# 11.4 battery monitoring w/ voltage divider?

I am well aware that a simple voltage divider circuit is all I need to monitor the voltage of an unregulated power source. My question is - can I use any R1 and R2 values that "work" for an 11.4 --> 5v voltage divider or do they need to be of some specific value according to how much current will/can go through the analog pins?

Ohms law says that V = IR. For a given resistor, either 11.4 volts or 5 volts will go through it. That will result in some current value.

The voltage times the current is the amount of power that the resistor must dissipate, in watts.

As long as the resistor is rated for that number of watts, the resistor can be used.

Hate to be a bother but it's not clicking... can you elaborate... I've tried to see this using ohms law, but i couldn't get it. You said that either 11.4 or 5 volts will go through, but which is it?

Basically, can you re-explain it, please, it's midterm week here and my brain is focused on newtonian physics atm, so it's taking me extra-long to comprehend this.

You canâ€™t use any values. The Atmel datasheet recommends a maximum source impedance of 10K. When you make a resistor divider you are increasing the source impedance.

If your battery has a very low series resistance then the impedance of the divider
is the dominant series resistance. The series resistance would then be R1|| R2.
If you keep one resistor at 10K then R1 || R2 will always be less than 10K.
If you want to draw less current then pick larger values keeping R1 || R2 < 10K

(* jcl *)

www: http://www.wiblocks.com
blog: http://luciani.org

Any value will "work" given the right ratio. For example (and to keep it simple) say you wanted to split the voltage in half, you need two resistors of equal value but what values? Well using two 100R resistors would work just as using two 1M resistors at the junction you would get half the voltage from both so which to choose? The lower the absolute values the more current this arrangement takes from power supply or battery. So does it matter?

That depends on what you want the half voltage for, that is what you are going to feed it into. For example if you are going to feed it into a high impedance load like the input of an operational amplifier then 1M will be fine. If however you want to light an LED then in order for the load not to "short out" the bottom resistor you need something much lower. As a good rule you need 10 times more current going down a potential divider than you take out at the junction.

In the case of the arduino's A/D then 10K is the optimum impedance for input although the actual input impedance is much higher 10K is low enough to reduce noise and pickup.

Perhaps a look at the basic would help.

Vout = R2/(R1+R2)*Vin

OR in this case the voltage is 11.4vdc - knowing that 5vdc is the max voltage a pin can handle, let's pick 15vdc as the max that will EVERY be on circuit. 10K is also the impedance desired by the Arduino, so let's say the sum of R1+R2 will be in the range of 10K.

Let's choose R2=3K, R1=7K (or numbers in that range - doesn't really matter. Nor does wattage really matter because you will NOT pull your current thru the voltage divider, this is for measuring voltage ONLY.

Vout = (3,000/10,000)*15 = 4.5vdc input to pin.

Now for 11.4 volts:

Vout = (3,000/10,000)*11.4 = 3.42vdc input to pin.

Just dig thru junk box and find resistors in that range, do the calculations to match and you've got your voltage measuring device.

Remember: DO NOT pull a load thru a voltage divider - that throws everthing off a good bit. connect load to input of R1 resistor, then to load, then to ground.

edit: I just found this webpage that should help explain also. http://www.kpsec.freeuk.com/vdivider.htm

Ken H>

Thanks a lot everyone, i'm starting to understand. ;)

Hi, Useful info here, but I am not sure what you mean by "10K is also the impedance desired by the Arduino, so let's say the sum of R1+R2 will be in the range of 10K.".

Is it possible to use the internal impedance? I thought this could be set between 20K and 50K so that this could form one side of the reisistance divider, or have I got that all wrong?

The internal impedance is very high. You should not plan on using it as part of a voltage divider. Look at this input pin diagram - the only resistance is the pullup at the top left, any voltage above 5V will just take the pin with it if there is much current available at all.

The original question was:-

I need to monitor the voltage of an unregulated power source.

Which meant using an analogue input. The impedance is high but it has to charge an internal capacitor for the sample and hold circuit on the front of the A/D. In order for this capacitor to charge quickly enough between changing the input multiplexer to the channel and taking the reading you are better off with a lowish source impedance of 10K. Higher ones will work but you have what appears to be cross talk if you try and use more than one channel. That is the voltage on one channel will appear to change the reading on another.

I think you are confusing internal impedance with the internal pull up resistor.