Is my PCB going to explode?

Aright. The PCB is a wireless Logitech PS2 pad that runs on two AA batteries. Bought a new battery holder to swap which has a built in 5 volt DC to DC step up converter in the circuit to give a Happ p360 power. So there will be 5 volts running through the whole PCB. Will resistors overload? Will capacitors blow up? Will PCB tracks melt? Will my PCB’s universe get fo’ shizzled?

Not sure if 3v to 5v is enough of a difference to cause a catastrophe but I know that people have gotten wired PCBs working with p360s and since the p360 is connected to the same circuit as everything else the +5v must be going through everything? Right? No?

Also while I’m asking, can I hook up a Sanwa Flash1 to a 3v circuit without a step up converter? I figure since it only needs another 0.3v it will just drain the battery faster or some shit.

Thank you in advance Toodles or whichever electronics genius bes my hero, this is a little over my head.

I never said anything about powering the PCB with 5v, so go back through and read whatever I wrote. I’d post it here but I cleaned out my inbox the other day.

Maybe. If the caps arent rated for 5v, they may pop. Resistors will probably be fine, and no way of knowing with ICs. But there is not logical reason to try and power the PCB with anything than the 3v it is expecting.

No. Everyone gather 'round for another damned Ohm’s law story. Let’s say that PCB has an LED on it. Everyone loves LEDs, but the principles hold true for every component. Most LEDs use about 20mA of power. THe guy who made the PCB knows this, and knows that he’s powering the pcb with a pair of batteries, for 3 volts of power. The LED has a voltage drop of say 1.4v. So the engineer puts in a resistor to make sure it gets 20mA.
R = V / I
Since the LEDs voltage drop is 1.4v, the actual voltage after the LED is 1.6v (3v from the batteries - 1.4v voltage drop = 1.6v) So we put in V into our equation
R = 1.6 / I
We know we want 20mA of current so we fill in I
R = 1.6 / 0.020
Solve for R.
80 = 1.6 / 0.020
R = 80

So he puts in a 80 Ohm resistor. All is well and good until someone decides they want to power the thing with a 5v step up converter for no reason. Now in our little equation, R is still the same 80 Ohms. But the voltage has changed and we need to see what the resulting current would be.
I = V / R
I = (5v from the converter - 1.4 voltage drop of the led) / 80
I = 3.6 / 80
I = 0.045
I = 45 mA
You’re now putting 225% of the current the LED is rated for through the LED.
Maybe it’ll be a much brighter LED. Maybe it will pop. But either way, you’re significantly reducing the life of that component.

You’ll be fine if you try and run it will a lower voltage; you wont hurt anything, it just may not work.

Power the pcb from the pair of batteries. Tap a point off of the 3v, and send that to the DC-DC converter to power the flash/p360 and only that.