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Currently, I'm working on a Invidia Jetson TX2 dev kit.

I'm trying to power a 12V 2A LED from a 3.3V GPIO. My current system uses a 12V relay with an optocoupler (generic Arduino One), but I'm trying to use a transistor instead.

  1. Using FQP30N60L MOSFET without a load and a 10kΩ pull-down resistor.
  2. Using 2N3904 for driving a 3V blue LED with a 390Ω on 12V (and 10kΩ on gpio<>base)

Both scenarios show that GPIO voltage drops from 3.3V to 1.3v~1.8v when connected. Is that normal or am I doing something wrong?

  • It's not very clear what current can the GPIO supply, but from my understanding it's 1mA (could be wrong).

Thanks!

EDIT: Thanks to 'amb' i understand my readings; thanks :) So after reading some more i have another question: GPIO port is an 'open-drain design' with 100k ohm internal pull-up (other GPIO ports have 20k and 47k), This port can provide 1ma of current.

What is the proper way of driving a 12v (2a max) LED using a MOSFET?

Note: i have added a simple drawing of my setup, i know it wont work, just trying so simulate my problem

EDIT 2: Internal pull-up is 100k ohm

EDIT 3: Is R2 needed here? as its an open drain with internal pull-up, pin wont be floating (i put it as i did in arduino, witch is not the case here)

enter image description here

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  • \$\begingroup\$ 2A LED, seriously? What does it do, light up the whole neighbourhood? \$\endgroup\$
    – Lundin
    Commented Aug 15, 2018 at 11:56
  • \$\begingroup\$ real time warning light in direct sunlight situations \$\endgroup\$
    – Moos
    Commented Aug 15, 2018 at 12:39
  • \$\begingroup\$ Still doesn't explain 2A. That's enough to drive all LEDs on a well-lit truck. Is it a light bulb? \$\endgroup\$
    – Lundin
    Commented Aug 15, 2018 at 12:43
  • \$\begingroup\$ Anyway, where do you measure the ~1.3V, at the MOSFET or at the CPU pin? Could there be any other components in between besides those in your schematic? \$\endgroup\$
    – Lundin
    Commented Aug 15, 2018 at 12:45
  • \$\begingroup\$ COB led 10w-20w \$\endgroup\$
    – Moos
    Commented Aug 15, 2018 at 12:46

1 Answer 1

Reset to default
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Sounds like the GPIO has a pullup (of about 10k) on the Jetson board. Your pulldown is forming a potential divider with this, hence the measured 1.3-1.8V.

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    \$\begingroup\$ Internal pull-ups should only matter if the port is set as input. In this case, it is an output, so I don't believe this is the problem. External pull-up could be. \$\endgroup\$
    – Lundin
    Commented Aug 15, 2018 at 12:00
  • \$\begingroup\$ I edited the typo in the scheme, R1 is 100K , therefore Vgate is ~0.3v \$\endgroup\$
    – Moos
    Commented Aug 15, 2018 at 12:48
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    \$\begingroup\$ @Moos But you said it was ~1.3V? Also the internal pull-up should not be active if the port is set as output. \$\endgroup\$
    – Lundin
    Commented Aug 15, 2018 at 14:07
  • \$\begingroup\$ Internal pull-up is always active as its an open-drain design(GPIO pin), otherwise the would be LOW\Floating. 1.3v is caused by internal pull up and R2 acting as a voltage divider (with 20K internal pull up from another pin) \$\endgroup\$
    – Moos
    Commented Aug 16, 2018 at 7:06

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