I know the 1mA only applies when input voltage is really low, but does that statement apply to when the output current (the load) is at 0mA or when the load is at the rated 200mA?
I’m interested in the current consumption when this regulator is used to power a wireless module. The module consumes 2uA in sleep mode. If the input voltage is 3.3V, and the output voltage is also 3.3V, the datasheet saids the efficiency is 80%. Does that mean the total power consumed is 1/0.8 * 2uA = 2.5uA? Am I reading that correctly? So the regulator is only imposing an extra .5uA of overhead?
Is it weird this table of voltage regulators below list “max INPUT current”? Isn’t OUTPUT current what’s important, since you generally know the current for the load, and you’re looking for a regulator that can supply the load? Or is that just semantics?
The quiescent current is the input current draw when there is no load connected to the output.
We have efficiency charts available in the “Typical Efficiency and Output Current” section on the NCP1402 step-up voltage regulator product page; however, we do not have the efficiency characterized for such low currents. The efficiency of the regulator is very low a those kinds of current draws since the regulator will always draw the quiescent current. Even if your device only draws a few microamps, the regulator will still draw around a milliamp.
The table for our step-up voltage regulators shows the max input current because the input current is usually the limiting factor of step-up regulators. The amount of current the regulator can output will depend on the input and output voltage.
If I use the enable pin to turn off the NCP1402 when I don’t need it, how much current does it consume? It looks like the enable pin needs to be HIGH to be in disable mode. How much current does that pin take?
According to the NCP1402 datasheet, the enable pin can draw up to 0.5 uA and the whole chip draws less than 1 uA when in the off-state. Please note, our carrier board does not bring out the enable pin, and as seen on the schematic, the enable pin is tied directly to VOUT, so it would probably be very difficult to control that pin (you might be able to cut around that pin on the board to gain access). Instead, you might consider using one of the other step-up voltage regulators we carry that has the enable pin broken out.
