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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?
Your effective input impedance is the parallel combination of 4k7 and 18k.
Formula is R1*R2/(R1 + R2).
HTH
BrianT
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)
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).Quote:
Originally Posted by mackrackit
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.
Consider BrianT answer confirmed :)
This may help, near the end....
http://www.kpsec.freeuk.com/imped.htm
Thanks BrianT and mackrackit.
It still seems counter-intuitive to me, but it means that I can have accurate battery readings with less wasted current, so I'm happy.
An easy way to understand it without going into thevenin equivalent circuits. :p
Consider the plus side and the 0 (ground) side of the power source as low impedance sources. In an ideal power source the + side and 0 side will each be zero ohms impedance and as such the PIC sees both the high side resistor and low side resistor going to to zero ohm sources making the value of the resistors parallel.
Nice explanation!Quote:
Consider the plus side and the 0 (ground) side of the power source as low impedance sources. In an ideal power source the + side and 0 side will each be zero ohms impedance and as such the PIC sees both the high side resistor and low side resistor going to to zero ohm sources making the value of the resistors parallel.
...or put a largish capacitor accross the pic pin and use the ESR of the cap as your source impedance!