Electrical – 12v to 5v buck converter to power rasberry pi on parallel circuit


Main question: will oversupplying current to a raspberry pi damage it? For example, will providing 5a at 5v be damaging when the pi requires only (up to) 2.5a at 5v?

Here is my backstory, assumptions, and sub-questions:

I am trying to power a raspberry pi in parallel with a 12v led lamp on the same circuit. My power supply for the lamp is a 1.5a, 12v wall wart. (The LEDs require 12v). The raspberry pi 3 power requirements are 5v and up to 2.5a.

I would prefer to use a buck converter to step down voltage to 5v for the pi rather than a linear regulator, such as the LM7805, so that I don't have to deal with a lot of heat output (and hopefully don't have to end up needing a high current power supply).

Therefore, my design has the LEDs on one branch and a buck converter + pi on the opposite branch of the parallel circuit.

I am considering using the 12v to 5v, 5a buck converter here:

I think power is conserved over the buck converter, and power = voltage * current.

Based on my ammeter, the LEDs of the lamp only require 0.5a of the 1.5a. I'm not sure how the remaining 1a will be divided on the two branches circuit.

Could I just stick a high resistance resistor on the LED side to push the remaining current over to the buck converter and pi?

If so, the power to the buck converter will be:

Power1 = voltage1 * current1

Power1 = 12v * 1a = 12w

The graph Pololu provides appears to indicate that the converter efficiency will be about 92% at Vin=12v and 2

0.92(Power2) = voltage2 * current2

0.92(12w) = 5v * current2

current2 = 2.2a

However, I'm guessing I should supply more current to the pi to be safe. If I got a 2a, 12v power supply I end up "pushing" 3.3 amps at the pi using the above assumptions.

Will this damage the pi or is it completely safe to oversupply current?

I am only a beginner, so let my know if my assumption are off. Thanks!

Best Answer

The amperage of your step-down regulator is just a description of capability to deliver such current. It can deliver 2A if required by circuit.
So if your circuit won't require more than 0.5A, your regulator will not push excessive current into your circuit.

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