0
\$\begingroup\$

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

Schematic

\$\endgroup\$
  • 4
    \$\begingroup\$ what is the load you are going to be driving with that? \$\endgroup\$ – ratchet freak Sep 15 '17 at 14:37
  • \$\begingroup\$ will I am making an arduino extension .. so basically arduino + DC motors + Servo Motors + Sensors maximum current by these parts is around less than 1A in total \$\endgroup\$ – ihsanogluu Sep 15 '17 at 14:38
  • 1
    \$\begingroup\$ A linear reg is a self regulating resistor. The voltage dropped over it multiplied with the current through it will be the power dissipation. \$\endgroup\$ – ratchet freak Sep 15 '17 at 14:45
  • 1
    \$\begingroup\$ You'll get better battery life with switching regulators... \$\endgroup\$ – bobflux Sep 15 '17 at 14:46
  • \$\begingroup\$ -1 for not including any of the information that we'd need to actually answer the question, even after being asked. Please edit the question to include the necessary information. \$\endgroup\$ – The Photon Sep 15 '17 at 16:03
5
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ 1A will be pulled from the 9V regulator only .. the 5V regulator is used for sensors and low current applications.. \$\endgroup\$ – ihsanogluu Sep 15 '17 at 14:48
  • \$\begingroup\$ thank you for your comment I would really appreciate it if you write a switching regulator name like 7805 \$\endgroup\$ – ihsanogluu Sep 15 '17 at 14:50
  • \$\begingroup\$ There are plenty - both Linear technology and TI have good offerings and both have selection tables that will help you decide. You pick one and tell me and I'll take a look but please embed the link to the data sheet. \$\endgroup\$ – Andy aka Sep 15 '17 at 15:17
  • \$\begingroup\$ thank you for your effort ... I will try to find one and make a schematic according to its datasheet and show it to you afterward. \$\endgroup\$ – ihsanogluu Sep 16 '17 at 11:00
2
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

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.

enter image description here

You could also consider replacing that series diode with a MOSFET to reduce the voltage drop for reverse-voltage protection.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.