would get if i were to abandon the 7805 and go with a battery pack.. I really don't want to do this as I realize the potential of the 7805 but I also want to get my first circuit working )

PS: Ic(min)=500 mA for the BC301

 

Hello,

You could have damaged the 7805 by connecting it the wrong way.
Also check the capacitors for a short.

Bertus

 

You have a buzzer connected directly to a PIC output pin. What is the power requirement for the buzzer? Voltage and current rating?

Is the input voltage at least 8v when you measure it?

Double check to see if you have connected the 7805 backwards.

The capacitor from input to ground is critical to prevent the regulator from oscillating. It should actually be 0.33uF. If the capacitor is bad (open), missing, or simply not connected properly, the regulator may oscillate at high frequencies, particularly if the wiring to the supply is more than a few inches. The output cap is similarly critical, although the datasheets claim that it is not.

The problem you can run into with the 78xx series is very small amounts of capacitance on the input or output, combined with the inductance of the wiring, can cause these high frequency oscillations. You wind up getting about 1.2v to 1.5v RMS out, but it is a rippled DC anywhere from 10kHz to 15MHz or so, depending on the length of your wiring.