Resistor Voltage Divider and ADC

[Kamikaze47]

Hi,

I need to measure the voltage of a battery, so i'm going to be using a resistor voltage divider to drop the voltage down to <5v that the ADC can handle. Like this:


Battery +ve ------- 18k Resistor ------- ADC Pin ------- 4.7k Resistor ------- Ground

My question is, what is my impedance for calculating the min ADC acquisition time in this instance? Is it just the impedance to ground (i.e. 4.7k)? Or do I have to consider the 18k resistor as well?

[BrianT]

Your effective input impedance is the parallel combination of 4k7 and 18k.

Formula is R1*R2/(R1 + R2).

HTH
BrianT

[Kamikaze47]

Oh, really? I cant say I understand why that is the case, but i'll take your word for it.

In that case I should be able to use even higher resistor values (to reduce the wasted current) without violating the PIC's specification of a 10k maximum impedance on ADC inputs?

Something like:

Battery +ve ------- 47k Resistor ------- ADC Pin ------- 12k Resistor ------- Ground

= 9.56k input impedance ~ 10uS ADC acquisition time (and only wasting fractions of a mA of battery current)

[mackrackit]

You might find this useful also.
http://www.pbpgroup.com/modules/wfse...p?articleid=25

[Kamikaze47]

You might find this useful also.
http://www.pbpgroup.com/modules/wfse...p?articleid=25

Thanks. I don't have any trouble working out the divisor so that it wont go over 5v for my input range. It's the impedance that I was worried about (but still want to minimise wasted current).

It's a shame that DT's calculator there doesn't have the impedance in the result. I have yet to find confirmation elsewhere that the input impedance is the parallel impedance of the 2 resistors. It seems a little counter-intuitive to me. If anything, I would have thought that it would be the series impedance.

[mackrackit]

Consider BrianT answer confirmed
This may help, near the end....
http://www.kpsec.freeuk.com/imped.htm