I'm trying to build a project on automatic liquid dispenser using the ATmega328P. I've made the circuit. Before making the PCB I've few issues to be solved.

  1. I'm powering the board using a 12V, 5A adapter. I have a 7805 regulator at the input side. I'm planning to run a small submersible 5V pump (5V,200mA) which is also connected to the same regulator. Is that OK?

  2. The entire board will consume around 150mA of power during idle. During pump actuation it will consume 600mA and the regulator will take 1A of load. Will it be ok during starting of the motor? (Inrush current, I'm using 5A supply. Pump motor actuation is discontinuous say 15 sec for a minute, heating of the regulator is not an issue I've calculated in detail.)

  3. I've chosen the filtering capacitors for regulator based on internet search. Is that it OK? Please suggest.

  4. I've made a common ground and planning to put ground plane. Will it be ok during motor and relay actuation?

Any other suggestion will be helpful.


  • 2
    \$\begingroup\$ Have you calculated how much watts the 7805 has to dissipate? Have you looked at 7805 datasheet how much temperature rises due to calculated heat dissipation? Will the temperature exceed rated maximum for the regulator? Will it exceed temperature of solder melting? \$\endgroup\$
    – Justme
    Oct 26, 2020 at 9:43
  • \$\begingroup\$ @Justme: yes i've calculated the power dissipation in detail . pump motor actuation is discontinuous . heating is not an issue . thanks \$\endgroup\$
    – Anil Suha
    Oct 26, 2020 at 10:25
  • 2
    \$\begingroup\$ OK, if you are sure you did the calculations right. My rough calculations about regulator heating indicates it will go into thermal shutdown, but you have not revealed the details or calculations, so I don't know which 7805 package is in question and whether there is a heat sink. \$\endgroup\$
    – Justme
    Oct 26, 2020 at 10:37
  • \$\begingroup\$ @Justme : yes i've calculated the max current to 150mA (idle) and max current to 600mA. so power dissipation is not a problem i'm using TO220 package and also providing provision for heat sink. \$\endgroup\$
    – Anil Suha
    Oct 26, 2020 at 10:49
  • \$\begingroup\$ Please show your calculations at 600 mA. \$\endgroup\$
    – winny
    Oct 26, 2020 at 11:58

1 Answer 1


I'm powering up the board using 12V,5A adapter. I have a 7805 regulator at the input side. I'm planning to run small submersible 5V pump(5V,200mA) which is also connected to the same regulator. is it OK?

Probably not.

That's 7 V drop you'll have in the regulator, and you're drawing 0.2 A; that makes 7 V · 0.2 A = 1.4 W of heat that you're producing in the regulator whenever your pump is running.

Depending on which variant of the LM7805 you're using, and how much cooling you add to that, this might either be on the edge of still working, or being so hot that your regulator gets damaged or stops working.

Really, nothing in your circuit uses 12 V – directly use a 5 V power supply would be my recommendation. If you were my neighbor, I'd just reach into my trashbag of old 5 V supplies and give you one. But you're not my neighbor (I think – else, come over for a coffee and say hello!), so you'd have to buy one (used or new), or find one that someone else threw away because the phone they used to charge with it no longer works.

If you really need to stick with the 12 V (I don't think so, honestly), you'll want to use a switch-mode step down (buck) converter to get from 12 V to 5 V. That doesn't get as hot.

By the way, there's no reason you'd have to use 5 V to drive your relay. You could just as well use a relay with a 12V-rated coil instead.


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