OK to operate at the absolute max rating?

[Tomás Ó hÉilidhe-2]

The 887 datasheet says each pin can give an absolute maximum of 25 mA,
but it warns against drawing that much.

I'm doing display multiplexing and I want to pump as much current as
possible through my LED's, while at the same time having a stable
reliable product.

Do you reckon it's OK to pull a full 25 mA from a pin if you're doing
display multiplexing?

I know you can probably draw 30 mA and get it way with it, (PIC's being
bulletproof and all that), but I'm trying to think in terms of a
developing a rock-solid product that has minimal faulty units, something
that will sell in the millions all over the world.

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[Robert Young-2]

> The 887 datasheet says each pin can give an absolute maximum of 25 mA,
> but it warns against drawing that much.
>
> I'm doing display multiplexing and I want to pump as much current as
> possible through my LED's, while at the same time having a stable
> reliable product.
>
> Do you reckon it's OK to pull a full 25 mA from a pin if you're doing
> display multiplexing?

No it is not OK.  There is a maximum power dissipation rating for the chip.  This sets a limit to the total current that can be drawn from ALL pins.

>
> I know you can probably draw 30 mA and get it way with it, (PIC's being
> bulletproof and all that), but I'm trying to think in terms of a
> developing a rock-solid product that has minimal faulty units, something
> that will sell in the millions all over the world.

Then don't exceed ANY rating.

Rob
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[Tomás Ó hÉilidhe-2]

Robert Young wrote:
>> Do you reckon it's OK to pull a full 25 mA from a pin if you're doing
>> display multiplexing?
>>    
>
> No it is not OK.  There is a maximum power dissipation rating for the chip.  This sets a limit to the total current that can be drawn from ALL pins.
>  


The total limit -- if we understand each other -- is 90 mA for all pins
combined. I've only got two pins sourcing current at a time, so I'm not
going over the limit for "ALL pins" as you call it.


>> I know you can probably draw 30 mA and get it way with it, (PIC's being
>> bulletproof and all that), but I'm trying to think in terms of a
>> developing a rock-solid product that has minimal faulty units, something
>> that will sell in the millions all over the world.
>>    
>
> Then don't exceed ANY rating.


I don't think anyone can be that pedantic about it, because after all
the PIC datasheet itself doesn't give a maximum rating. It tells you "25
mA" but then it kind of shies away and says "yeah but you probably don't
wanna draw 25 mA from it". So then what's a safe current to draw? 24.9
mA? 23 mA? 20 mA? We're left in the dark.

Judging by the amount of abuse I've seen PIC chips being able to take, I
get the feeling I'd be A-OK to draw 25 mA when multiplexing my display.

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[Apptech]

> I'm doing display multiplexing and I want to pump as much
> current as
> possible through my LED's, while at the same time having a
> stable
> reliable product.

> Do you reckon it's OK to pull a full 25 mA from a pin if
> you're doing
> display multiplexing?>

No! See below



 The 887 datasheet says each pin can give an absolute
maximum of 25 mA,
> but it warns against drawing that much.

I assume you mean the PIC 16F877.

    http://ww1.microchip.com/downloads/en/DeviceDoc/41291E.pdf

You need to mentally integrate all the available information
in a data sheet to get a proper impression of true
capabilties.

In the above data sheet.

- Page 245.
ABSOLUTE maximum Vdd or Vss current is 95 mA.
So you could operate no more than 3 I/O pins at 25 mA
without exceeding this.

- Also page 245.
Absolute maximum readings are just that. As they note,
proper operation of the IC is not guaranteed at that level
AND the IC may be damaged if you run it above this level. ie
this is the outer limit at which damage to the IC probably
won't happen.

- Also p245.
Max pin currents individually and together. As above re
damage and proper operation.

The data sheet seems VERY uninformative about allowable
operating current per I/O pin, but it can be very safely
assumed that they are no greater than the limits set above.

For high current drive of multiple LEDs at once you are
almost certainly going to need a driver IC.

___________

> I know you can probably draw 30 mA and get it way with it,
> (PIC's being
> bulletproof and all that)

Maybe you can. But the absolute ratings spec given above
says explicitly that the IC may be damaged if you exced 25
mA on any pin.



   Russell


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[Cedric Chang-2]

The issue with pulling 25ma from each output most likely will be a  
problem with the 887 heating up and failing catastrophically.  If you  
heat portions of the chip up quickly enough, the chip will fail  
without the entire chip heating up.  Localized heating will do it in.

A rock solid product will be defined differently by different engineers.
If it were me, I would stay away from max values by a factor of 2:1.  
I would monitor the 887 for temperature rise at room temperature ( or  
whatever environment it will be in ).  I wouldn't want the chip to  
have a temperature of more than the midpoint between 30C and the  
maximum temperature.   I would also test the design in a test chamber  
where the ambient was 50C.   Testing the design means testing more  
than one board.  The more boards you test, the better handle you have  
on typical performance.

No doubt there are geniuses on this list who know how to  
statistically analyze MTBF ( mean time between failure ) after a  
number of boards have been tested.  And they probably want to be  
paid.  ( I would )

cc

> On Jul 13, 2008, at 5:42 PM, Tomás Ó hÉilidhe wrote:
>
>
> The 887 datasheet says each pin can give an absolute maximum of 25 mA,
> but it warns against drawing that much.
>
> I'm doing display multiplexing and I want to pump as much current as
> possible through my LED's, while at the same time having a stable
> reliable product.
>
> Do you reckon it's OK to pull a full 25 mA from a pin if you're doing
> display multiplexing?
>
> I know you can probably draw 30 mA and get it way with it, (PIC's  
> being
> bulletproof and all that), but I'm trying to think in terms of a
> developing a rock-solid product that has minimal faulty units,  
> something
> that will sell in the millions all over the world.
>
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>
>

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[Mark Rages]

On Sun, Jul 13, 2008 at 6:42 PM, Tomás Ó hÉilidhe <toe@...> wrote:

>
> The 887 datasheet says each pin can give an absolute maximum of 25 mA,
> but it warns against drawing that much.
>
> I'm doing display multiplexing and I want to pump as much current as
> possible through my LED's, while at the same time having a stable
> reliable product.
>
> Do you reckon it's OK to pull a full 25 mA from a pin if you're doing
> display multiplexing?
>

Did you read my reply that I sent you under "Pin current limits"?

What do you think the pin's output voltage will be when you source
25mA from it?  What will the PIC's dissipation be?

Regards,
Mark
markrages@gmail
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[Tomás Ó hÉilidhe-2]

Mark Rages wrote:
> What do you think the pin's output voltage will be when you source
> 25mA from it?  What will the PIC's dissipation be?
>  


I'm powering the 887 off 3 volts.

When a pin is set to high, I guess it will be between 2.8 volts and 3
volts maybe.

This pin will go to an LED whose voltage drop is 2.1, and thence to a 33
ohm resistor to ground.

Current will be about 25 mA or so.

Power dissipation in LED = 2.1 V x 25 mA = 53 mW or so
Power dissipation in resistor = 900 mV x 25 mA = 23 mW or so

Total power dissipation = 3 V x 25 mA = 75 mW or so

As far as I'm aware -- and I'm open to correction here -- the only
figure I need to keep an eye on is the current drawn from the PIC pin
(i.e. I should keep it less than 25 mA).

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[Sean Breheny]

Hi Tomas,

ICs typically have two max ratings for very fundamental things (like
temperature, voltage, current, etc.). Absolute max, and typical max.

Absolute max means "It may be permanently damaged if you go above this"
Typical max means "It may not work properly if you go above this, and
life may be shortened too"

Good engineering design practice is never to exceed the typical max,
since you need the device to operate correctly, not just remain
undamaged.

Sean


On Sun, Jul 13, 2008 at 8:23 PM, Tomás Ó hÉilidhe <toe@...> wrote:

>
>
> Robert Young wrote:
>>> Do you reckon it's OK to pull a full 25 mA from a pin if you're doing
>>> display multiplexing?
>>>
>>
>> No it is not OK.  There is a maximum power dissipation rating for the chip.  This sets a limit to the total current that can be drawn from ALL pins.
>>
>
>
> The total limit -- if we understand each other -- is 90 mA for all pins
> combined. I've only got two pins sourcing current at a time, so I'm not
> going over the limit for "ALL pins" as you call it.
>
>
>>> I know you can probably draw 30 mA and get it way with it, (PIC's being
>>> bulletproof and all that), but I'm trying to think in terms of a
>>> developing a rock-solid product that has minimal faulty units, something
>>> that will sell in the millions all over the world.
>>>
>>
>> Then don't exceed ANY rating.
>
>
> I don't think anyone can be that pedantic about it, because after all
> the PIC datasheet itself doesn't give a maximum rating. It tells you "25
> mA" but then it kind of shies away and says "yeah but you probably don't
> wanna draw 25 mA from it". So then what's a safe current to draw? 24.9
> mA? 23 mA? 20 mA? We're left in the dark.
>
> Judging by the amount of abuse I've seen PIC chips being able to take, I
> get the feeling I'd be A-OK to draw 25 mA when multiplexing my display.
>
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[Apptech]

> The total limit -- if we understand each other -- is 90 mA
> for all pins
> combined.

Yes. 90 mA ABSOLUTE MAXIMUM. Beyond that damage may occur.
Also 25 mA drive per single pin, above which damage may
occur.
And, proper operation of the IC is not guaranteed at these
levels - these are just "stress ratings" (their words more
or less)

> I've only got two pins sourcing current at a time, so I'm
> not
> going over the limit for "ALL pins" as you call it.

I really is the limit for 2 pins.
And knowing that you are talking about only 2 pins is
helpful. If you give as much information as possible in the
original query (such as how many pins you want to use at
once) it helps people direct the answers appropriately and
avoid blind alleys.

>>> I know you can probably draw 30 mA and get it way with
>>> it, (PIC's being
>>> bulletproof and all that), but I'm trying to think in
>>> terms of a
>>> developing a rock-solid product that has minimal faulty
>>> units, something
>>> that will sell in the millions all over the world.
>>>
>>
>> Then don't exceed ANY rating.

This is crucial.
Exceeding the absolute maximum ratings MAY dstroy the IC.
Operating AT the maximum ratings may lead to improper
operation.
Both those points are explicitly stated in the data sheet.

> I don't think anyone can be that pedantic about it,
> because after all ...

The "ASBSOLUTE MAXIMUM section of any data sheet should be
read with an utterly pedantic mindset.
It means just what those two words say.

> the PIC datasheet itself doesn't give a maximum rating.

It does. As above. Absolute Maximum is the maximum rating
under ANY conditions to guarantee IC survival.
You can sometimes destrioy it with lower combinations but
the are saying that it may die if even any one of these
ratings is exceeded at one time.

> It tells you "25
> mA" but then it kind of shies away and says "yeah but you
> probably don't
> wanna draw 25 mA from it". So then what's a safe current
> to draw? 24.9
> mA? 23 mA? 20 mA? We're left in the dark.

The apparent "shieing away" is a result of the two sections
concerned dealing with different things.
Absolute maximum deals with IC death
Operating conditions deal with actually doing useful work.

As you draw more current from a pin (or sink it) the voltage
is pulled further from the supply rail due to internal
voltage drop. A point is reached where the logic level
becomes invalid. If your external devices have tighter logic
level requirements then you MAY not be able to draw full
rated current.

A made up example only.
IC has pin current of 25 mA abs max, 20 mA max operating.
Vout is 5V at 0 mA and say 3.5 V at 20 mA.
If your external device needs to see 4V or more to see a
logic high then a 20 mA load would pull the voltage below
the level where it was valid for your external device.

> Judging by the amount of abuse I've seen PIC chips being
> able to take, I
> get the feeling I'd be A-OK to draw 25 mA when
> multiplexing my display.

Depends on what you mean by A-OK.
The data sheet says that 25.001 mA MAY destroy the IC.
It also says that the IC MAY NOT work properly at 24.999 mA
but will survive.
The max allowable OPERATING current is elsewhere in the data
sheet (usually) and may be 20 mA.

While many PICs may well sink 25 or 30 mA per I/O pin and
work OK, if you design for 25 mA, and build a million
products and some source 26 mA you may find you have
thousands of customers with games of "Connect 0".


        Russell


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[Tomás Ó hÉilidhe-2]

Take the following simple circuit:

PIC pin set high  -->  resistor -->  LED  -->  ground

Now here's what I'm curious about:

The datasheet cites the maximum current for an I/O pin as 25 mA.

What I think is a little strange about this is that you can power the
PIC at different voltages; I mean if the PIC's powered by 5 V, then 25
mA equates to a power dissipation of 125 mW. However if you run the PIC
at 3 V, then 25 mA means a power dissipation of only 75 mW.

If you're allowed source 25 mA when running at 5 V, then should you not
be allowed to source 41 mA when running at 3 V? (41 = 25 * 5 / 3)

Just a thought.

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[Sean Breheny]

Hi Tomas,

The power dissipated by the PIC in this case is NOT (Vdd-Vss)*Current.
It is (Vdd-Vpin)*Current. So, if you are supplying the PIC with 5V,
and the voltage under a 25mA load on the pin is 3V, then the
contribution to the total PIC power dissipation from this pin is
(5-3)*0.025=50mW. This is all being dissipated by a tiny FET driver
inside the IC.

At 3V supply, the situation would be something like the following: 1V
on the pin at a 25mA load, giving (3-1)*.025=50mW still. The voltage
drop between the PIC's Vdd pin and the output pin comes from the
current (25mA) times the on resistance of the FET driver for that pin.

Another problem which Mark alluded to is that the voltage on the
output pin may drop considerably (as in the above example, 2V drop) at
such high currents. This is because you are beginning to take the FET
driver out of the ohmic region into the constant current region.

I suspect that if you shorted one of the output pins to ground and
drove it high, it would probably source about 100mA. This is the
reason why some people don't feel the need to use current limiting
resistors with LEDs on PIC pins - it will often work out that you will
get something like 20mA out of the pin when you force it to the Vf of
a typical LED. The problem with this is that PIC output pin resistance
varies from IC to IC, and with temperature, and the same is true of
the Vf of an LED, so that perhaps 10% of the products produced this
way will have out of spec currents which will damage either the LED or
the PIC output driver.

You are extremely ambitious, which is good to a point. You remind me
of myself at about 16 years old. That's about the time when I joined
the PICLIST (I'm 28 now) and I have learned a LOT. Please try to be
patient and learn from the good people here - even those who can be
gruff at times. If you really do take your product to production in
the millions, you will learn many things along the way, one of them
being that you WILL see almost every failure mode imaginable. Stuff
that you would never see doing hobby pic projects (because it has a
probability of 0.01%) will show up in some of your production units.
You therefore must rely on a good deal of planning and testing, and
re-design using info from failed units, to perfect your design.
Anything which is slopped together and just "works" will probably work
in 80% of production units, but there WILL also be that extra 20%
fallout or more due to something you didn't see in your single
prototype.

Sean


On Sun, Jul 13, 2008 at 9:49 PM, Tomás Ó hÉilidhe <toe@...> wrote:

>
> Take the following simple circuit:
>
> PIC pin set high  -->  resistor -->  LED  -->  ground
>
> Now here's what I'm curious about:
>
> The datasheet cites the maximum current for an I/O pin as 25 mA.
>
> What I think is a little strange about this is that you can power the
> PIC at different voltages; I mean if the PIC's powered by 5 V, then 25
> mA equates to a power dissipation of 125 mW. However if you run the PIC
> at 3 V, then 25 mA means a power dissipation of only 75 mW.
>
> If you're allowed source 25 mA when running at 5 V, then should you not
> be allowed to source 41 mA when running at 3 V? (41 = 25 * 5 / 3)
>
> Just a thought.
>
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[Jinx-4]

> If you really do take your product to production in the millions,
> you will learn many things along the way, one of them being that
> you WILL see almost every failure mode imaginable

Assuming Hasbro give (sell) a license, they're sure to demand
performance and a return. Their accountants aren't schmucks


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[Mark Rages]

On Sun, Jul 13, 2008 at 7:56 PM, Tomás Ó hÉilidhe <toe@...> wrote:

>
>
> Mark Rages wrote:
>> What do you think the pin's output voltage will be when you source
>> 25mA from it?  What will the PIC's dissipation be?
>>
>
>
> I'm powering the 887 off 3 volts.
>
> When a pin is set to high, I guess it will be between 2.8 volts and 3
> volts maybe.
>

No need to guess, see figure 18-28 in the datasheet.  (well, you may
need to extrapolate a little.  The graph only goes to 4 mA output
current.)

Regards,
Mark
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[Apptech]

> When a pin is set to high, I guess it will be between 2.8
> volts and 3
> volts maybe.

Only at VERY low current.

>
No need to guess, see figure 18-28 in the datasheet.  (well,
you may
need to extrapolate a little.  The graph only goes to 4 mA
output
current.)
/>

Page 287.
IF the internal resistance stays the same then that graph
can be VERY roughly translated to

    Vout = 2.9 - 0.125V/mA

ie 1 V drop at 8 mA or about 24 MA out with pin grounded.

If you drove a 2.1V load directly with no resistor you might
hope for about 6 mA.

As the IC warms up this is liable to decrease.

*** An LED driver is looking to be essential ***

BUT this is exceedingly cheap and easy to achieve. An
emitter follower consisting of a single transistor will
suffice. NPN. Collector to V+. Emitter = output. Base to
output pin.
Vout ~= Vdd - 0.7V.
Given the closeness of your desired load voltage to Vdd a 2
transistor driver would be preferred if cathode to ground
connected LED is required. Done properly this takes 4
resistors and 2 transistors. One less resistor if one FET
used. 2 less resistors if 2 FETs used.

One transistor plus two resistor driver if anode to Vdd LED
connection OK


        Russell McMahon







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[Vitaliy-2]

Apptech wrote:

> *** An LED driver is looking to be essential ***
>
> BUT this is exceedingly cheap and easy to achieve. An
> emitter follower consisting of a single transistor will
> suffice. NPN. Collector to V+. Emitter = output. Base to
> output pin.
> Vout ~= Vdd - 0.7V.
> Given the closeness of your desired load voltage to Vdd a 2
> transistor driver would be preferred if cathode to ground
> connected LED is required. Done properly this takes 4
> resistors and 2 transistors. One less resistor if one FET
> used. 2 less resistors if 2 FETs used.

Why "emitter follower"?
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[Apptech]

>> *** An LED driver is looking to be essential ***

>> BUT this is exceedingly cheap and easy to achieve. An
>> emitter follower consisting of a single transistor will
>> suffice. NPN. Collector to V+. Emitter = output. Base to
>> output pin.
>> Vout ~= Vdd - 0.7V.
>> Given the closeness of your desired load voltage to Vdd a
>> 2
>> transistor driver would be preferred if cathode to ground
>> connected LED is required. Done properly this takes 4
>> resistors and 2 transistors. One less resistor if one FET
>> used. 2 less resistors if 2 FETs used.
>
> Why "emitter follower"?

Good question.

He has previously emphasised low parts count.

Emitter follower allows same drive sense with a transistor
only added and LED still common cathode.

The one transistor common emitter driver mentioned at the
end uses a little more but requires common anode led
connection which may be an issue in a multiplexed scheme.

The 2 transistor driver preserves both drive sense and
common cathode connection but uses (a few) more parts.



        Russell McMahon
 

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[Wouter van Ooijen]

> The 887 datasheet says each pin can give an absolute maximum of 25 mA,
> but it warns against drawing that much.

I'd be surprised. Where (on what page) does it say so?

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