Usually, pull-down resistors are say 10k or less but when I try and calculate the value for the TI LMR62421 I keep getting 3M ohm.
Data sheet here: https://www.ti.com/lit/ds/symlink/lmr62421.pdf
The Enable pin current, sink/source, is shown to be 'typically' 100 nA, (there is no min/max value?)
Also, the Enable shutdown voltage threshold is 0.4v max.
Using 0.3v / 100 x 10^-9 I am getting 3,000,000 Ohm!
Does 3Mohm sound correct? thanks for any help.
I added a snippet from the datasheet.
ENABLE PIN / SHUTDOWN MODE
The LMR62421 has a shutdown mode that is controlled by the Enable pin (EN). When a logic low voltage is
applied to EN, the part is in shutdown mode and its quiescent current drops to typically 80 nA. Switch leakage
adds up to another 1 μA from the input supply. The voltage at this pin should never exceed VIN + 0.3V.
What is lacking in the datasheet is the maximum current source/drain of the enable pin (4) to calculate a more 'normal' source/drain current.
You can tie it ground or Vin. Your calculations are probably correct, just use a value less then that. It is a MOS based device.
rtek1000:
Yes 'rtek1000', I have read that. It doesn't seem to answer my question of the pull-down resistor value for the Enable pin.
This quote says that the pin can be connected to the same potential as the power supply, just look at the diagrams in the examples.
The fact that there is a consumption current at the entrance does not necessarily mean that this current is mandatory. Except in special cases, where the same pin has multiple functions, but this is not the case.
It has a minimum voltage acceptable as activated and another for disabled, just read the VEN_TH part again, where it has MIN and MAX. MIN means it works if it is above, and MAX means it must be below. It is so for any digital input, even on Arduino.
