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GPIO Sinking Resistance

edwardsbobg

I am powering an LED / current limiting resistor by sinking current into an IO pin. Where can I find the resistance of the IOs when sinking current?

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    This is really the wrong question. The GPIO doesn't have a fixed resistance, it does have a current limiter which limits current to an amount depending on drive mode (standard/high). So as long as your LED/resistor combination, when the pin is low, has enough resistance that current sunk is less than the maximum drive mode current, the GPIO has basically zero resistance. If however your current limiting resistor is too small and you attempt to sink more than the drive mode current allowed, the 'resistance' of the GPIO will rise to whatever level is necessary, in series with your LED and resistor, to keep the current to the maximum allowed. 

    eg say you're using standard 2mA drive, your LED has Vf of 2.1v and you're powering off 3.3v. (3.3-2.1) = 1.2v you need to drop across the resistor, so you need 600 ohms. Provide a resistor of that or higher, the GPIO 'resistance' will be 0. Provide anything lower and the GPIO effective resistance will be (600 - your resistor) ohms. If you don't even provide a resistor the current will still be limited to 2mA and the GPIO effective resistance will be 600 ohms. However you're then asking the nRF to dissipate 2.4mW of power, so you're better off with an external resistor. 

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  • 0

    This is really the wrong question. The GPIO doesn't have a fixed resistance, it does have a current limiter which limits current to an amount depending on drive mode (standard/high). So as long as your LED/resistor combination, when the pin is low, has enough resistance that current sunk is less than the maximum drive mode current, the GPIO has basically zero resistance. If however your current limiting resistor is too small and you attempt to sink more than the drive mode current allowed, the 'resistance' of the GPIO will rise to whatever level is necessary, in series with your LED and resistor, to keep the current to the maximum allowed. 

    eg say you're using standard 2mA drive, your LED has Vf of 2.1v and you're powering off 3.3v. (3.3-2.1) = 1.2v you need to drop across the resistor, so you need 600 ohms. Provide a resistor of that or higher, the GPIO 'resistance' will be 0. Provide anything lower and the GPIO effective resistance will be (600 - your resistor) ohms. If you don't even provide a resistor the current will still be limited to 2mA and the GPIO effective resistance will be 600 ohms. However you're then asking the nRF to dissipate 2.4mW of power, so you're better off with an external resistor. 

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