I am making a new design which I use a 4S 18650 batteries with total voltage of 14.8V
in my design I need to have 9V and 5V using 7809 and 7805 regulators
are these regulators going to overheat ?
here is a schematic of my design
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1 amp from a 5 volt linear regulator fed with a supply of 14.8 volts quite simply means the power dissipated by the device will be 1 x 9.8 volts = 9.8 watts. Power dissipation in the device is defined by the volt drop across it (9.8 volts) and the 1 amp flowing through it.
I'd definitely consider a switching regulator - power efficiency is going to be about 90% so, with 5 watts out, there will be about 5.5 watts in and, from a 14.8 volt supply this means it will take a current of about 375 mA i.e. a lot smaller drain on your battery.
1A will be pulled from the 9V regulator only .. the 5V regulator is used for sensors and low current applications..
That is 5.8W of heat to dissipate in U1. The L7809ABD2T that you are using is a surface mount part with a practical power limit of around 1W. Therefore it will overheat, badly. If you are to use a linear regulator you need to use the TO-220 part (L7809ABV) with a small heatsink. The better solution is to use a small DC-DC switching regulator which is far more efficient, will dissipate far less heat and use less battery power. If you like ST parts try a L5973A.
You may be OK using the L7805ACD2T for the 5V supply so long as you draw less than 100mA.
Assuming this is a hobby one-off, suggest you use an adjustable LM2596-based module for the 9V 1A supply and hang an LM7805 off the 9V input for the low current 5V supply. You can get 150mA @5V without a heatsink conservatively.
You could also consider replacing that series diode with a MOSFET to reduce the voltage drop for reverse-voltage protection.