How does input buffer disable work for GPIO on K10 micro?

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How does input buffer disable work for GPIO on K10 micro?

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JewelHolt
Contributor I

Hello, I read other posts about leaving disabled GPIO pins floating for devices like K64. Wondering if that persists on the K10? The reason given in other posts is that the buffer is disabled so no shoot through currents can occur, however how exactly is it disabled? Is the path to Vcc and GND disconnected or is the input to the cmos gates isolated from external voltages by the input buffer being disabled? In the latter case wouldn't that mean the gates could still be biased according to the internal circuitry and still have the ability to shoot through?

Thank you for your insight in advance!

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947 次查看
xiangjun_rong
NXP TechSupport
NXP TechSupport

Hi,

For Kinetis K10 family, some pins are multiplexed with analog channel, if you configure the MUX as 0, the pad is configured as analog channel, the digital input and output buffer are disabled automatically.

 

xiangjun_rong_1-1725246663111.png

If you configure the MUX as non-zero(digital mode), the digital input/output are enabled automatically, if the pin is in digital input mode, the pad is in high-impedance, the input leakage current is in "uA" level, so the pad has high impedance.

 

xiangjun_rong_3-1725247557174.png

 

Hope it can help you

BR

XiangJun Rong

 

 

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JewelHolt
Contributor I
Hello,
Thank you for your response. So for example, pin 6 (PTE5) which does not default to analog but rather to 'disabled' state. If I leave that disabled pin floating, should I expect any shoot through currents? What am I to expect from a disabled state? I am assuming that disabled means the input and output buffer is disabled. But that statement is a bit too vague to draw any conclusion about possible EMI issues if its left floating. It's not very clear from the manual.
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xiangjun_rong
NXP TechSupport
NXP TechSupport

Hi,

For unused GPIO pins, we suggest you connect external pull-up or pull-down resistor rather than float so that the pin has a fixed high/low logic because the pin is in input state with high impedance, in this way, the chip can reduce current, reduce EMI. Of course, if the pin has internal pull-up or down resistor, you can use software to enable the pull-up or down resistor.

Another method is to configure the GPIO in output mode and set/clear the pin logic.

Hope it can help you

BR

XiangJun Rong

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