uoficowboy
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Switching the bottom resistor will hardly have the intended effect, because the top resistor continues to draw current from the battery with the divider output clamped at the ADC input. Disconnecting the high side is the only reasonable way, usually with a PMOSFET. Besides some minimal leakage currents, there should be no problems involved. Particularly no offset voltages as BJT switches do.
Hi Alex - most of my professional work is on battery powered devices. Ultra low power battery powered devices. In sleep mode (where they spend most of their lives) their current draw is typically less than 1ua. So you can understand why I don't want a voltage divider always on my batteryThe project involves an mcu which will consume several mA, if we assume two 100K resistors in the divider then the consumption will be 9v/200K=45uA, this current compared with the mcu consumption would probably be about 1% (if we assume 4.5mA for the mcu) so the battery will last 1% less time, doesn't seem a big deal to me.
The OP said that this is a battery powered application so I guess that the mcu circuit consumption will be far worse than the above assumption of 4.5mA so the difference that the divider will make will be even less.
Alex
Hi Alex - I recognize that the P-FET on top would be best in the original situation I described - but my original question remains somewhat: would there be no voltage drop across the FET? (assuming the on resistance of the FET is negligible in comparison to the voltage divider resistance)In that case you have to use the Pmosfet as FvM has suggested but this mosfet will be connected to 9v (mosfet source) so you will need an additional control device to turn the mosfet on/off.
If you use a pullup resistor for the gate to have 9v to the gate so that the mosfet is off by default then you will need a device (like a Nmosfet) controlled by the 3.3v of the mcu that will be able to ground the gate of the Pmosfet to turn it on.
Alex
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