I would like to integrate a Si4703 breakout board into a project. I’m only going to use its two-wire interface, without much intelligence to drive any of the other pins. Therefore, I need to initiate the startup sequence with the most primitive means; my intention is to initiate it every time the circuit is powered up. (Resetting the thing during operation is not really a requirement—if at all, it’s OK to require a power cycle for that.)
As part of the startup sequence, I need to keeep RST_n low for at least 300 ns 30 ns after powerup, then pull it high (and keep it high). The board has an internal 10 kΩ pull-down resistor on that pin, with an extra 330 Ω resistor between the pin and the header (see linked schematic).
While some sources suggest playing with the impedance of the pull-up resistor, increasing it to introduce a delay before the threshold is reached, I am wondering if a low-pass filter is maybe a better idea.
I figured that for a delay of t, I would need to aim for a cutoff frequency of f = 1 / (2 t).
Assuming I’d use an RC filter with a 1 kΩ resistor (1/10 of the internal pull-down resistor), I would then need a 10 pF capacitor to get within the neighborhood of the delay I aimed for.
Is this going to work as intended, or am I making a mistake here? Especially, are my calculations quantitatively correct? (Admittedly, I’m more a software than a hardware guy…)
