Electronics > Projects, Designs, and Technical Stuff
Turning off an LDO using a mosfet
OM222O:
The application is a milliohm meter, so switching regulators could not be used. preferably I would use an external source, but it needs to be portable, so that was out of the question. 6AAAs also physically won't fit inside the box I have for this project. 9V battery was a decent compromise. Also for the voltage drop:
Atmega 328P-AU is rated to work down to 1.8V! the thing I'm most worried about is the 24bit ADC which has a differential input voltage of maximum 2.048V. Even if the regulator drops to about 3 volt, I would be fine :D
And at that point the MCU will kick in and display a warning or low battery symbol, telling you the 1A range is not usable anymore. That means you can still use the device to measure higher resistance values, or if you desperately need it, carry a separate 9V battery in the box, it can fit 2 9v batteries.
Edit: I will create a new thread shortly, posting all the details about the project I have so far and asking for some feedback, feel free to comment about my newbie mistakes there :D
schmitt trigger:
"I was wondering what would happen if I were to disconnect the GND pin of the LDO using that fet? would that damage something or have some weird side effects? "
The ground pin carries the quiescent current for the regulator's internal circuitry.
Remove it and the regulator will cease to operate.
Krampmeier:
If you want to use your N-channel FET, you can use a photodiode coupler like the TLP3905 to control it. It provides a floating gate voltage for the FET and just needs a few milliamps control current for the LED.
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