I am designing an embedded system having following components:

  1. One arduino uno (~300-500mA 5V=2.5W)
  2. One nodemcu (esp8266) (~300-500mA 3.3v=~2W)
  3. One PAM8403 based audio amp board (3W output, so on 5V it should be ~600mA)
  4. One custom made mic preamp (2n2222A+LM358) (don't know, should be within 200 mA=1W)
  5. One car door lock actuator (12V, running for 500ms while activated, assuming 1A required so power is 12W).

Total power=2.5+2+3+1+12=~21W

All of the above systems are connected in a same system. I want to use one PSU that would provide power to all of the modules in the system. I bought a 12V 2A switching power supply, assuming it would be enough to run all systems together.

So I made a regulator board using two 7805 (one for arduino and preamp, another for audio amp), and one lms111733 (for esp8266) (All TO220 package for high power handling), feeding them by the 12V PSU. Problem is while I connect all systems all regulators are running hot, and most hot is the one connected to the audio amp, its so hot it burnt my hand (ran only for a minute or so).

I read through couple of forum posts and found that the 7805 linear would run hot as it would dissipate the power (input power-output power) inside itself. For example if the output is 5V-1A then it would dissipate 12V*1A-5V*1A=7W in itself so the heat.

In such situation I can use one 12V PSU along with 1 5V mobile charger adapter kind of things, but I want to stick to using one PSU. Can anyone suggest how to design the power supply for such system.

  • 1
    \$\begingroup\$ Use a switching regulator to do as much of the drop as you can and still supply enough voltage to the 7805, and use the 7805 to get you to 5v \$\endgroup\$ Aug 19, 2016 at 10:39
  • \$\begingroup\$ The TI website has a custom solution for multiple outputs say using a small laptop charger \$\endgroup\$ Aug 19, 2016 at 20:29

2 Answers 2


Like Scott Seidman already said in his comment you can use a switching regulator. But i would suggest to replace the 7805 with the regulator. You could use for example the LM2575. This one comes in your preffered TO220 package, too.

Another sideeffect is that your circuit wouldn't consume as much power as it does right now, because the power won't go into heat anymore.

Edit: Altered my answer according to FiddlyOhm's comment. I researched too less, sorry for that. My intention was to say that you don't neet a fixed voltage regulator anymore if you have an switching regulator like the PT5101. I hope it is more clear now.

  • \$\begingroup\$ Your answer is ambiguous. DO you mean the OP should replace the entire 5 volt regulator chain with the BA05CC0FP? Or, should do you mean he should still incorporate the switcher suggested by Scott Seidman, PLUS use the BA05 in place of the final 7805 regulator? \$\endgroup\$
    – FiddyOhm
    Aug 19, 2016 at 11:31
  • \$\begingroup\$ I mean he should replace every 7805 with a switching regulator. \$\endgroup\$ Aug 19, 2016 at 11:51
  • \$\begingroup\$ Hi, Thanks for the response. What I understood is: LDO, or normal linear regulators dissipate excess power as heat given by the expression (Vin-Vout)*Iout, and the efficiency is given by Vout/Vin. In my case it is pretty low(5/12). I should use a switching regulator to reduce the voltage close to linear regulator input (for 7805 it is 6.2v), and then feed it to linear regulator. I have a lm2596S regulator board, probably it can help. Kindly confirm. Thanks. \$\endgroup\$ Aug 19, 2016 at 20:22

Have you considered a surplus PC PSU with 12,5,3.3V ?

Then mod. the cable harness to suit your power connections and check power sequence issues , if they exist.


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