The datasheet (https://www.st.com/resource/en/datasheet/st25dv16k.pdf) states that "The impedance on LDP pin when set high doesn't exceed 5kΩ.", which means that a lot of power is wasted just signalling the IC to power down. Am I interpreting this wrong or is the datasheet wrong?
Hello Christian,
Fortunately, this does not mean that when LPD is high, it is charged by a 5kohm impedance.
In the sentence "The impedance on LDP pin when set high doesn't exceed 5kΩ.", I think that the "doesn't" is confusing, and shall by replaced by "should not".
What this sentence really means is that the impedance that is driving the LPD pin should be less than 5Kohm.
The reason is, internally, the LPD signal is pulled-down by a resistor. When LPD pin is driven high, this internal resistor is automatically disengaged in order not to consume power. But in order to "disengaged" it, the driving impedance must be several scales lower than this pull-down resistor. This is why it should be less than 5kohm.
For instance, a simple wire coming from a GPIO pin of a microcontroller does usually have a impedance much lower than 5kohm and is ok.
When LPD is driven high, there is no more internal pull-down resistor, and the only power consumed is some leakage current, which is less than 1uA (usually about 400nA at 25C from my measurements).
Hope it clarifies your point.
Best regards.
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