-2
\$\begingroup\$

Is there a way to alter the 5V 2A wall wart into 5V 5A? I want to power neopixel LEDs but I just need slightly more current supply.

| improve this question | | | | |
\$\endgroup\$
  • 1
    \$\begingroup\$ Slightly? It's more than double. No, there is no way. \$\endgroup\$ – Eugene Sh. Oct 3 '18 at 19:13
  • 1
    \$\begingroup\$ No. "slightly more" isn't 2.5 times more. \$\endgroup\$ – brhans Oct 3 '18 at 19:13
  • \$\begingroup\$ Please read Olin's subtle answer. \$\endgroup\$ – Sparky256 Oct 3 '18 at 21:06
4
\$\begingroup\$

Converting voltages and currents requires the total maximum power to not increase. In other words, input V * I >= output V * I (this is the concept of conservation of power). Therefore, it would be impossible to convert a 5V * 2A = 10W input to a 5V * 5A = 25W output because the 10W input is not greater than the 25W output.

| improve this answer | | | | |
\$\endgroup\$
  • 1
    \$\begingroup\$ I would say input V * I >= output V * I. Given that nothing is 100% efficient, maybe just input V * I > output V * I \$\endgroup\$ – Kevin Kruse Oct 3 '18 at 19:18
  • \$\begingroup\$ yes, good point, updated answer \$\endgroup\$ – mith Oct 3 '18 at 19:18
  • 1
    \$\begingroup\$ I guess the question is a bit different and can be rephrased as "is it possible to squeeze more power from a PS than it is rated for?" \$\endgroup\$ – Eugene Sh. Oct 3 '18 at 19:19
  • \$\begingroup\$ The question is about altering a wall wart to produce more power, not how to convert its output. However, your point is still valid and a good one, especially considering the OP might come back and ask about converting the output once he realizes that modifying the wall wart isn't going to work. \$\endgroup\$ – Olin Lathrop Oct 3 '18 at 20:25
2
\$\begingroup\$
Is there a way to alter the 5V 2A wall wart into 5V 5A?

No. Not gonna happen. Fahgeddaboudit.

I just need slightly more current

A factor of 2.5 is a lot more than "slightly" higher.

If the 2 A wall wart could produce 5 A, then it would be sold as such and therefore fetch a higher price. The manufacturer had to pick something as the spec, and they picked 2 A.

These things are manufactured in high volume, so saving parts cost probably got significant design attention. Parts to support 5 A output would definitely cost more, and would take more space and weigh more too. You can't just tweak a resistor somewhere and have the device output 2.5 times more power.

If you want 5 V at 5 A, go buy a 5 V 5 A wall wart or power supply. These things are quite cheap nowadays. You can buy a ready-made unit for less than you can buy the parts for separately in low volumes.

What you are asking to do just doesn't make sense. Not even close.

| improve this answer | | | | |
\$\endgroup\$

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