Attached are schematics and an image of a circuit I made for a project. I asked this question months ago with an earlier iteration of the circuit and finally got around to assembling it on a single, new PCB. There is one important problem:

The wall-wart (12V 1A) simply blinks rather than work normally. It is a nameless wholesale wall-wart, but still worked reliably. I suspect I may have a short somewhere, or hopefully I simply added one through shoddy assembly.

Schematic of PCB seen in the footprint below Schematic of PCB seen in the footprint below

Footprint of PCB used Footprint of PCB used

Schematic of the PCB with the addition of a jumper wire Schematic of the PCB with the addition of a jumper wire

Image of the PCB as assembled and tested Image of the PCB as assembled and tested

Top view of blank PCB Top view of blank PCB

Bottom view of blank PCB. The orange jumper (seen in pictures above) runs from the "INPUT +" to "OUTPUT +" Bottom view of blank PCB. The orange jumper (seen in pictures above) runs from the "INPUT +" to "OUTPUT +"

Possible errors and how I have addressed them (with the intent of eliminating them in the next iteration):

a. holes for photoresistor pins too small–– fixed by filing down pins until they fit and then adjusting the footprint for next version

b. LED+ does not connect to input+ 12V –– fixed through a jumper visible only in the pictures. Next version will include that connection

c. Solder-joints poor–– admittedly KitchenAid is not known for their reflow solder systems, nonetheless this time around the solder paste (same paste as last time) barely melted after 8 minutes and it was only by centering the PCB on the oven rack that by minute 13 it had melted. (280F convection bake, I think normal bake works better) I touched up the capacitors later, manually soldered all THT parts, and the MCU was soldered and desoldered many times manually. An incorrect footprint for the LED also contributed to this, but it should still work based on the dimensions.

BOM: https://docs.google.com/spreadsheets/d/1vrnyYMaL4LTUEdR-Fa35R20bMgI528D4ixzeeOuktbI/edit?usp=sharing

  • 2
    \$\begingroup\$ Do you have a bench power supply? Something that will show you the current draw of a load? Whenever I test a prototype of anything, the rule is to current-limit the supply and monitor current draw. If it's not within expected limits, there may be a short/fault. If you don't current-limit, you could release magic smoke, rendering the diagnosis of failure much more complicated (since now you may have multiple failures). \$\endgroup\$
    – JYelton
    Commented Jun 28, 2022 at 22:58
  • 1
    \$\begingroup\$ Did you check for power shorts with a DMM ohmmeter. Power off of course. \$\endgroup\$
    – Mattman944
    Commented Jun 28, 2022 at 23:37
  • 2
    \$\begingroup\$ The PCB drawing looks like the red traces short out a lot of pins. Please post a photo of an empty PCB. Zoom in so that there is less table and more PCB compared to the current picture. \$\endgroup\$
    – Justme
    Commented Jun 28, 2022 at 23:48
  • 1
    \$\begingroup\$ I would cut or desolder the input pin of the 7805, so only the input caps and wiring are in circuit, and check again. If the power still fails, then the short is in the input wiring. If the power works, the short is in the 5V circuit. Solder the regulator pin back together afterwards. \$\endgroup\$
    – emrys57
    Commented Jun 29, 2022 at 14:21
  • 1
    \$\begingroup\$ @Lorenzo The reasoning is if you expect your board to draw 100 mA, when you connect it to a current-limited supply (say at 200 mA) and it immediately reaches 200 mA and stays there, you know you have a short, or a component failure, etc. If you were to connect it to a supply that's not current limited, and it can deliver a lot more, then you can potentially destroy any component affected by the fault. \$\endgroup\$
    – JYelton
    Commented Jun 29, 2022 at 14:53

1 Answer 1


Do things step by step. Start with a fresh board. Connect to a lab power supply. Ensure there's no current draw. Then add the components in front of the 7805. Try again. Then add the 7805. Try again. And so on.

The quality of the assembly process seen on the photograph is likely the source of all the trouble: there's no proper reflow of the joints, and there likely are solder whiskers that bridge things together.

For example, just a 7805 with capacitors should draw under 10mA.

  • \$\begingroup\$ Sorry, but could you please explain the process behind ensuring there's no current draw? Is the idea that I build another copy of the board only without the power regulation components, and then measure the current draw using an external power supply? By "in front of" do you mean a specific group of components or just that I should generally test chunks of the board at a time? Thank you for your answer and I apologize if I am asking you to repeat something straightforward. \$\endgroup\$
    – Lorenzo
    Commented Jun 29, 2022 at 2:10
  • 3
    \$\begingroup\$ @Lorenzo The goal here is to check that each functionality of your board is OK before soldering the next one. Thus the suggestion is to start by the power supply block of your board; Check it; and then solder the next part and so on. Why we are talking about the power drawn ? it's because it's an indication of good or bad behavior of the board and it's easy to measure. If you solder your 7805 with its surrounding caps, you know that, by design, this should not draws hundreds of mili-amps but only a small current. Thus if you measure too much current: you have a problem and you know where it is \$\endgroup\$
    – Blup1980
    Commented Jun 29, 2022 at 7:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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