# Electronic – Does the value of a resistor matter when doing voltage division to cut it into a half

resistorsvoltagevoltage divider

I have a 7.4V (8.4V when fully charged) battery that is connected to an Arduino and I need to monitor the battery. To do so, I need to plug the battery into an analog pin. However, that pin only accepts voltages from 0-5V, and returns a value from 0-1023. Therefore I want to cut that 8.4V when fully charged into 4.2V. I understand I need to use voltage division where the resistors are both equal, does it matter which resistors? \$10\text{k}\Omega-10\text{k}\Omega\$ vs \$1\text{k}\Omega-1\text{k}\Omega\$?

The Atmel data sheet says "The ADC is optimized for analog signals with an output impedance of approximately 10KΩ or less. If such a source is used, the sampling time will be negligible".

To have an impedance of 10K\$\Omega\$ or less, the resistors in the divider should be 20K or less. As others have pointed out, lowering the resistors consumes more power, so using 20K resistors makes sense to me.

simulate this circuit – Schematic created using CircuitLab

Edit: To explain the source impedance looking into the "middle" of the divider and the top:

If the top of the divider goes to a 'stiff' voltage (a battery in this case), the impedance looking into the center point is 20K||20K. You can think of it as 20K||(20K+Rs) where Rs is the source resistance of the battery (or whatever the top of the divider is connected to). Since Rs << 20K, it's very close to 20K||20K = 10K. If you were to disconnect the battery, (Rs \$\rightarrow\infty\$) it would be 20K.

The impedance from the point of view of the battery (looking down into the divider) is about 20+20 = 40K, so the drain is only a couple hundred uA. That is because the input impedance of the ADC is very high and is in parallel with 20K, so it's about equal to 20K, and it's in series with another 20K.