user69282: I need to make voltage divider, so I could read a battery voltage up to 14.4 V. Could I use very high resistance resistors for that, so the current would be very low (to not to discharge battery)?
On the Internet there are examples with 500 - 2000 ohm resistors, but could I use hundreds of thousands to reduce the current to minimum? Would Arduino's analog input still read the value?

user16591: I was able to implement it with 3 MΩ resistors.
The problem is that the readanalog() function executes 10 reads. So the first one is the right value and then you will not measure the right value anymore.
Solution: A voltage divider with 2.5 MΩ and 500 kΩ with a 470 nF capacitor on the 500 kΩ resistor. The solution with the MOSFET is better. This is easier, and it works (but only for 20 years ).

user49068: Yes, you can use very high value resistors for your voltage divider.  But the recommended maximum is 10 kohms for the ADC pins.  So the combined resistance of your two resistors should be maximum 10 kohms.

it is recommended that whatever you connect to the A/D have an output impedance of 10k or less for best accuracy. This is to allow the A/D input capacitor on the sample and hold to charge up in the time allotted to it between switching the input multiplexer over and starting the conversion.

From: Input Impedance of A0-A5

user4143: The most efficient is not to have a voltage divider at all. Or at least to only have a voltage divider when you actually need one.
My preferred schematic is this:

simulate this circuit – Schematic created using CircuitLab
With this M1 is normally kept turned off by R3. When you want to take a measurement you turn on M2 by setting D3 HIGH (or whatever pin you attached it to) which pulls down the gate of M1 turning it on. This then connects the batter to the voltage divider R1/R2 and allows you to read the voltage at A0.
The voltage divider of 20k / 10k allows you to read up to 15V on the battery, and gives you an output impedance of 6.67kΩ which is within the recommended maximum of 10kΩ for the ATMega's ADC.
During "idle" times when M1 and M2 are both off the only current flowing is the minuscule leakage current through M2 tempered by the large 100kΩ resistor R3 (which itself pales into insignificance compared to the "off" resistance of M2 anyway).
Selecting M1 so that the ON resistance is very small (<0.1Ω) allows you to essentially ignore it in your calculations since that will be swamped by the tolerance of R1 and R2 anyway.  M2 needs to be selected so that it is "logic level" (i.e, with a gate threshold V_GS well below 5V).
This double-FET arrangement should be used rather than just using an N-channel FET in the ground connection of the divider because when that simpler arrangement is turned off the Arduino will be seeing 12V directly connected to A0 which really is not good - you will kill the Arduino - so it is important that you switch the "high" side of the network not the "low" side, and to do that requires two FETs.

69282: <p>I need to make voltage divider, so I could read a battery voltage up to 14.4 V. Could I use very high resistance resistors for that, so the current would be very low (to not to discharge battery)?</p>
<p>On the Internet there are examples with 500 - 2000 ohm resistors, but could I use hundreds of thousands to reduce the current to minimum? Would Arduino's analog input still read the value?</p>
 16591: <p>I was able to implement it with 3 MΩ resistors.
The problem is that the readanalog() function executes 10 reads. So the first one is the right value and then you will not measure the right value anymore.</p>
<p>Solution: A voltage divider with 2.5 MΩ and 500 kΩ with a 470 nF capacitor on the 500 kΩ resistor. The solution with the MOSFET is better. This is easier, and it works (but only for 20 years ).</p>
 49068: <p>Yes, you can use very high value resistors for your voltage divider.  But the recommended maximum is 10 kohms for the ADC pins.  So the combined resistance of your two resistors should be maximum 10 kohms.</p>
<blockquote>
<p>it is recommended that whatever you connect to the A/D have an output impedance of 10k or less for best accuracy. This is to allow the A/D input capacitor on the sample and hold to charge up in the time allotted to it between switching the input multiplexer over and starting the conversion.</p>
</blockquote>
<p>From: <em><a href="https://forum.arduino.cc/index.php?topic=65134.0" rel="nofollow noreferrer">Input Impedance of A0-A5</a></em></p>
 4143: <p>The most efficient is not to have a voltage divider at all. Or at least to only have a voltage divider when you actually need one.</p>
<p>My preferred schematic is this:</p>

<p><img src="https://i.stack.imgur.com/hLS3G.png" alt="schematic" /></p>

<p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fhLS3G.png">simulate this circuit</a> &ndash; Schematic created using <a href="https://www.circuitlab.com/" rel="noreferrer">CircuitLab</a></sup></p>
<p>With this M1 is normally kept turned off by R3. When you want to take a measurement you turn on M2 by setting D3 HIGH (or whatever pin you attached it to) which pulls down the gate of M1 turning it on. This then connects the batter to the voltage divider R1/R2 and allows you to read the voltage at A0.</p>
<p>The voltage divider of 20k / 10k allows you to read up to 15V on the battery, and gives you an output impedance of 6.67kΩ which is within the recommended maximum of 10kΩ for the ATMega's ADC.</p>
<p>During &quot;idle&quot; times when M1 and M2 are both off the only current flowing is the minuscule leakage current through M2 tempered by the large 100kΩ resistor R3 (which itself pales into insignificance compared to the &quot;off&quot; resistance of M2 anyway).</p>
<p>Selecting M1 so that the ON resistance is very small (&lt;0.1Ω) allows you to essentially ignore it in your calculations since that will be swamped by the tolerance of R1 and R2 anyway.  M2 needs to be selected so that it is &quot;logic level&quot; (i.e, with a gate threshold V_GS well below 5V).</p>
<p>This double-FET arrangement should be used rather than just using an N-channel FET in the ground connection of the divider because when that simpler arrangement is turned off the Arduino will be seeing 12V directly connected to A0 which really is not good - you <em>will</em> kill the Arduino - so it is important that you switch the &quot;high&quot; side of the network not the &quot;low&quot; side, and to do that requires two FETs.</p>
