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'[EE]: Help! Data Flash corruption'
2001\08\30@163730 by Thomas N

picon face
Dear Dan,

Thank you for your quick reply.  I operate the data flash at min supply
2.7V.  The IC is specs at 2.7 to 3.6V.  I don't think the data flash will
work at 2.3V (0.4V down).

I actually put the cap at the input of a 2.7V regulator to keep the voltage
at the flash regulated after the unit is powered down.  I already tried with
470uF cap, but sometimes the flash is still corrupted (less though).  Is
there any thing I can do in software?  Anyting else?

Thanks again,
Thomas


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2001\08\30@200324 by Thomas N

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Richard,

I use a simple transistor circuit to detect the voltage drop to shut down to
MCU.  However, the voltage drop down too fast and in some cases, it screws
up the flash if it's in a write operation.  I use a separate regulator for
the flash.  The input of the regulator is connected to 5V and with a 470uF
cap to ground.  The output of the regulator is connected to 10uF cap.
Still, the 470uF cap can't supply enough energy.  I checked out the specs,
and the flash consumes up to 35mA during the write operation.  I need to
retain its power for at least 20ms, 40ms is refer...  Guess I am stuck and
have to live with it :(

Thomas


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2001\08\30@205231 by hard Prosser

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face
Thomas
Lets see, you need 35mA at 2.7V(minimum) for 40mS. This corresponds to 95mW
for 40mS or 3.8mJ.
Allowing for 1V drop on the storage cap (5V to 4V) - enough for an LDO
regulator to easily maintain 2.7V on the output - then E=1/2C(Vh-Vl)^2
or 3.8e-3 = 0.5 x C x (5-4) x (5-4)  i.e. C=3.8e-3/(0.5  x 1) = 8mF
(8000uF) !

Something of a problem.

If you could increase the supply voltage to the regulator to , say 9V, the
the cap would only need to be about 3.8e-3/(0.5 x 5 x 5) = 304uF,   i.e.
your 470uF cap should be OK

Or support the Flash from a supercap or in-board battery backup?

Richard











Richard,

I use a simple transistor circuit to detect the voltage drop to shut down
to
MCU.  However, the voltage drop down too fast and in some cases, it screws
up the flash if it's in a write operation.  I use a separate regulator for
the flash.  The input of the regulator is connected to 5V and with a 470uF
cap to ground.  The output of the regulator is connected to 10uF cap.
Still, the 470uF cap can't supply enough energy.  I checked out the specs,
and the flash consumes up to 35mA during the write operation.  I need to
retain its power for at least 20ms, 40ms is refer...  Guess I am stuck and
have to live with it :(

Thomas


{Quote hidden}

fix
>is to make sure you have enough on-board capacity (caps or battery) to
hold
>the supply up until the EEPROM write is complete.
>Since the micro won't be drawing all of the 10mA (I guess) -  can you
>separate the micro supply from the rest of the circuit using a diode and
>use a cap to hold the micro supply up until the write is compete? - You
may
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2001\08\30@213142 by Thomasn101

picon face
I guess I have a problem then!  8000uF cap is too costly for my low-cost
product and I don't have 9V voltage rail.  Actually, I have 8V rail but it's
noisy like hell.

Thanks for your help!
Thomas
{Original Message removed}

2001\08\31@090548 by Olin Lathrop

face picon face
> Thank you for your quick reply.  I operate the data flash at min supply
> 2.7V.  The IC is specs at 2.7 to 3.6V.

That may be the problem right there if you are operating the flash at the
exact edge of its supply spec.  The nominal supply should be high enough so
that under worst case conditions and part variations it doesn't go below the
minimum required by the flash.  Running a 2.7 to 3.6 volt part with a 2.7
volt regulator sounds like a bad idea.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, spamBeGoneolinspamBeGonespamembedinc.com, http://www.embedinc.com

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