Kat, errrr, 9.23uA x 10 ohms = 92uV (9.23e-5), not 10uV. :) ----- Original Message ----- From: "Kathy Quinlan" <kat-yahoo@kaqelectronics.dyndns.org> To: <AVR-Chat@yahoogroups.com> Sent: Friday, May 21, 2004 12:07 AM Subject: Re: [AVR-Chat] low-battery detection, ADC, and ICP of mega16 > jay marante wrote: > > > are you sure you got your formulas right?! > > that's just a voltage divider and im pretty sure you can't generate > > 650kV from a 12V battery. im sorry but i can't remember the formula for > > the voltage divider circuit. > > maybe someone might confirm that the multisim was right with its output. > > as far as i know, current can be computed by dividing Vbatt over the > > total resistance, which is 1,300,010 ohms. pretty low current there! > > > > > > > battery > > > | > > > < > > > > 1.3 Mega ohm > > > < > > > | > > > |------------------- ADC7 ~ 5V as simulated in multisim > > > | > > > < > > > > 10ohm > > > < > > > | > > > GND > > > > > > Work it the other way, I = V/R = 12/1,300,010 = > 9.2306982253982661671833293590049e-6 A > > > Now we know that V=IxR = 9.2306982253982661671833293590049e-6 x 10 = > 9.2306982253982661671833293590049e-5V or ~10uV at the ADC7 input... > > > I think the 10 ohm resistor needs to be higher say 10K and the 1.3M > resistor lower say 13K which would make the ADC voltage 5.21 V now for > saftey I would use a 15K resistor on the top. > > > All this is hypothetical if you are using the internal Aref as it is > only 2.56V (haaaaaa why be so presise with the voltage ie to 2 decimal > places when heat and supply variations make it nowhere near that acurate) > > I would use an external Vref something like a 4.096V reference chip. > > Regards, > > Kat. > > -- --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.688 / Virus Database: 449 - Release Date: 5/18/2004
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Re: [AVR-Chat] low-battery detection, ADC, and ICP of mega16
2004-05-24 by wagner
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