ADC accuracy degrades quickly with battery

zonemikel · Joined: 13 Oct 2007 Posts: 53 Location: Texas

Hello ppl,
Thanks again for your previous help. I have a nice module that reads a analogue value and transmits it via serial and displays it on a screen.

I wanted to test to see how long it would last on a 9v battery, the 9v goes to a 5v regulator and to the pic/7 seg/ leds/ etc. I bought the cheepest batteries they were 2 9v for $1. I got the cheep ones because i was just going to run them till they ran out.

Anyway it reads the ADC conversion fine for about a hour then the adc conversion starts climbing. It will just slowly continue to go up. The reading should be about .80v and it gets up to like 1.19v. At this point if i plug in a good battery it works fine. If you guys need the code ill post it.

What im thinking is since the conversion goes from 10 bit adc it assumes that 5v = 1023 so when this value decreases the value of whatever your reading gets inflated. I believe the vref is the power that goes into the pic.

One more thing, the module stayed on for a good 12hours then i turned it off. So it will continue to function for a very long time, but the ADC reading is off.

tia
-zonemikel

bwhiten · Joined: 26 Nov 2003 Posts: 151 Location: Grayson, GA

What PIC are you using?
Did you monitor the output of the regulator? If that is still a constant 5 volts you shouldn't have a problem. Using a 3.3V PIC and LDO style regulator with the 9V input would allow for much longer operating time IMO.

newguy · Joined: 24 Jun 2004 Posts: 1909

There's a way to compensate for the falling supply rail. Create a zener diode voltage reference circuit (say approx 2.5V or something like that) and monitor this voltage with a spare A/D input on your PIC.

Just before you take an A/D reading (of whatever you are reading), read the voltage from the zener reference. When that reading starts to rise, the Vdd supply rail is falling. Based on the full power zener reading and the amount it has risen from that value, you can calculate what the actual supply rail is. You can use this value to compensate your other reading.

You can also put a FET in the zener reference circuit and use the PIC to turn the FET on/off to conserve battery power.

PCM programmer · Joined: 06 Sep 2003 Posts: 21708

zonemikel · Joined: 13 Oct 2007 Posts: 53 Location: Texas

ckielstra · Joined: 18 Mar 2004 Posts: 3680 Location: The Netherlands

bwhiten · Joined: 26 Nov 2003 Posts: 151 Location: Grayson, GA

Most importantly to me is you don't want the output from the regulator to be dropping below it's rated output. If it is that means you are no longer operating the device within it's specifications. At that point, most of the rest of it's specs are also out the window, such as ripple, load regulation, operating temperature, power dissipation, etc. An LDO regulator will let your input voltage drop much closer to the rated output before failing. I don't know what else is running on the regulator output but an unstable (drooping) output will have a negative impact on most circuitry.

newguy · Joined: 24 Jun 2004 Posts: 1909

Be careful with the calculation. It's not an add or subtract issue, it's a scaling issue (multiply/divide).