Powering an element that draws more current/voltage than GPIO can supply?

Question

I'm a little over my head here and I'm hoping someone might be able to help. I'm trying to control this Heating Pad using Raspberry Pi GPIO pins, turning it on and off via GPIO. It takes 5V DC and seems to draw around 0.5A of current, though slightly less may be fine as well (I don't need it to get THAT hot – just comfortably warm. It's for a wearable)

The Raspberry Pi isn't turning the heating pad on – I've been doing research and it seems like I need to do something with transistors, but I'm confused exactly what. Also I thought the RPi outputted 5V from GPIO, but it may output 3.3V? I can remedy this by sticking a coin battery in series with the GPIO which is no big deal, but I'd like to get 5V if I can. Any help would be VERY appreciated!

I've seen a lot of vague answers online such as "use a MOSFET" and I have those, but I have no idea how to wire them up or how to decide exactly what to do.

Answer 1

first thing first, you cant extract that much power from the pi, you need an extra batery and mosfet for that.

heres the deal, in order to heat your pad you need and extra batery conected to the pad, but you must use a mosfet, that would behave like a on/off switch.

 --- batery --- Heating pad---
 |                            |
 ------Mosfet------------------
         |
         |
    Raspberry pi
       GPIO

Basically you are using the mosfet to avoid bigger currents been drawn from the RPi, the batery and the heating pad only "see" the mosfet, but never the RPi.

This is the main way of conecting you mosfet, you only replace the TTL for your RPi or any other Microcontroler that you like, on the part where it said Dc motor, you only replace it with your Heating pad, and the Vdd its your batery, it could even be a power bank if it have enough power to run the pad

Answer 2

Think of a MOSFET as a remote control switch, one that isn't so remote, but is certainly a switch.

Your device with the high current requirements will have its own power source with appropriate leads to activate it. Plan to interrupt the power lead with one pair of contacts on the MOSFET called the load.

The other pair of contacts on the MOSFET may be called trigger or similar. You would run from the ground of the Pi to the ground of the trigger with one lead, and from one of your GPIO pins to the other contact of the trigger.

When you have the Pi send a signal to that pin, it causes the MOSFET to close the switch on the load side.

Answer 3

Whatever you do DON'T "stick a coin battery in series with the GPIO".

You will destroy the Pi!

See https://elinux.org/RPi_GPIO_Interface_Circuits

A MOSFET would seem most appropriate, but you need one with a low threshold which can be switched by 3.3V (the majority of power MOSFET need more).