0

i am running a low side switch with the 3.3V gpio output of the pi, and for the particular transistor i’m using (irf520 ) the 3.3V is barely enough to turn it on - is there some clever positive feedback i can use to force the transistor ‘very on’ when on and off when off?

How abt this, I put the gpio into high-impedance input mode to turn transistor on, and set output=0 to turn transistor off. current into gpio pin when output=0 is 1.2mA which i suppose is ok. Is 12V after 10Kohms ok on a raspi input pin?

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

jeremy_rutman
  • 125
  • 1
  • 3

3 Answers3

1

Firstly you are applying 6V to the GPIO and are likely to damage the Pi.

There is nothing you can do to make a MOSFET which needs more than 3.3V miraculously work. Although you could use a transistor to drive the gate.

There ARE MOSFETs which switch with 3.3V Vgs although most need more.

Milliways
  • 62,573
  • 32
  • 113
  • 225
1

You have to drive this MOSFET with another transistor(a BJT perhaps). Or you could replace the MOSFET with a BJT. The are good for driving transistors.

Sohan Arafat
  • 1,852
  • 2
  • 12
  • 43
1

Your schematic is not a good idea: It will put 6V on your GPIO pin, and fry your RPi. You could save yourself some money by avoiding the obvious mistakes :)

The best solution for driving a MOSFET from your 3.3V GPIO is a Darlington Pair. Here's why I say this.

If you want to drive an n-channel MOSFET as you've shown in your schematic, simply swap the npn Darlington device in this answer for a pnp Darlington. Similarly, you could use the npn Darlington to drive a high-side p-channel MOSFET.

I'll stop here. If you're interested in this approach, mod your question & post a comment here. I'll try to help.

Seamus
  • 23,558
  • 5
  • 42
  • 83