HI, I've designed my power supply circuit as shown below:
First, if I use 1N5392GP diode as the bridge rectifier, then can I use 1A transformer? or I need more than 1A?
Second, are the values of those components (capacitor and resistor) correct?
Third, when I use it to supply my AVR, LCD, and optocoupler, the 7805 regulator is getting hot (i dont know why). So I modify it to become like this:
I add TIP32C to 7805 regulating circuit. Well, I know by adding this transistor will provide a higher current to the load, then do I need to change my transformer and diodes (related to the current)?
You should have at least 3V higher before 12V 7812 regulator, I mean when load is present. After bridge rectification you will get 12V x 1,414, but on load this will fall down for few volts.
You dont need D4 diode.
Use heatsink for both regulators.
1A is fine for most regular projects with 7-seg LED displays, LCD displays, uC,....
You should have at least 3V higher before 12V 7812 regulator, I mean when load is present. After bridge rectification you will get 12V x 1,414, but on load this will fall down for few volts.
You dont need D4 diode.
Use heatsink for both regulators.
1A is fine for most regular projects with 7-seg LED displays, LCD displays, uC,....
You have 7V across your 7805 which is dissipated as head.
If you put an intermediate voltage regulator between the 7812 and 7805, say a 7809, the power dissipation will be split between two ICs and the heating would not be so much.
Are you using the backlight for the LCD? If so this is would be consuming some unnecessary current from the 7805 output which is heating it up.