# Am I risking a short circuit/gaping situation?

I am rather new to the DIY/hobbyist electrical engineering scene. I am working with an arduino, a 9V power supply, and a few 7805 voltage regulators. I bought some sort of run of the mill PCB board with many many copper squares and pin holes on it, and I have been using that to construct my project. In the end I am trying to drive up to about 100 LEDs.

The copper squares on the pcb board seem really close. And yes, I am sure the wiring/soldering job I have done is rather horrid, but I have a limited budget and knowledge.

What I would like to know, as I have had a hard time tracking down an answer, is whether or not my solder joints and the contacts of my various components are too close together and I am risking a situation where an arching might occur. I am really not all too sure on how much distance it takes for 5 volts to gap. So far I haven't had any issues, but since I want to make a gift for someone I care about, making sure its relatively safe for them to possess would make me feel comfortable giving it to them.

I have linked to two pictures, one of the underside of my board, and the other of the side of the board with the 7805s in sight, so that hopefully some rough idea of scale here is possible.

Thanks much in advance!

• Get yourself some flux. The solder isn't wetting the pads properly. Or even some sandpaper would help. – Ignacio Vazquez-Abrams Sep 11 '14 at 3:39
• Okay, can do. What about the distance though? Would these pads ever gap at 5 volts? – user53006 Sep 11 '14 at 3:46
• at 5v and the power levels you're dealing with, you don't really need to worry about arcing. Only worry about accidentally soldering shorts. Just improve your soldering skills so the joints look nice, and you'll be in the clear. – whatsisname Sep 11 '14 at 3:46
• I estimate my circuit could draw up to 2.5 amps, but again, only at 5 volts. @whatsisname do you feel that the arching still isn't an issue? – user53006 Sep 11 '14 at 3:47
• Arc distance is dependent on voltage. – Ignacio Vazquez-Abrams Sep 11 '14 at 3:49

The breakdown voltage of air is about 3kV/mm, so there will be no problem. With 5V this will be 1.7µm.

And if you leave always 2 rows of copper squares alone, there shouldn't be a problem with short circuits.

But: If you want to use about 100 LEDs (each drawing 20mA i assume?) each of the 7805 (you use 3 all in all?) will need about 700mA. I don't know your input voltage. But depending on it and Ohm's Law the 7805 will be pretty hot... You can calculate the power dissipation with $$P_W = (U_{in} - U_{out}) \cdot I_{out}$$

In your datasheet you will find a diagramm showing IC temperature over P_W.

• +1 for adding about temperatures...too often overlooked. – Joshpbarron Sep 11 '14 at 8:08
• I like the use of the real brakedown voltage, but it might also be worth pointing to the IPC-2221 standard on safe conductor spacing (or a summary of it such as this one) which gives a nice table for higher voltages. – Xcodo Sep 11 '14 at 8:50
• Regarding the heat dissipation, the OP may find My linear voltage regulator is overheating very fast useful. – Andrew Morton Sep 11 '14 at 10:32
• I have a 9 volt power supply rated at 3000mA. I don't really understand what you mean when you say 'leave always 2 rows of copper square alone'. There isn't two rows of copper in between the contacts of my 7805s. I also tried looking on the datasheet for the heat dissipation properties, but was unable to find it. Maybe I am not reading the datasheet correctly? sparkfun.com/datasheets/Components/LM7805.pdf – user53006 Sep 11 '14 at 11:59
• About the copper rows: You have to be sure that the solder won't connect two circuits for a short circuit. Use a multimeter while soldering to ensure that :) – GER_Moki Sep 11 '14 at 12:50