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battery level monitor circuit

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wanchope

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Hi,
I have an 12 bit ADC to monitor a lithium polymer battery. The useful voltage ranges from around 8V to 12V. What is the better way to monitor this range using my 5 volt powered ADC?
I was using an opamp voltage translater to detect it, but it gives me 0.1 to 0.2 volt difference for different boards. (FYI, I am using 1% tolorance resistor.) And hence the ADC values are not accurate.
Anyone has a better circuit to do this job with minimum board-to-board difference?

regards
 

You can use better resitors, 0.1%, for example ..

If you can't, allow for variable resistor on one of the opamp's inputs ..

Some pcbs are designed in such a way that there are additional spots for rsistors which, when fitted, will compensate for resistors' tolerances: a big value in parallel or a small value in series ..

Regards,
IanP
 

Thanks, IanP.

Not sure 0402 type resistor has 0.1% kind of resistor. So do you think it is the resistor tolorance causing the difference?
Would it be the amplifier itself?
If every piece needs to go through the calibration process, then it would be costly also. Normally what other circuits you will use?

regards
 

Hi Wanchope,

Although 0402 resistors with 0,1 % tolerance do exist, getting some in small
quantities might be a problem.

**broken link removed**

But I don't think that the resistors are the only cause for the differences
in measurement, there are a more possible causes.
Maybe you should take a closer look at the two 10 µF capacitors and replace
them with tantalum or solid aluminium types, perhaps also with a higher
working voltage.
You could also install an offset correction by connecting an 100 ohm trimmer
between the two 10 k resistors. This could cancel the differences between the
two 10 k resistors and correct the offset voltage off the opamp at the same time.
You can also try to replace the LMC6482 with a LMC6082.
But I'm only guessing, you should compare different boards and measure the voltage at the non-inverting input of the opamp and also the voltage between the two 10 k resistors. The error you introduce with your multimeter will be the same for different boards. This might give you a clue of what you are looking for.

on1aag.
 

Hi... one thing i may add is that this is common difereence amplifier circuit.
What usually is done is the resistors are set so that R2=R4, R1=R3 (refer to the atached picture) so that the output expression simplifies to

Vout = (V1 - V2) (R4/R3)

i notices that this is not the case in your circuit. Maybe that's why you are experiencing those problems.
 

You are using 5 volts power supply as a voltage reference, but it can introduce an error greater than 0.1 volts unless fully stabilized.
 

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