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'[PIC]: Measuring Current Consumption During Sleep'
2002\01\21@140234 by John Hansen

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I am currently doing my first project where current consumption is an
issue.  I've configured a keypad on portB of a 16F628 and have my program
set up so that I'm using the internal port B pullups.  I have the PIC sleep
except when a key is pressed.  The program works great.   However, I am a
little concerned by the current consumption.  In measuring the current
drawn by the circuit I'm getting 85 microamps.  I've checked my meter by
measuring the current consumed by a circuit that simply has a 1 megaohm
resistor in it, and I get the expected 5 microamps.

That's pretty good, but Microchip seems to indicate the current consumption
of this device during sleep should be much lower.  So I set up a simpler
circuit that has only a PIC, all pins configured as outputs and all pins
tied to ground.  I still get about 85 microamps.  I tried it with all pins
configured as inputs.  Still 85 microamps.

Then I read application note AN606, where it talks about a circuit that
measures current consumption of the PIC by putting 100 ohm resistors on the
input and output and alternately shorting them and measuring the current
differential.  This would seem to imply that there is something wrong with
my technique of simply using a multimeter to measure the current
consumption of the device.

As I said, I'm pretty new to very low power circuits.  Is my device
actually consuming 85 microamps, or is there something wrong with my
measurement technique?


John Hansen

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2002\01\21@150405 by Don Hyde

You should be able to get it down to a couple of microamps.

It is very easy to consume a few extra microamps.  To get down to the
minumum, you need to turn off the brownout detector, make sure than you have
not left any peripherals running -- such as analog voltage reference, and
take great care with each and every I/O pin.

An input pin that's left to float will (sometimes, and not always repatable)
float into the linear region, which will cause both of its CMOS transistors
to turn on, wasting a few microamps.  Either connect it to ground or Vcc
with a resistor (1M is enough), or if it's not connected to anything at all,
make it an output.  It doesn't matter whether it's high or low, as long as
it's not somewhere in the middle.

An output pin that is driving any sort of load will consume power.  This
includes pullup/down resistors and devices which you have powered down.  You
need to look carefully at each one and make sure it's not sourcing current
to something.

I often find it helpful to separately measure the current through Vcc and
GND to the PIC.  This will give you a clue as to whether you have a pin
that's pulling up or down.  If you have an output that's pulling down
against a pullup resistor, you'll see more current through Vcc than Gnd.  If
you have a high output going to a powered-down chip, then you'll see more
current through Vcc.

> {Original Message removed}

2002\01\21@155903 by John Hansen

picon face
Thank you!   I had assumed it had something to do with the I/O pins.  It
seems I was barking up the wrong tree.  I turned the brownout detector off
and current consumption dropped to less than 1 microamp.  Thanks so much, I
should now be able to run this on a battery for the shelf life of the battery.


John Hansen

At 02:02 PM 1/21/2002 -0600, you wrote:
{Quote hidden}

> > {Original Message removed}

2002\01\22@003214 by Russell McMahon

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> You should be able to get it down to a couple of microamps.
> It is very easy to consume a few extra microamps.

Presumably you are using a crystal or resonator.
When using an RC clock with external R the oscillator tends to add about
Vcc/2R to the current drain.


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