HelloDoes any one know how much current is needed on the DSOP pin to pull it high. I can't find this in the datasheet?
I have a signal with only 10µA. Will this be possible.
Kind regards Stef
It consumes about 109 uA while setting DSOP High at 5V. ADL5375-05 IC at this measurement consumes 195 mA at normal operation(DSOP=low) and 128mA(DSOP=High) with 5V Power Supply. The output is disabled at this IC on an evaluation board when it is above about 2.164V. And at this case, it is 36.6 uA to disable the output(means DSOP high). So 10 uA is not enough to make DSOP high. The impedance of DSOP pin is about 58.7Kohm and the current(Supply voltage/57.8Kohm) for DSOP high goes through roughly 58.7Kohm.
Theoretically , DSOP High at 5V : 5V/58.7Kohm -->85.2uA
DSOP High at 2.164V : 2.164V/58.7Kohm --> 36.9uA (Minimum requirement at least at this measurement)
If 10uA drives through 58.7Kohm, it will set DSOP pin at 0.587V that is not enough to make DSOP pin High enough to disable the output.
Please find it at the table 1(see page 6). I copy it for your information. The DSOP pin high disables the output stage of the ADL5375, not whole. So 10uA is too low.
The currents in those table 1 are the current that the whole ic will use. When DSOP is high the output is disabled why the ic use less current.
The value I need is how much current is needed to pull the DSOP correctly high.
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