0
\$\begingroup\$

I have battery source which provides 12V supply and I need to feed 5V to the MCU. The electronic in which it will be use, will be running continuously for 10-12 hours per day. I have very less current requirement of around 100 mA.

The battery has high capacity as used in motorcycle. So battery life is not of much issue. But the heat dissipation is.

I am planning to use L78L05 positive regulator from ST. Here is the part :

I could have use the switching regulator but it will be expensive, I have to keep BOM cost as less as possible. So is it a good idea to use above part? Any other part can I use?

I know the heat dissipation will be higher when using the linear regulator, but using switching regulator will make huge impact on the cost.

Any other alternative to step down supply from 12V to 5V with less heat dissipation and low BOM cost.?

Thermal Data of the part : enter image description here

\$\endgroup\$

closed as off-topic by Arsenal, Eugene Sh., JRE, Bimpelrekkie, Leon Heller May 4 '17 at 15:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ Switching adjustable PS modules are available from online marketplaces for as low as couple of bucks. \$\endgroup\$ – Eugene Sh. May 4 '17 at 14:52
  • \$\begingroup\$ Please suggest. Any Link for reference ? \$\endgroup\$ – RS System May 4 '17 at 14:52
  • 1
    \$\begingroup\$ A linear regulator will draw about 1 A-hr per day from the battery. A switching regulator will draw half that. This means the the switching regulator will double the life of the battery between charges. If you don't care about battery life, that's your choice. Plus, the 78L05 has a maximum current of 100 mA per your data sheet. Running any component at maximum is likely to shorten its operating life, even if you can dissipate the heat. Which is more important, low BOM cost or product reliability? Again, your choice. \$\endgroup\$ – WhatRoughBeast May 4 '17 at 15:06
  • 1
    \$\begingroup\$ @FakeMoustache No nonsense! The part I am using is worth 0.2$ - 0.3$ the extra 0.7$ will make huge impact on the cost. \$\endgroup\$ – RS System May 4 '17 at 15:18
  • 1
    \$\begingroup\$ It's far worse than that - you should really read up on the subject of automotive electrical systems. \$\endgroup\$ – Chris Stratton May 4 '17 at 15:39
5
\$\begingroup\$

Any linear regulator will waste most of the battery energy as heat in this application.

With 12 V in and 5 V out, the regulator drops 7 V. That times the 100 mA current is 700 mW. That may be above what a 78L05 can handle. Check the datasheet. A 7805 in TO-222 or DPAK package should be able to handle it, though.

A switching regulator will cost a little more, but also give you significantly more run time from the same battery. With the linear regulator, the battery will see a drain of 100 mA. With a 90% efficient buck switcher, the battery will see a drain of 46 mA. The switcher more than doubles the battery life.

Look at the whole system cost. You haven't given us any particulars, but a extra dollar or two for a switcher may well be more than offset by allowing for half the battery. That also makes the final product smaller, and there isn't 700 mW of gratuitous heat to get rid of.

\$\endgroup\$
  • 1
    \$\begingroup\$ The battery life is not of much issue because, the electronic will be fitted into the motorcycle which has high capacity and gets charged everytime the motorcycle is running. I am only scared of the heat dissipation and the life of regulator. As you mentioned the heat dissipation will be higher. The cost is the important factor and also reliability of the design has to be taken into consideration, adding switching regulator to the board will make the electronic expensive, but reliable. I have updated the question with thermal data of the part. \$\endgroup\$ – RS System May 4 '17 at 15:16
  • \$\begingroup\$ I would expect a motorcycle battery to be of fairly low capacity - it will be much smaller/lower capacity than a car battery - so you should be concerned with the power draw of your device. \$\endgroup\$ – Peter Bennett May 4 '17 at 15:41

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