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'[PIC]: 16F627 brown out? question'
2002\11\14@162816 by Scott Touchton

picon face
Hello List,

I have a 16F627 that appears to be sitting in some form of brownout.  The oscillator is running, but she ceased executing code.

Unfortunately, none of the fancy power up controls are enabled.  MCLR is disabled in the program fuses, along with the brown out detect and the power up timers. (Don't shoot me... I inherited the design.  Did I remember to tell you that the power supply  powers up and down twice a second!!!!!)

>From this I gather she has gone into a brown out mode.  Here is my exact question:   If in brown out, will the WDT cease to function?  Or should it apply a reset and get everything running again???


Sincerely,

Perplexed in Pennsylvania

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2002\11\15@073239 by Scott Touchton

picon face
Ok.... did I:       1. stump the list (doubt it)     2.  ask such a stupid
question that it does not deserve an answer  (probably)   or   3.  upset
someone without any knowledge of doing so???
{Original Message removed}

2002\11\15@080027 by Michael Rigby-Jones

picon face
> > Hello List,
> >
> > I have a 16F627 that appears to be sitting in some form of brownout.
> The
> oscillator is running, but she ceased executing code.
> >
> > Unfortunately, none of the fancy power up controls are enabled.  MCLR is
> disabled in the program fuses, along with the brown out detect and the
> power
> up timers. (Don't shoot me... I inherited the design.  Did I remember to
> tell you that the power supply  powers up and down twice a second!!!!!)
> >
> > From this I gather she has gone into a brown out mode.  Here is my exact
> question:   If in brown out, will the WDT cease to function?  Or should it
> apply a reset and get everything running again???
>
If the chip is in a brown-out state, nothing will happen until the supply
voltage is raised past the brown-out threshold.  The brownout should itself
cause a full reset.

Mike

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2002\11\15@082255 by Olin Lathrop

face picon face
> Ok.... did I:       1. stump the list (doubt it)     2.  ask such a
stupid
> question that it does not deserve an answer  (probably)   or   3.  upset
> someone without any knowledge of doing so???

I didn't answer because you left out too much important information and I
didn't feel like playing twenty questions.  It was easier to just hit
delete and move on.

> > I have a 16F627 that appears to be sitting in some form of brownout.
The
> oscillator is running, but she ceased executing code.

What oscillator mode?  What frequency?  How do you know it's in
"brownout"?  What voltage exactly it it running at?

> > Unfortunately, none of the fancy power up controls are enabled.  MCLR
is
> disabled in the program fuses, along with the brown out detect and the
power
> up timers.

If brownout detect is disabled, it can't be in "brownout mode", although
the power voltage can be too low for proper operation.  Again, what
exactly is the power voltage?

> Did I remember to
> tell you that the power supply  powers up and down twice a second!!!!!)

No, you didn't.  So this condition only last 1/2 second?  What does the
chip do when power comes back?  Does the power go all the way to 0 and 5V?
What is the rise time?

> > From this I gather she has gone into a brown out mode.

I can't gather that from the information you supplied.

> Here is my exact
> question:   If in brown out, will the WDT cease to function?  Or should
it
> apply a reset and get everything running again???

The WDT should function normally as long as the power voltage is at the
minimum operating level.  The entire chip, including the WDT, may act out
of spec below that.


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2002\11\15@092405 by Scott Touchton

picon face
Thank you Olin.  You are 100% right about my lack of info.  I  probably
would have nailed the delete key myself.  I really needed help thinking
through this and appreciate your response.


I have been through this before, but never with the WDT enabled.


>
> What oscillator mode?  What frequency?  How do you know it's in
> "brownout"?  What voltage exactly it it running at?
>
LP, 32.768Khz, 3.015V
The oscillator is alive, though execution has ceased based upon observance
of the IO pins.



> If brownout detect is disabled, it can't be in "brownout mode", although
> the power voltage can be too low for proper operation.  Again, what
> exactly is the power voltage?

I would disagree on this point, but I am confused and it is probably
semantics.  When I had a brown out problem many years ago, once the
condition was reached power supply voltage was not a factor.  In exact terms
(and this is on the 16C54) when the power supply dipped to 0.705V and then
returned to >3V the processor would not execute code though the oscillator
was running.  Microchip stated the part "was in brown out".  We shot the
engineer and installed a voltage detector to correct.

>
> No, you didn't.  So this condition only last 1/2 second?  What does the
> chip do when power comes back?  Does the power go all the way to 0 and 5V?
> What is the rise time?

Here is the normal course of events (which I intend to get rid of, but must
live with at the moment).  Processor powers up (0 to 3V in under 150uS).
Processor reads operating parameters from internal EPROM.  Processor takes
150ms scanning an IO port for signal and giving commands to an amplifier.
If no signals are found it sends an IO pin high to initiate power shut down.
The power supply is terminated, using an RC network to re-apply power 300 ms
later.  Power supply goes all the way down to 0, and all the way back to 3V.
This repeats indefinitely.


>
> I can't gather that from the information you supplied.

Based on Microchips definition to me of "brown out"  (oscillator running, no
code execution observable).  Maybe this definition does not hold true to the
interpretation of the list.



> The WDT should function normally as long as the power voltage is at the
> minimum operating level.  The entire chip, including the WDT, may act out
> of spec below that.

This was exactly what I was wondering.... seems as if the WDT is not
functioning although I am at the minimum operating level (3V).

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2002\11\15@094011 by Alan B. Pearce

face picon face
>Based on Microchips definition to me of "brown out"
>(oscillator running, no code execution observable).
>Maybe this definition does not hold true to the
>interpretation of the list.

I do not think this is the Microchip definition of Brownout, but I may be
wrong, and am prepared to be corrected.

IIRC the Microchip brownout detector operates if the supply drops below
4.something volts, and is designed to detect sag in a 5V supply, and stop
operation on a chip designed for 5V only operation. The detector is still
fitted to chips that will work on lower voltages, but needs to be disabled
for the low voltage operation.

For what you are doing, I suspect you would be better off using the :LF
version of the chip, which has the added advantage of being specified to 2V
supply. This will give you some margin over what your current chip can do.

As an aside, are you using the power up timer??? This will take 1024 cycles
before the chip starts running its program, and at 32kHz that starts to
become a significant part of your 150mS.

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2002\11\15@095836 by Olin Lathrop

face picon face
> LP, 32.768Khz, 3.015V
> The oscillator is alive, though execution has ceased based upon
observance
> of the IO pins.
>
> ...
>
> Here is the normal course of events (which I intend to get rid of, but
must
> live with at the moment).  Processor powers up (0 to 3V in under 150uS).
> Processor reads operating parameters from internal EPROM.  Processor
takes
> 150ms scanning an IO port for signal and giving commands to an
amplifier.
> If no signals are found it sends an IO pin high to initiate power shut
down.
> The power supply is terminated, using an RC network to re-apply power
300 ms
> later.  Power supply goes all the way down to 0, and all the way back to
3V.
> This repeats indefinitely.

As an experiment, try making the 300mS time much larger so that the
processor is held at 0V power longer.  Perhaps 300mS is not enough to
drain some internal capacitance in the reset logic.  Also note that the
crystal oscillator takes a while (1000 cycles if I remember right) before
the chip is allowed to run.  This is independent of the powerup timer, and
would be over 30mS in your case.

The 16F627 should draw under 10uA with a 32KHz crystal.  Is it really
worth it to shut down the processor completely for such short periods of
time?  You may be causing more effective supply current draw by charging
and discharging capacitances in the system every time the supply goes up
and down.  For example, to get 10uA effective average current drain only
requires a 1.7uF capacitor to be charged to 3V twice per second.


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2002\11\15@104216 by Scott Touchton

picon face
> As an experiment, try making the 300mS time much larger so that the
> processor is held at 0V power longer

There be the other aspect of the problem.  I have had about 50 of these fail
in the field.  I have not been able to duplicate the problem in my lab.  The
customer removes the battery, thus clearing the problem.  This probably
means you are right... receiver was at 0V much longer than 300mS.   I could
change the timing and put a bunch in the field to see what happens.

I have one on my bench that got returned with the battery in it.  Managed to
remove it from its case without disturbing power so I can make observations
and try some things.

I like your suggestion, just can't duplicate the problem to test.  I know if
I power down the on my bench it will power back up just fine (at least with
my luck it will).

I have tried temperature and battery voltage variation to duplicate the
issue with no luck.


> crystal oscillator takes a while (1000 cycles if I remember right) before
> the chip is allowed to run.  This is independent of the powerup timer, and
> would be over 30mS in your case.

So that means the oscillator should be stabilized before the program is
released, right?

> The 16F627 should draw under 10uA with a 32KHz crystal.  Is it really
> worth it to shut down the processor completely for such short periods of
> time?

Nope... total waste to power it down.  This is all going away in my
redesign.  I don't know why the engineer before me did it this way.



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> (978) 742-9014, http://www.embedinc.com
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2002\11\15@104858 by Olin Lathrop

face picon face
> As an aside, are you using the power up timer??? This will take 1024
cycles
> before the chip starts running its program, and at 32kHz that starts to
> become a significant part of your 150mS.

The powerup timer and the crystal oscillator timer are two separate
things.  The powerup timer provides a fixed 72mS (typical) delay, and can
be disabled.  The crystal timer is always enabled in crystal oscillator
modes.


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2002\11\15@105520 by Olin Lathrop

face picon face
> > crystal oscillator takes a while (1000 cycles if I remember right)
before
> > the chip is allowed to run.  This is independent of the powerup timer,
and
> > would be over 30mS in your case.
>
> So that means the oscillator should be stabilized before the program is
> released, right?

Yes.  That's the reason for the 1024 cycles.


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2002\11\15@111151 by Alan B. Pearce

face picon face
>The powerup timer and the crystal oscillator timer are
>two separate things.  The powerup timer provides a fixed
>72mS (typical) delay, and can be disabled.  The crystal
>timer is always enabled in crystal oscillator modes.

Well that may be so, but my understanding is that the chip is being powered
down, and then not running on power up. Now if the power up timer is
enabled, then he has a timeout of 72ms, which is half the power on time of
150mS.

But thank you for pointing out something I had not realised about the
16F627.

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2002\11\15@135113 by Dennis J. Murray

flavicon
face
If I can offer my 2 cents worth (you get what you pay for)!  I also ran into
problems with a very low power PIC running @ 32KHz. that included short
power supply shutdowns.  In my case, part of the problem was the bypass cap
across the chip (0.1 uf) held the chip running a split second longer than
desired.  Once that was resolved, I put the debugged chip in the final
circuit and it wouldn't shut down at all on it's own, even WITH the battery
disconnected!  Turned out the PIC was getting enough power from the camera
whose electronic shutter the PIC was trying to trip!  Once the battery was
removed, the PIC got its power thru the internal protection diode of the
PIC's output port I was driving the camera with - and that was enough to
keep the PIC running.  (I know, Microchip has documented this as something
to be careful with, but I missed it!)

This is probably not your problem, just thought it may help.  Good luck!
Dennis

{Original Message removed}

2002\11\15@142849 by Scott Touchton

picon face
> This is probably not your problem, just thought it may help.  Good luck!
> Dennis


Thanks Dennis.  I did look at this (got burned in the past).  Power wise,
everything looks OK.  I think Olin hit the nail on the head.  Probably 300mS
isn't enough time to guarantee complete, orderly shutdown.

Thanks also to Olin and the rest who responded.  Short term I have converted
production over to the LF part.  However, it is still doing the power up and
down.  Long term I will remove that (within the next month or so).

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2002\11\15@171832 by Dwayne Reid

flavicon
face
At 04:18 PM 11/14/02 -0500, Scott Touchton wrote:
>Hello List,
>
>I have a 16F627 that appears to be sitting in some form of brownout.  The
>oscillator is running, but she ceased executing code.
>
>Unfortunately, none of the fancy power up controls are enabled.  MCLR is
>disabled in the program fuses, along with the brown out detect and the
>power up timers. (Don't shoot me... I inherited the design.  Did I
>remember to tell you that the power supply  powers up and down twice a
>second!!!!!)
>
> From this I gather she has gone into a brown out mode.  Here is my exact
> question:   If in brown out, will the WDT cease to function?  Or should
> it apply a reset and get everything running again???

Since you are not using any form of brown-out detector, I'd say YES, you
have a problem.

I've had major problems occur if I allow a PIC's power supply to drop to
about 0.7V, then restore back to normal.  The chip does not function
correctly and the internal watchdog does NOT correct the problem.  The only
sure cure is to allow Vdd to collapse all the way to 0V, then re-apply power.

I've verified that I can cause this to happen on the 16c71, 16c73,
12c508.  I have not had occasion to test other chips for the same problem
(yet).

My solution is to ALWAYS have an external power-up reset controller.  I use
a variety of techniques, ranging from dedicated supervisor chips (high end
projects) through to simple power supply controllers (hold Vdd @ 0V until
supply is high enough, then apply Vdd to the circuit).

Most of my stuff is used in noisy, nasty environments.  I spend a lot of
time ensuring that the power supply is clean and turns on
monotonically.  The extra parts cost hardly anything and I don't have to
worry about having thousands or tens of thousands of units coming back.

The easy way for you to test if this is the problem is to do whatever it
takes to get the chip to misbehave, then briefly short Vdd to Gnd.  If the
chip comes to life, you have found the probable cause.

The fix may be as simple as adding a single J-FET and 1 resistor.  Grab a
J176 or J177 fet, connect S to GND, connect D to Vdd and feed G from the
input of your regulator circuit via a 1M0 resistor.  While the supply is
less than the gate threshold of the FET, the FET looks like a low value
resistance load on the supply.  Once the supply is above the threshold, the
FET turns off.

dwayne

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2002\11\16@040650 by Peter L. Peres

picon face
Maybe a bleeder resistor between vcc and vdd is needed to make sure the
voltage goes to 0V on every turn off ? Try 2k7 for 3V. This may or may not
upset the power budget. You could hot-connect the resistor to the returned
unit. Just use a soldering iron with the cord unplugged.

Peter

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2002\11\16@081458 by Olin Lathrop

face picon face
> Maybe a bleeder resistor between vcc and vdd is needed to make sure the
> voltage goes to 0V on every turn off ?

It already does according to his description.  The problem may be that it
doesn't stay there long enough for internal capacitances to get drained.

> Try 2k7 for 3V. This may or may not
> upset the power budget.

It certainly would!  Remember, this thing shuts itself off for 300mS at a
time to save a few uA of current drain (and probably causes more in the
process, but that's another story).


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