linear voltage regulator (power supply for MCU)

[nikens]

I am building simple PIC based PCB and I need proper power supply for it.
Can I use L78M05CV 500mA, 5V votage regulator?
Can I connect this IC like below schematics from datasheet shows?
Do I need any additional capacitors...?

 

[svhb]

The usefulness depends also on the input voltage you apply to the 7805. And if 0.5A is enough for your system.

The input must be above 7.5Volt for correct regulation, and below a certain voltage depending on the maximum allowable power dissipation (heatsink). (P = (Vin-5)*Iout)

For the capacitors, I should add a big electrolytic capacitor (100µF) at the input. But this depends also on how stable is the input voltage is, and what peak currents you expect.

Stefaan

 

[dick_freebird]

In addition to the caps shown, you would put whatever the
suggested decoupling is from the MCU vendor close-in to that
device. What is shows for the regulator is its minimum
for self-happiness and does not speak to the load's needs.

 

[nikens]

The usefulness depends also on the input voltage you apply to the 7805. And if 0.5A is enough for your system.

The input must be above 7.5Volt for correct regulation, and below a certain voltage depending on the maximum allowable power dissipation (heatsink). (P = (Vin-5)*Iout)

For the capacitors, I should add a big electrolytic capacitor (100µF) at the input. But this depends also on how stable is the input voltage is, and what peak currents you expect.

Stefaan

input will be 12V car's battery.
I intened to power up one PIC16Fxxx, two ILQ optocouplers, one ULN2003A darlington and thre relays (each 40mA @ 5V).

About electrolytic capacitor; isn't this needed when you connect this IC to the rectified AC source?

About power disipation 12-5V * 400mA = 2.8W. Is this true that at 400mA (as I predicted the max current will be) this IC will produce almost 3W of hetat?
I can live with 2W (5V * 400mA) max power flow (normaly 300mW), but additinal 3watts will drawm my car's 60Ah battery in 10 days! The application is meant to be used as a car alarm unit and must be able to run in standby mode as long as the battery isn't self emptied.

 

[svhb]

the total power dissipation will be 12*0.4 = approx 5W, where 3W is lost in the linear regulator (with added heatsink).

You can use an SMPS-design instead, or a simple replacement like **broken link removed**

Stefaan

 

[pranam77]

Untill the regulator is intended to supply only the MCU, the given device will suffice. Else you may use LM7805 with better current capacity than the later. The input of the regulator may be filtered with 1000Mfd 25V and output with 100Mfd 25V capacitors for best results....cheers