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atmega16 adc question

The data sheet is clear that inputs should not exceed Vcc + 0.5V.

There are two things

  1. As you are feeding the input from a divider, you are simply current limiting the input drain through the input protection diode, which should be ok for small currents (what are the divider values/battery voltage?), which leads to:

  1. You are draining the battery through the measurement circuitry while the equipment is off. The drain is, quite simply Vbatt - 0.5V/ (R div) where 0.5V is a typical forward bias voltage at low currents for the input protection diode and Rdiv is the value of the resistor from the battery to the ADC input.

So is it a problem? Electrically to the controller, probably not, but if the battery is always connected it will discharge faster than you expect.

Cheers

PeteS

Reply to
PeteS
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Assuming you are feeding the battery to the ADC channel via the 15k, ( I assume the battery is effectively paralleled on the main supply for this) then 18V/15k ~ 1.2mA into the pin while still powered down, which is for the order of tens of milliseconds, one would think. All that current is flowing through the input protection diode(s) when powered down.

I wouldn't personally think that should give you any trouble on the MCU.

Cheers

PeteS

Reply to
PeteS

I'm working on a project using an Atmega16 mcu. It's adc is being used to measure the battery voltage via a divider. A potential problem is that when the board is switched on there is a short time that the voltage is applied to the adc pin before the switching power supply brings the mcu up. My question is whether this is a problem, and if so how to solve it. Thanks for your assistance.

Mark

Reply to
Mark X

Sorry, but I really didn't describe the circuit fully - there is a pmos switch between the battery and the switcher. Pressing a pushbutton switch turns the pmos switch on, and as soon as the mcu comes up, it latches it on via a diode. The battery voltage is read from a point after the pmos switch, so it isn't draining the battery. The divider is a 15K and a 4.99K - the input voltage is 10-18 vdc (sometimes battery, sometimes wall wart).

Mark

Reply to
Mark X

^^^^^^^^^ nyuck...

How about a schottky diode from the ADC input to the VCC pin.

--
Regards,
  Bob Monsen

Nature does not at once disclose all Her mysteries.
- Lucius Seneca (Roman philosopher)
Reply to
Bob Monsen

Thanks, Pete - I appreciate the input!

Mark

Reply to
Mark X

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