Analysis on the selection of external pull-up resistors for AVR pins

The I/O port of the AVR microcontroller is a bidirectional port. It has the following characteristics:

AVR IO has multiple IO modes:

1 High impedance state, mostly used for high impedance analog signal input, such as ADC digital-to-analog converter input, analog comparator input

2 Weak pull-up state (Rup=20K~50K), for input. Optimized for low-level signal input, eliminating the need for external pull-up resistors, such as key input, low-level interrupt trigger signal input

3. Push-pull strong output state, with extremely strong driving capability (>20mA), can directly drive LED, and the high and low driving capabilities are symmetrical.

In actual application, I used a 1M external pull-up to measure the pulse of the Hall device. It turned out that the pulse was measured even when there was no action. So I made the following measurements:

1. The pin is set as input, and the internal pull-up is disabled. Instead, an external 1M resistor is used to pull the pin up (PORTX is 0). The voltage of the pin is measured to be approximately 3.06 (the power supply voltage is 3.37).

2. The pin is set as input, and the internal pull-up is enabled (with an external pull-up resistor of 1M). The voltage of the pin is measured to be about 3.36 (the power supply voltage is 3.37).

3. The pin is set as input, and the internal pull-up is not enabled. Instead, an external adjustable resistor is used for pull-up (PORTX is 0). The pin voltage increases as the resistance of the external pull-up resistor decreases. When the value of the external adjustable resistor is about 40K, the pin voltage is close to the Vcc supply voltage, and the voltage no longer increases by reducing the resistance.

To sum up a rule, if an external pull-up resistor is used, the resistance should not exceed 40K, otherwise the voltage of the pin will not reach the voltage of Vcc.

correct:

The above measurement is problematic. The voltage measured in 1 is 3.06, which is due to the influence of the internal resistance of the multimeter. In other words, if the internal resistance of the multimeter is M level, the external 1M resistor is large enough compared to the multimeter, so there will be a relatively large voltage drop, resulting in a lower voltage on the pin (the actual voltage value of the pin should be close to the power supply voltage). The original system does not work because the external resistance is too large. What is the specific reason?