good day!!

my batteries current rating is 2300mah, and my 7805 cant handle it, it gets to hot when the power is turn on..

my question is it posibble to parallel the two 7805 to minimize and share the current rating of 2300mah between the two voltage reg ic's?..

if this is possible can you share to me the schematic on how doing this. thank you...

 

What is your batteries voltage? For this problem it is more important than their AH rating.

 

sir my batteries voltage is 7.8 when i measured it with digital vom..

 

Hello,

Can you post a schematic of how you have connected the battery to the 7805 ?
Are there capacitors at in and output of the 7805 (about 0.1 uF ) ?
If your battery is the input of the 7805, the voltage will be to low.
The 7805 needs minimum 3 volts difference between in and output.
The minimum input voltage should be 8 Volts.

Greetings,
Bertus

 

That and you need to know current (which AH is not) that is flowing through this device.

 

Hello,

The capacitors I told you about are there to prevent oscillation of the 7805.
If the 7805 is oscillating it can get hot to.

Greetings,
Bertus

 

The heating of the regulator is caused by the voltage across it times the current through it.
The input voltage is too low so the regulator might not regulate properly. The 7.8V lithium battery will drop to 6V when it needs recharging.
You did not tell us the load current which might be too high for a 7805 regulator that has a max allowed current of 1A.

Two regulators can be connected in parallel if each has a low value resistor in series with its output and the output of the resistors are the combined output. But then the voltage regulation is ruined.
The datasheet for most regulators show how a current-boosting transistor circuit can be added. But then additional parts are needed for current-limiting.

An LM317 regulator has an output current of 1.5A.
An LM350 regulator has an output current of 3A.
An LM338 regulator has an output current of 5A.
Do you need more current?