GPIO Pulldown IMX6SL

[stephangm]

How can I set a GPIO pin to be a pulldown without using an external resistor?

[art · Accepted solution]

An internal 100kOhm pull-down resistor can be enabled on most of GPIO pins by appropriately configuring the corresponding IOMUXC_SW_PAD_CTL_PAD_<xxx> register, where <xxx> is the corresponding pin name. The sequence is as follows.

1. Set the IOMUXC_SW_PAD_CTL_PAD_<xxx>[PUS] bits to 0b00 to select the 100kOhm pull-down resistor.

2. Set the IOMUXC_SW_PAD_CTL_PAD_<xxx>[PUE] bit to 1 to enable the pull resistor (not a keeper circuit).

3. Set the IOMUXC_SW_PAD_CTL_PAD_<xxx>[PKE] bit to 1 to physically enable the pull resistor.

For more information, refer to the Chapter 30 "IOMUX Controller (IOMUXC)" of the i.MX6SoloLite Reference Manual Rev.2 document, available on the i.MX6SoloLite Documentation web page:

http://www.nxp.com/products/microcontrollers-and-processors/arm-processors/i.mx-applications-process...

Have a great day,
Artur

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[Carlos_Musich]

Hi Stephan,

do you want to do it in bareboard code? Modifying directly the registers? In this case you must check the Pad Control Register (IOMUXC_SW_PAD_CTL_PAD_AUD_MCLK). Here you will find the following options for a pin:

00 100K_OHM_PD — 100K Ohm Pull Down

01 47K_OHM_PU — 47K Ohm Pull Up

10 100K_OHM_PU — 100K Ohm Pull Up

11 22K_OHM_PU — 22K Ohm Pull Up

If this is not what you are looking for please let me know details of what you need to configure so I can provide more details.

Regads,
Carlos

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