looking for a 12v precision voltage reference or something else to do the job

hey guys. im using an adc on a vellman k8055 to measure battery voltages.
my battery voltage range is 12 (flat)-14.4 volts (charged). my adc range is 0-5v and its only 8 bit.

to maximise the resolution i raise the ground voltage with a 12v regulator so the input on the adc will be from 0-4.4v. problem is that the regulator has a variance of about .05v on the output which is stuffing up my measurements considerably.

so what i believe im looking for is a precision 12v reference.
I cant find any 12 references anywhere though. i found this 10v one though - **broken link removed**

close but no cigar.

is there something else that can do the job here? thanks for any tips.

Try a TL431, it's an adjustable precision reference IC, 2.5V upwards, either adjust it for 12V, or use an opamp to increase the voltage.

that looks like it might do the job
Voltage Reference - Shunt Voltage Reference - TL431 - TI.com

looking at the pdf, im wondering which circuit i would use? figure 17 doesn't look right because the voltage divider would vary with the input source, which would defeat the purpose of the reference wouldn't it?

thanks for the suggestion!

the other thing I'm wondering is if i have vout set to 12v, what does the input voltage have to be? would i get away with a 12v battery charged between 12.3 and 13v?

As all you're wanting to do is drop the voltage to the ADC range, then just use a simple attenuator, messing about with regulators is pointless - your multimeter does it with attenuators, your scope does it with attenuators, EVERYTHING does it with attenuators.

im not familiar with attenuators, so i may be wrong here, forgive me if i am - but after reading wikipedia it appears attenuators will drop the voltage range by a factor. it appears they will scale the whole range, and not just clip the bottom like i am after.

eg 12-14.4 volts will become (divide by 3) will become 4-4.8 volts.

this is undesirable as it is not taking advantage of the full resolution of the adc.

with a precision reference that range is instead brought down to 0-2.4 volts which can be multiplied by a circuit included in the adc to become 0-5 volts, taking advantage of the full 8 bits.

correct me if im wrong though, im a newb to these circuits. thanks for the help. if this circuit actually does the right thing, how would i calculate the parts required to simply subtract 12v from the input?

You are correct, an attenuator simply scales it down - but your resolution isn't going to be grossly affected - and it's still likely to be more accurate than trying to subtract 12V from it.

If yoiu do want to, use an opamp as a differential amplifier, and subtract that way (using a precision reference on the other input).