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Discussion Groups | Piclist | Is 24 mA too much for PIC I/O pin if it has a duty cyle?

A discussion group for the PICMicro microcontroller. Also called the Microchip PIC, this list is dedicated to the use and abuse of this fine, simple, microcontroller. Close to topic posts are welcome, ie. general electronics.

Is 24 mA too much for PIC I/O pin if it has a duty cyle? - ransom peek - May 6 12:49:00 2004

Is 24 mA too much for PIC I/O pin if it has a duty cyle?

Using a PIC 16F688, the maximum current for an I/O pin is 20 mA.  Is it ok to exceed this if the drain has a duty cycle as follows:


I/O pin feeds 120 ohm 5% resistor (114 ohms at low end of tolerance).  The resistor is part of a 6 bit binary ladder generating audio direct feed to headphones (6 resistors to six I/O pins, then tied together  -  this pin has the lowest value resistor, so I am concerned).  The ladder has a 10 uf series cap (dc block) and then to a 10K pot to ground.  Center of pot to headphones.

3 volt circuit with 2N5819 schotky barrier diode, so new batteries give (1.53 volts x 2) - 0.3 volt drop  =  2.76 volts max.

2.76 volts / 114 ohms = 24.2 mA !

But in generating the audio sine wave the pin has a duty cycle.  And, the headphones add some resistance, maybe 16 ohms.

At 300 Hz, with approx 40 points for each sine wave, this pin is on for approx .002 second, then off .001 sec, then on again .002 sec then off .001 sec, etc.  So, about 67% duty cycle.

Is there some risk that the PIC will fail?



Ransom Peek

Fort Collins CO


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Re: Is 24 mA too much for PIC I/O pin if it has a duty cyle? - Scott Lee - May 6 14:15:00 2004

This is not answering your question but pointing something out that
you probably didn't realise...

Remember that sources have internal impedance too. (And in this case
it is likely dynamic and changes with the amount of current being
drawn by the sink.)

Thus, I wouldn't expect the PIC to source the max rated current at
the supply voltage. I would expect that as the load increases the
actual voltage you see at the IO pin will drop.

In other words, you might find that your current calculations are
wrong because you won't be seeing 2.76V across the load resistor.

--Scott

--- In , "ransom peek" <ransom.peek@c...>
wrote:
> Using a PIC 16F688, the maximum current for an I/O pin is 20 mA.
Is it
> ok to exceed this if the drain has a duty cycle as follows: > I/O pin feeds 120 ohm 5% resistor (114 ohms at low end of
tolerance).
> The resistor is part of a 6 bit binary ladder generating audio
direct
> feed to headphones (6 resistors to six I/O pins, then tied
together -
> this pin has the lowest value resistor, so I am concerned). The
ladder
> has a 10 uf series cap (dc block) and then to a 10K pot to ground.
> Center of pot to headphones.
>
> 3 volt circuit with 2N5819 schotky barrier diode, so new batteries
give
> (1.53 volts x 2) - 0.3 volt drop = 2.76 volts max.
>
> 2.76 volts / 114 ohms = 24.2 mA !
>
> But in generating the audio sine wave the pin has a duty cycle.
And,
> the headphones add some resistance, maybe 16 ohms.
>
> At 300 Hz, with approx 40 points for each sine wave, this pin is on
for
> approx .002 second, then off .001 sec, then on again .002 sec then
off
> .001 sec, etc. So, about 67% duty cycle.
>
> Is there some risk that the PIC will fail? >
> Ransom Peek
> Fort Collins CO




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Re: Is 24 mA too much for PIC I/O pin if it has a duty cyle? - Phil - May 6 16:40:00 2004

I agree with scott and have seen this with 3.2V battery powered PICs.
I was trying to power a 2.3V LED from a PIC pin. Given a 0.7 voltage
drop from the PIC, I should have gotten 2.5V, full current and
brightness from the LED with no dropping resistor. yes, I was
somewhat over voltage but was PWMing the LED, same theory as you. I
was actually seeing, iirc, something less than 2V and the entire PIC
circuit pulled around 18 mA. Whether this is enough for your circuit
to work is hard to say. I wound up using a driver transistor for the
LED and it worked quite well.

My guess is that you'll probably be ok - current-wise. since the
chip is pretty cheap, why dont you just try it? set up a stress test
and see if the chip overheats or burns out. I believe the 20mA limit
is because of heating and thus a duty cycle that brings the average
below 20mA will probably be ok. Obviously, we aren't talking about
huge current spikes. --- In , "Scott Lee" <midl_man@y...> wrote:
> This is not answering your question but pointing something out that
> you probably didn't realise...
>
> Remember that sources have internal impedance too. (And in this
case
> it is likely dynamic and changes with the amount of current being
> drawn by the sink.)
>
> Thus, I wouldn't expect the PIC to source the max rated current at
> the supply voltage. I would expect that as the load increases the
> actual voltage you see at the IO pin will drop.
>
> In other words, you might find that your current calculations are
> wrong because you won't be seeing 2.76V across the load resistor.
>
> --Scott
>
> --- In , "ransom peek" <ransom.peek@c...>
> wrote:
> > Using a PIC 16F688, the maximum current for an I/O pin is 20 mA.
> Is it
> > ok to exceed this if the drain has a duty cycle as follows:
> >
> >
> > I/O pin feeds 120 ohm 5% resistor (114 ohms at low end of
> tolerance).
> > The resistor is part of a 6 bit binary ladder generating audio
> direct
> > feed to headphones (6 resistors to six I/O pins, then tied
> together -
> > this pin has the lowest value resistor, so I am concerned). The
> ladder
> > has a 10 uf series cap (dc block) and then to a 10K pot to ground.
> > Center of pot to headphones.
> >
> > 3 volt circuit with 2N5819 schotky barrier diode, so new
batteries
> give
> > (1.53 volts x 2) - 0.3 volt drop = 2.76 volts max.
> >
> > 2.76 volts / 114 ohms = 24.2 mA !
> >
> > But in generating the audio sine wave the pin has a duty cycle.
> And,
> > the headphones add some resistance, maybe 16 ohms.
> >
> > At 300 Hz, with approx 40 points for each sine wave, this pin is
on
> for
> > approx .002 second, then off .001 sec, then on again .002 sec
then
> off
> > .001 sec, etc. So, about 67% duty cycle.
> >
> > Is there some risk that the PIC will fail?
> >
> >
> >
> > Ransom Peek
> > Fort Collins CO


______________________________
Stellaris® MCU Family: New Parts, New Package, New Price.


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Phil's Driving Transistor - Is 24 mA too much for PIC I/O pin if it has a duty cyle? - ransom peek - May 7 10:21:00 2004

Phil – your experience of needing to use a transistor to drive the LEDs is key.  Since I have 0.3V diode drop and battery voltage varies from 1.53V(new) to 1.30V(old), I also will need a transistor to drive each LED.

 

 

Voltage could be:

 

     (1.53V x 2) - 0.3V (diode) - 0.7V (PIC I/O drop) = 2.06V

or  (1.30V x 2) - 0.3V - 0.7V = 1.6V

 

 

Should I use a SO2222A or MMBT3904 (surface mount) NPN transistor? 

Is this how to do it?

  

                            Collector---- +1.6V - 2.06V

                            | 

PIC---/\/\/\/\/\---Base

           R1             |

                            Emitter----/\/\/\/\/----LED-----GND

 

What is value for R1?

I’ll need three transistors but can’t find any quad smt transistor packs in the mouser catalog.  Do they exist?

 

 

Ransom

 

 

 

 

-----Original Message-----
From: Phil [mailto:p...@yahoo.com]
Sent: Thursday, May 06, 2004 3:40 PM
To: p...@yahoogroups.com
Subject: [piclist] Re: Is 24 mA too much for PIC I/O pin if it has a duty cyle?

 

I agree with scott and have seen this with 3.2V battery powered PICs.
I was trying to power a 2.3V LED from a PIC pin.  Given a 0.7 voltage
drop from the PIC, I should have gotten 2.5V, full current and
brightness from the LED with no dropping resistor.  yes, I was
somewhat over voltage but was PWMing the LED, same theory as you. I
was actually seeing, iirc, something less than 2V and the entire PIC
circuit pulled around 18 mA.  Whether this is enough for your circuit
to work is hard to say.  I wound up using a driver transistor for the
LED and it worked quite well.

My guess is that you'll probably be ok - current-wise.  since the
chip is pretty cheap, why dont you just try it?  set up a stress test
and see if the chip overheats or burns out.  I believe the 20mA limit
is because of heating and thus a duty cycle that brings the average
below 20mA will probably be ok.  Obviously, we aren't talking about
huge current spikes. --- In p...@yahoogroups.com, "Scott Lee" <midl_man@y...> wrote:
> This is not answering your question but pointing something out that
> you probably didn't realise...
>
> Remember that sources have internal impedance too.  (And in this
case
> it is likely dynamic and changes with the amount of current being
> drawn by the sink.)
>
> Thus, I wouldn't expect the PIC to source the max rated current at
> the supply voltage.  I would expect that as the load increases the
> actual voltage you see at the IO pin will drop. 
>
> In other words, you might find that your current calculations are
> wrong because you won't be seeing 2.76V across the load resistor.
>
> --Scott
>
> --- In p...@yahoogroups.com, "ransom peek" <ransom.peek@c...>
> wrote:
> > Using a PIC 16F688, the maximum current for an I/O pin is 20 mA. 
> Is it
> > ok to exceed this if the drain has a duty cycle as follows:
> >
> >
> > I/O pin feeds 120 ohm 5% resistor (114 ohms at low end of
> tolerance).
> > The resistor is part of a 6 bit binary ladder generating audio
> direct
> > feed to headphones (6 resistors to six I/O pins, then tied
> together  -
> > this pin has the lowest value resistor, so I am concerned).  The
> ladder
> > has a 10 uf series cap (dc block) and then to a 10K pot to ground.
> > Center of pot to headphones.
> >
> > 3 volt circuit with 2N5819 schotky barrier diode, so new
batteries
> give
> > (1.53 volts x 2) - 0.3 volt drop  =  2.76 volts max.
> >
> > 2.76 volts / 114 ohms = 24.2 mA !
> >
> > But in generating the audio sine wave the pin has a duty cycle. 
> And,
> > the headphones add some resistance, maybe 16 ohms.
> >
> > At 300 Hz, with approx 40 points for each sine wave, this pin is
on
> for
> > approx .002 second, then off .001 sec, then on again .002 sec
then
> off
> > .001 sec, etc.  So, about 67% duty cycle.
> >
> > Is there some risk that the PIC will fail?
> >
> >
> >
> > Ransom Peek
> > Fort Collins CO



to unsubscribe, go to http://www.yahoogroups.com and follow the instructions






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Re: Phil's Driving Transistor - Is 24 mA too much for PIC I/O pin if it has a duty cyle? - Phil - May 7 13:19:00 2004

I used an mmbt3904 but have used 2n2222 and 2n4401. they will all
work just fine. You dont need R1. I posted a snippet of the
schematic to the photos section. I show a 12 ohm dropping resistor
but eliminated it from the final design. Total draw is slightly more
than 20 mA at full power but I'm using an 80% PWM duty cycle for full
brightness and it pulls around 18mA. well within limits.

--- In , "ransom peek" <ransom.peek@c...>
wrote:
> Phil - your experience of needing to use a transistor to drive the
LEDs
> is key. Since I have 0.3V diode drop and battery voltage varies
from
> 1.53V(new) to 1.30V(old), I also will need a transistor to drive
each
> LED. > Voltage could be: >
> (1.53V x 2) - 0.3V (diode) - 0.7V (PIC I/O drop) = 2.06V
>
> or (1.30V x 2) - 0.3V - 0.7V = 1.6V > Should I use a SO2222A or MMBT3904 (surface mount) NPN transistor?
>
> Is this how to do it? >
> Collector---- +1.6V - 2.06V
>
> |
>
> PIC---/\/\/\/\/\---Base
>
> R1 |
>
> Emitter----/\/\/\/\/----LED-----GND >
> What is value for R1?
>
> I'll need three transistors but can't find any quad smt transistor
packs
> in the mouser catalog. Do they exist? > Ransom >
> -----Original Message-----
> From: Phil [mailto:phil1960us@y...]
> Sent: Thursday, May 06, 2004 3:40 PM
> To:
> Subject: [piclist] Re: Is 24 mA too much for PIC I/O pin if it has a
> duty cyle? >
> I agree with scott and have seen this with 3.2V battery powered
PICs.
> I was trying to power a 2.3V LED from a PIC pin. Given a 0.7
voltage
> drop from the PIC, I should have gotten 2.5V, full current and
> brightness from the LED with no dropping resistor. yes, I was
> somewhat over voltage but was PWMing the LED, same theory as you. I
> was actually seeing, iirc, something less than 2V and the entire
PIC
> circuit pulled around 18 mA. Whether this is enough for your
circuit
> to work is hard to say. I wound up using a driver transistor for
the
> LED and it worked quite well.
>
> My guess is that you'll probably be ok - current-wise. since the
> chip is pretty cheap, why dont you just try it? set up a stress
test
> and see if the chip overheats or burns out. I believe the 20mA
limit
> is because of heating and thus a duty cycle that brings the average
> below 20mA will probably be ok. Obviously, we aren't talking about
> huge current spikes. > --- In , "Scott Lee" <midl_man@y...> wrote:
> > This is not answering your question but pointing something out
that
> > you probably didn't realise...
> >
> > Remember that sources have internal impedance too. (And in this
> case
> > it is likely dynamic and changes with the amount of current being
> > drawn by the sink.)
> >
> > Thus, I wouldn't expect the PIC to source the max rated current
at
> > the supply voltage. I would expect that as the load increases
the
> > actual voltage you see at the IO pin will drop.
> >
> > In other words, you might find that your current calculations are
> > wrong because you won't be seeing 2.76V across the load resistor.
> >
> > --Scott
> >
> > --- In , "ransom peek" <ransom.peek@c...>
> > wrote:
> > > Using a PIC 16F688, the maximum current for an I/O pin is 20
mA.
> > Is it
> > > ok to exceed this if the drain has a duty cycle as follows:
> > >
> > >
> > > I/O pin feeds 120 ohm 5% resistor (114 ohms at low end of
> > tolerance).
> > > The resistor is part of a 6 bit binary ladder generating audio
> > direct
> > > feed to headphones (6 resistors to six I/O pins, then tied
> > together -
> > > this pin has the lowest value resistor, so I am concerned).
The
> > ladder
> > > has a 10 uf series cap (dc block) and then to a 10K pot to
ground.
> > > Center of pot to headphones.
> > >
> > > 3 volt circuit with 2N5819 schotky barrier diode, so new
> batteries
> > give
> > > (1.53 volts x 2) - 0.3 volt drop = 2.76 volts max.
> > >
> > > 2.76 volts / 114 ohms = 24.2 mA !
> > >
> > > But in generating the audio sine wave the pin has a duty
cycle.
> > And,
> > > the headphones add some resistance, maybe 16 ohms.
> > >
> > > At 300 Hz, with approx 40 points for each sine wave, this pin
is
> on
> > for
> > > approx .002 second, then off .001 sec, then on again .002 sec
> then
> > off
> > > .001 sec, etc. So, about 67% duty cycle.
> > >
> > > Is there some risk that the PIC will fail?
> > >
> > >
> > >
> > > Ransom Peek
> > > Fort Collins CO >
> to unsubscribe, go to http://www.yahoogroups.com and follow the
> instructions >
> Yahoo! Groups Sponsor >
> ADVERTISEMENT <http://rd.yahoo.com/SIG=129our7br/M=295196.4901138.6071305.3001176/D=
gr
>
oups/S=1706554205:HM/EXP=1083966028/A=2128215/R=0/SIG=10se96mf6/*http:
/c
> ompanion.yahoo.com> click here >
> <http://us.adserver.yahoo.com/l?
M=295196.4901138.6071305.3001176/D=group
> s/S=:HM/A=2128215/rand=407049941 >
> _____
>
> > Service.



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Re: Phil's Driving Transistor - Is 24 mA too much for PIC I/O pin if it has a duty cyle? - Phil - May 7 13:23:00 2004

sorry, I forgot your last question. I've look for bipolar arrays but
haven't found them. You could use a darlington array but i think they
are pretty power inefficient.



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Re: Phil's Driving Transistor - Is 24 mA too much for PIC I/O pin if it has a duty cyle? - rtstofer - May 7 13:28:00 2004


I wouldn't put the resistor and LED in the emitter. They normally
get install between Vcc and the collector and the emitter is
grounded. Otherwise the emitter voltage rises and the base voltage
has to keep rising and it can never get to Vcc - Vce (sat) + Vbe
(sat) - in other words the collector voltage minus the drop across
the transistor (collector to emitter) when saturated plus the
voltage across the base emitter junction at saturation.

You can get an estimate of R1 if you assume an emitter current of
say 20 mA and a DC current gain of say 100 (pick your own values).
You need base current of 200 uA (20 mA / 100). So, figure the
voltage across the resistor when the PIC output is logic 1 (Voh = ?)
and the transistor is saturated (Vbe = 0.7). Given that voltage and
a current of 200 uA you can determine the resistor.

Now, you might want to lower the value by half just in case the
transistor doesn't quite get to saturation.

Or you could just try 10K which is where I would start.

--- In , "ransom peek" <ransom.peek@c...>
wrote:
> Phil - your experience of needing to use a transistor to drive the
LEDs
> is key. Since I have 0.3V diode drop and battery voltage varies
from
> 1.53V(new) to 1.30V(old), I also will need a transistor to drive
each
> LED. > Voltage could be: >
> (1.53V x 2) - 0.3V (diode) - 0.7V (PIC I/O drop) = 2.06V
>
> or (1.30V x 2) - 0.3V - 0.7V = 1.6V > Should I use a SO2222A or MMBT3904 (surface mount) NPN
transistor?
>
> Is this how to do it? >
> Collector---- +1.6V - 2.06V
>
> |
>
> PIC---/\/\/\/\/\---Base
>
> R1 |
>
> Emitter----/\/\/\/\/----LED-----GND >
> What is value for R1?
>
> I'll need three transistors but can't find any quad smt transistor
packs
> in the mouser catalog. Do they exist? > Ransom >
> -----Original Message-----
> From: Phil [mailto:phil1960us@y...]
> Sent: Thursday, May 06, 2004 3:40 PM
> To:
> Subject: [piclist] Re: Is 24 mA too much for PIC I/O pin if it has
a
> duty cyle? >
> I agree with scott and have seen this with 3.2V battery powered
PICs.
> I was trying to power a 2.3V LED from a PIC pin. Given a 0.7
voltage
> drop from the PIC, I should have gotten 2.5V, full current and
> brightness from the LED with no dropping resistor. yes, I was
> somewhat over voltage but was PWMing the LED, same theory as you.
I
> was actually seeing, iirc, something less than 2V and the entire
PIC
> circuit pulled around 18 mA. Whether this is enough for your
circuit
> to work is hard to say. I wound up using a driver transistor for
the
> LED and it worked quite well.
>
> My guess is that you'll probably be ok - current-wise. since the
> chip is pretty cheap, why dont you just try it? set up a stress
test
> and see if the chip overheats or burns out. I believe the 20mA
limit
> is because of heating and thus a duty cycle that brings the
average
> below 20mA will probably be ok. Obviously, we aren't talking
about
> huge current spikes. > --- In , "Scott Lee" <midl_man@y...> wrote:
> > This is not answering your question but pointing something out
that
> > you probably didn't realise...
> >
> > Remember that sources have internal impedance too. (And in this
> case
> > it is likely dynamic and changes with the amount of current
being
> > drawn by the sink.)
> >
> > Thus, I wouldn't expect the PIC to source the max rated current
at
> > the supply voltage. I would expect that as the load increases
the
> > actual voltage you see at the IO pin will drop.
> >
> > In other words, you might find that your current calculations
are
> > wrong because you won't be seeing 2.76V across the load resistor.
> >
> > --Scott
> >
> > --- In , "ransom peek" <ransom.peek@c...>
> > wrote:
> > > Using a PIC 16F688, the maximum current for an I/O pin is 20
mA.
> > Is it
> > > ok to exceed this if the drain has a duty cycle as follows:
> > >
> > >
> > > I/O pin feeds 120 ohm 5% resistor (114 ohms at low end of
> > tolerance).
> > > The resistor is part of a 6 bit binary ladder generating audio
> > direct
> > > feed to headphones (6 resistors to six I/O pins, then tied
> > together -
> > > this pin has the lowest value resistor, so I am concerned).
The
> > ladder
> > > has a 10 uf series cap (dc block) and then to a 10K pot to
ground.
> > > Center of pot to headphones.
> > >
> > > 3 volt circuit with 2N5819 schotky barrier diode, so new
> batteries
> > give
> > > (1.53 volts x 2) - 0.3 volt drop = 2.76 volts max.
> > >
> > > 2.76 volts / 114 ohms = 24.2 mA !
> > >
> > > But in generating the audio sine wave the pin has a duty
cycle.
> > And,
> > > the headphones add some resistance, maybe 16 ohms.
> > >
> > > At 300 Hz, with approx 40 points for each sine wave, this pin
is
> on
> > for
> > > approx .002 second, then off .001 sec, then on again .002 sec
> then
> > off
> > > .001 sec, etc. So, about 67% duty cycle.
> > >
> > > Is there some risk that the PIC will fail?
> > >
> > >
> > >
> > > Ransom Peek
> > > Fort Collins CO >
> to unsubscribe, go to http://www.yahoogroups.com and follow the
> instructions >
> Yahoo! Groups Sponsor >
> ADVERTISEMENT <http://rd.yahoo.com/SIG=129our7br/M=295196.4901138.6071305.3001176/D
=gr
>
oups/S=1706554205:HM/EXP=1083966028/A=2128215/R=0/SIG=10se96mf6/*http
:/c
> ompanion.yahoo.com> click here >
> <http://us.adserver.yahoo.com/l?
M=295196.4901138.6071305.3001176/D=group
> s/S=:HM/A=2128215/rand=407049941 >
> _____
>
> > Service.




(You need to be a member of piclist -- send a blank email to piclist-subscribe@yahoogroups.com )

Re: Phil's Driving Transistor - Is 24 mA too much for PIC I/O pin if it has a duty cyle? - Phil - May 7 13:50:00 2004

I dunno, seems to work well for me. I have no trouble getting a
couple 100 mV over the LED's Vf. Your approach will, of course, also
work.

--- In , "rtstofer" <rstofer@p...> wrote:
>
> I wouldn't put the resistor and LED in the emitter. They normally
> get install between Vcc and the collector and the emitter is
> grounded. Otherwise the emitter voltage rises and the base voltage
> has to keep rising and it can never get to Vcc - Vce (sat) + Vbe
> (sat) - in other words the collector voltage minus the drop across
> the transistor (collector to emitter) when saturated plus the
> voltage across the base emitter junction at saturation.
>
> You can get an estimate of R1 if you assume an emitter current of
> say 20 mA and a DC current gain of say 100 (pick your own values).
> You need base current of 200 uA (20 mA / 100). So, figure the
> voltage across the resistor when the PIC output is logic 1 (Voh
= ?)
> and the transistor is saturated (Vbe = 0.7). Given that voltage
and
> a current of 200 uA you can determine the resistor.
>
> Now, you might want to lower the value by half just in case the
> transistor doesn't quite get to saturation.
>
> Or you could just try 10K which is where I would start.
>
> --- In , "ransom peek" <ransom.peek@c...>
> wrote:
> > Phil - your experience of needing to use a transistor to drive
the
> LEDs
> > is key. Since I have 0.3V diode drop and battery voltage varies
> from
> > 1.53V(new) to 1.30V(old), I also will need a transistor to drive
> each
> > LED.
> >
> >
> >
> >
> >
> > Voltage could be:
> >
> >
> >
> > (1.53V x 2) - 0.3V (diode) - 0.7V (PIC I/O drop) = 2.06V
> >
> > or (1.30V x 2) - 0.3V - 0.7V = 1.6V
> >
> >
> >
> >
> >
> > Should I use a SO2222A or MMBT3904 (surface mount) NPN
> transistor?
> >
> > Is this how to do it?
> >
> >
> >
> > Collector---- +1.6V - 2.06V
> >
> > |
> >
> > PIC---/\/\/\/\/\---Base
> >
> > R1 |
> >
> > Emitter----/\/\/\/\/----LED-----GND
> >
> >
> >
> > What is value for R1?
> >
> > I'll need three transistors but can't find any quad smt
transistor
> packs
> > in the mouser catalog. Do they exist?
> >
> >
> >
> >
> >
> > Ransom
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > -----Original Message-----
> > From: Phil [mailto:phil1960us@y...]
> > Sent: Thursday, May 06, 2004 3:40 PM
> > To:
> > Subject: [piclist] Re: Is 24 mA too much for PIC I/O pin if it
has
> a
> > duty cyle?
> >
> >
> >
> > I agree with scott and have seen this with 3.2V battery powered
> PICs.
> > I was trying to power a 2.3V LED from a PIC pin. Given a 0.7
> voltage
> > drop from the PIC, I should have gotten 2.5V, full current and
> > brightness from the LED with no dropping resistor. yes, I was
> > somewhat over voltage but was PWMing the LED, same theory as you.
> I
> > was actually seeing, iirc, something less than 2V and the entire
> PIC
> > circuit pulled around 18 mA. Whether this is enough for your
> circuit
> > to work is hard to say. I wound up using a driver transistor for
> the
> > LED and it worked quite well.
> >
> > My guess is that you'll probably be ok - current-wise. since the
> > chip is pretty cheap, why dont you just try it? set up a stress
> test
> > and see if the chip overheats or burns out. I believe the 20mA
> limit
> > is because of heating and thus a duty cycle that brings the
> average
> > below 20mA will probably be ok. Obviously, we aren't talking
> about
> > huge current spikes.
> >
> >
> > --- In , "Scott Lee" <midl_man@y...> wrote:
> > > This is not answering your question but pointing something out
> that
> > > you probably didn't realise...
> > >
> > > Remember that sources have internal impedance too. (And in
this
> > case
> > > it is likely dynamic and changes with the amount of current
> being
> > > drawn by the sink.)
> > >
> > > Thus, I wouldn't expect the PIC to source the max rated current
> at
> > > the supply voltage. I would expect that as the load increases
> the
> > > actual voltage you see at the IO pin will drop.
> > >
> > > In other words, you might find that your current calculations
> are
> > > wrong because you won't be seeing 2.76V across the load
resistor.
> > >
> > > --Scott
> > >
> > > --- In , "ransom peek"
<ransom.peek@c...>
> > > wrote:
> > > > Using a PIC 16F688, the maximum current for an I/O pin is 20
> mA.
> > > Is it
> > > > ok to exceed this if the drain has a duty cycle as follows:
> > > >
> > > >
> > > > I/O pin feeds 120 ohm 5% resistor (114 ohms at low end of
> > > tolerance).
> > > > The resistor is part of a 6 bit binary ladder generating
audio
> > > direct
> > > > feed to headphones (6 resistors to six I/O pins, then tied
> > > together -
> > > > this pin has the lowest value resistor, so I am concerned).
> The
> > > ladder
> > > > has a 10 uf series cap (dc block) and then to a 10K pot to
> ground.
> > > > Center of pot to headphones.
> > > >
> > > > 3 volt circuit with 2N5819 schotky barrier diode, so new
> > batteries
> > > give
> > > > (1.53 volts x 2) - 0.3 volt drop = 2.76 volts max.
> > > >
> > > > 2.76 volts / 114 ohms = 24.2 mA !
> > > >
> > > > But in generating the audio sine wave the pin has a duty
> cycle.
> > > And,
> > > > the headphones add some resistance, maybe 16 ohms.
> > > >
> > > > At 300 Hz, with approx 40 points for each sine wave, this pin
> is
> > on
> > > for
> > > > approx .002 second, then off .001 sec, then on again .002 sec
> > then
> > > off
> > > > .001 sec, etc. So, about 67% duty cycle.
> > > >
> > > > Is there some risk that the PIC will fail?
> > > >
> > > >
> > > >
> > > > Ransom Peek
> > > > Fort Collins CO
> >
> >
> >
> > to unsubscribe, go to http://www.yahoogroups.com and follow the
> > instructions
> >
> >
> >
> >
> >
> >
> > Yahoo! Groups Sponsor
> >
> >
> >
> > ADVERTISEMENT
> >
> >
>
<http://rd.yahoo.com/SIG=129our7br/M=295196.4901138.6071305.3001176/D
> =gr
> >
>
oups/S=1706554205:HM/EXP=1083966028/A=2128215/R=0/SIG=10se96mf6/*http
> :/c
> > ompanion.yahoo.com> click here
> >
> >
> >
> > <http://us.adserver.yahoo.com/l?
> M=295196.4901138.6071305.3001176/D=group
> > s/S=:HM/A=2128215/rand=407049941>
> >
> >
> >
> > _____
> >
> > > Service.




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RSTOFER - it wont work as you say... Is 24 mA too much for PIC I/O pin if it has a duty cyle? - ransom peek - May 7 14:45:00 2004

Your suggestion of 10K series resistor to drive npn transistor fo rlighting LED (VDD to LED, to resistor to collector, and emitter to ground is the perfect solution for the single LED.  But also have a dual LED with common ground, so need another way.  Would a PNP resistor work for this?

 

Ransom

 

-----Original Message-----
From: rtstofer [mailto:r...@pacbell.net]
Sent: Friday, May 07, 2004 12:28 PM
To: p...@yahoogroups.com
Subject: [piclist] Re: Phil's Driving Transistor - Is 24 mA too much for PIC I/O pin if it has a duty cyle?

 


I wouldn't put the resistor and LED in the emitter.  They normally
get install between Vcc and the collector and the emitter is
grounded.  Otherwise the emitter voltage rises and the base voltage
has to keep rising and it can never get to Vcc - Vce (sat) + Vbe
(sat) - in other words the collector voltage minus the drop across
the transistor (collector to emitter) when saturated plus the
voltage across the base emitter junction at saturation.

You can get an estimate of R1 if you assume an emitter current of
say 20 mA and a DC current gain of say 100 (pick your own values). 
You need base current of 200 uA (20 mA / 100).  So, figure the
voltage across the resistor when the PIC output is logic 1 (Voh = ?)
and the transistor is saturated (Vbe = 0.7).  Given that voltage and
a current of 200 uA you can determine the resistor.

Now, you might want to lower the value by half just in case the
transistor doesn't quite get to saturation.

Or you could just try 10K which is where I would start.

--- In p...@yahoogroups.com, "ransom peek" <ransom.peek@c...>
wrote:
> Phil - your experience of needing to use a transistor to drive the
LEDs
> is key.  Since I have 0.3V diode drop and battery voltage varies
from
> 1.53V(new) to 1.30V(old), I also will need a transistor to drive
each
> LED.
>
> 
>
> 
>
> Voltage could be:
>
> 
>
>      (1.53V x 2) - 0.3V (diode) - 0.7V (PIC I/O drop) = 2.06V
>
> or  (1.30V x 2) - 0.3V - 0.7V = 1.6V
>
> 
>
> 
>
> Should I use a SO2222A or MMBT3904 (surface mount) NPN
transistor? 
>
> Is this how to do it?
>
>   
>
>                             Collector---- +1.6V - 2.06V
>
>                             | 
>
> PIC---/\/\/\/\/\---Base
>
>            R1             |
>
>                             Emitter----/\/\/\/\/----LED-----GND
>
> 
>
> What is value for R1?
>
> I'll need three transistors but can't find any quad smt transistor
packs
> in the mouser catalog.  Do they exist?
>
> 
>
> 
>
> Ransom
>
> 
>
> 
>
> 
>
> 
>
> -----Original Message-----
> From: Phil [mailto:phil1960us@y...]
> Sent: Thursday, May 06, 2004 3:40 PM
> To: p...@yahoogroups.com
> Subject: [piclist] Re: Is 24 mA too much for PIC I/O pin if it has
a
> duty cyle?
>
> 
>
> I agree with scott and have seen this with 3.2V battery powered
PICs.
> I was trying to power a 2.3V LED from a PIC pin.  Given a 0.7
voltage
> drop from the PIC, I should have gotten 2.5V, full current and
> brightness from the LED with no dropping resistor.  yes, I was
> somewhat over voltage but was PWMing the LED, same theory as you.
I
> was actually seeing, iirc, something less than 2V and the entire
PIC
> circuit pulled around 18 mA.  Whether this is enough for your
circuit
> to work is hard to say.  I wound up using a driver transistor for
the
> LED and it worked quite well.
>
> My guess is that you'll probably be ok - current-wise.  since the
> chip is pretty cheap, why dont you just try it?  set up a stress
test
> and see if the chip overheats or burns out.  I believe the 20mA
limit
> is because of heating and thus a duty cycle that brings the
average
> below 20mA will probably be ok.  Obviously, we aren't talking
about
> huge current spikes.
>
>
> --- In p...@yahoogroups.com, "Scott Lee" <midl_man@y...> wrote:
> > This is not answering your question but pointing something out
that
> > you probably didn't realise...
> >
> > Remember that sources have internal impedance too.  (And in this
> case
> > it is likely dynamic and changes with the amount of current
being
> > drawn by the sink.)
> >
> > Thus, I wouldn't expect the PIC to source the max rated current
at
> > the supply voltage.  I would expect that as the load increases
the
> > actual voltage you see at the IO pin will drop. 
> >
> > In other words, you might find that your current calculations
are
> > wrong because you won't be seeing 2.76V across the load resistor.
> >
> > --Scott
> >
> > --- In p...@yahoogroups.com, "ransom peek" <ransom.peek@c...>
> > wrote:
> > > Using a PIC 16F688, the maximum current for an I/O pin is 20
mA. 
> > Is it
> > > ok to exceed this if the drain has a duty cycle as follows:
> > >
> > >
> > > I/O pin feeds 120 ohm 5% resistor (114 ohms at low end of
> > tolerance).
> > > The resistor is part of a 6 bit binary ladder generating audio
> > direct
> > > feed to headphones (6 resistors to six I/O pins, then tied
> > together  -
> > > this pin has the lowest value resistor, so I am concerned). 
The
> > ladder
> > > has a 10 uf series cap (dc block) and then to a 10K pot to
ground.
> > > Center of pot to headphones.
> > >
> > > 3 volt circuit with 2N5819 schotky barrier diode, so new
> batteries
> > give
> > > (1.53 volts x 2) - 0.3 volt drop  =  2.76 volts max.
> > >
> > > 2.76 volts / 114 ohms = 24.2 mA !
> > >
> > > But in generating the audio sine wave the pin has a duty
cycle. 
> > And,
> > > the headphones add some resistance, maybe 16 ohms.
> > >
> > > At 300 Hz, with approx 40 points for each sine wave, this pin
is
> on
> > for
> > > approx .002 second, then off .001 sec, then on again .002 sec
> then
> > off
> > > .001 sec, etc.  So, about 67% duty cycle.
> > >
> > > Is there some risk that the PIC will fail?
> > >
> > >
> > >
> > > Ransom Peek
> > > Fort Collins CO
>
>
>
> to unsubscribe, go to http://www.yahoogroups.com and follow the
> instructions
>
>
>
>
>
>
> Yahoo! Groups Sponsor
>
>
>
> ADVERTISEMENT
> 
>
<http://rd.yahoo.com/SIG=129our7br/M=295196.4901138.6071305.3001176/D
=gr
>
oups/S=1706554205:HM/EXP=1083966028/A=2128215/R=0/SIG=10se96mf6/*http
:/c
> ompanion.yahoo.com> click here
>
>
> 
> <http://us.adserver.yahoo.com/l?
M=295196.4901138.6071305.3001176/D=group
> s/S=:HM/A=2128215/rand=407049941>
>
> 
>
>   _____ 
>
> >  Service.



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Re: RSTOFER - it wont work as you say... Is 24 mA too much for PIC I/O pin if it has a duty cyle? - rtstofer - May 7 15:31:00 2004


Yes, if you have a common ground you need to use PNP transistors.
Connect the collector to Vdd and the resistor and one side of LED to
the emitter.

Now your problem for base current and R1 calculations are the same.
What is the PIC output at logic 0 and what is the voltage drop
across the base-emitter junction at saturation? Combined these are
probably 0.9V. So, Vdd - 0.9 dropped across the resistor at say 200
uA. Again, perhaps half this much resistance will be appropriate -
still headed toward 10k. This will connect between the base and the
PIC output port.

The problem with this design is keeping the transistor turned off.
With the NPN design the logic 0 output would never forward bias the
transistor. With the PNP design the logic 1 voltage may not be high
enough to keep the transistor turned off. An ugly solution is to
run another 10k resistor from the base to Vdd. This will keep the
transistor off at logic 1 but it changes the current flow through R1
at logic 0. So, you need to recalculate R1 with more current (
(Vdd - Vol) / 10k) from the pull-up resistor. This pull-up resistor
could also be much larger - perhaps 100k and this will have less
effect on R1.

Another way to do the pull-up is to add the 10k resistor at the PIC
port. This won't influence the calculation of R1 and the PIC output
is looking at a 5k ohm load. About 1 mA if Vdd is 5V so no real
concern about loading.

You just need to be sure the transistor is not conducting when the
PIC output bit is logic 1. So, the voltage at the emitter wants to
be 0. You also should be certain the transistor is saturated at
logic 0. The voltage across the collector and emitter should be
close to 0.2V but check the Vce(sat) spec for your transistor. --- In , "ransom peek" <ransom.peek@c...>
wrote:
> Your suggestion of 10K series resistor to drive npn transistor fo
> rlighting LED (VDD to LED, to resistor to collector, and emitter to
> ground is the perfect solution for the single LED. But also have
a dual
> LED with common ground, so need another way. Would a PNP resistor
work
> for this? >
> Ransom >
> -----Original Message-----
> From: rtstofer [mailto:rstofer@p...]
> Sent: Friday, May 07, 2004 12:28 PM
> To:
> Subject: [piclist] Re: Phil's Driving Transistor - Is 24 mA too
much for
> PIC I/O pin if it has a duty cyle? >
>
> I wouldn't put the resistor and LED in the emitter. They normally
> get install between Vcc and the collector and the emitter is
> grounded. Otherwise the emitter voltage rises and the base
voltage
> has to keep rising and it can never get to Vcc - Vce (sat) + Vbe
> (sat) - in other words the collector voltage minus the drop across
> the transistor (collector to emitter) when saturated plus the
> voltage across the base emitter junction at saturation.
>
> You can get an estimate of R1 if you assume an emitter current of
> say 20 mA and a DC current gain of say 100 (pick your own
values).
> You need base current of 200 uA (20 mA / 100). So, figure the
> voltage across the resistor when the PIC output is logic 1 (Voh
= ?)
> and the transistor is saturated (Vbe = 0.7). Given that voltage
and
> a current of 200 uA you can determine the resistor.
>
> Now, you might want to lower the value by half just in case the
> transistor doesn't quite get to saturation.
>
> Or you could just try 10K which is where I would start.
>
> --- In , "ransom peek" <ransom.peek@c...>
> wrote:
> > Phil - your experience of needing to use a transistor to drive
the
> LEDs
> > is key. Since I have 0.3V diode drop and battery voltage varies
> from
> > 1.53V(new) to 1.30V(old), I also will need a transistor to drive
> each
> > LED.
> >
> >
> >
> >
> >
> > Voltage could be:
> >
> >
> >
> > (1.53V x 2) - 0.3V (diode) - 0.7V (PIC I/O drop) = 2.06V
> >
> > or (1.30V x 2) - 0.3V - 0.7V = 1.6V
> >
> >
> >
> >
> >
> > Should I use a SO2222A or MMBT3904 (surface mount) NPN
> transistor?
> >
> > Is this how to do it?
> >
> >
> >
> > Collector---- +1.6V - 2.06V
> >
> > |
> >
> > PIC---/\/\/\/\/\---Base
> >
> > R1 |
> >
> > Emitter----/\/\/\/\/----LED-----GND
> >
> >
> >
> > What is value for R1?
> >
> > I'll need three transistors but can't find any quad smt
transistor
> packs
> > in the mouser catalog. Do they exist?
> >
> >
> >
> >
> >
> > Ransom
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > -----Original Message-----
> > From: Phil [mailto:phil1960us@y...]
> > Sent: Thursday, May 06, 2004 3:40 PM
> > To:
> > Subject: [piclist] Re: Is 24 mA too much for PIC I/O pin if it
has
> a
> > duty cyle?
> >
> >
> >
> > I agree with scott and have seen this with 3.2V battery powered
> PICs.
> > I was trying to power a 2.3V LED from a PIC pin. Given a 0.7
> voltage
> > drop from the PIC, I should have gotten 2.5V, full current and
> > brightness from the LED with no dropping resistor. yes, I was
> > somewhat over voltage but was PWMing the LED, same theory as
you.
> I
> > was actually seeing, iirc, something less than 2V and the entire
> PIC
> > circuit pulled around 18 mA. Whether this is enough for your
> circuit
> > to work is hard to say. I wound up using a driver transistor
for
> the
> > LED and it worked quite well.
> >
> > My guess is that you'll probably be ok - current-wise. since
the
> > chip is pretty cheap, why dont you just try it? set up a stress
> test
> > and see if the chip overheats or burns out. I believe the 20mA
> limit
> > is because of heating and thus a duty cycle that brings the
> average
> > below 20mA will probably be ok. Obviously, we aren't talking
> about
> > huge current spikes.
> >
> >
> > --- In , "Scott Lee" <midl_man@y...>
wrote:
> > > This is not answering your question but pointing something out
> that
> > > you probably didn't realise...
> > >
> > > Remember that sources have internal impedance too. (And in
this
> > case
> > > it is likely dynamic and changes with the amount of current
> being
> > > drawn by the sink.)
> > >
> > > Thus, I wouldn't expect the PIC to source the max rated
current
> at
> > > the supply voltage. I would expect that as the load increases
> the
> > > actual voltage you see at the IO pin will drop.
> > >
> > > In other words, you might find that your current calculations
> are
> > > wrong because you won't be seeing 2.76V across the load
resistor.
> > >
> > > --Scott
> > >
> > > --- In , "ransom peek"
<ransom.peek@c...>
> > > wrote:
> > > > Using a PIC 16F688, the maximum current for an I/O pin is 20
> mA.
> > > Is it
> > > > ok to exceed this if the drain has a duty cycle as follows:
> > > >
> > > >
> > > > I/O pin feeds 120 ohm 5% resistor (114 ohms at low end of
> > > tolerance).
> > > > The resistor is part of a 6 bit binary ladder generating
audio
> > > direct
> > > > feed to headphones (6 resistors to six I/O pins, then tied
> > > together -
> > > > this pin has the lowest value resistor, so I am concerned).
> The
> > > ladder
> > > > has a 10 uf series cap (dc block) and then to a 10K pot to
> ground.
> > > > Center of pot to headphones.
> > > >
> > > > 3 volt circuit with 2N5819 schotky barrier diode, so new
> > batteries
> > > give
> > > > (1.53 volts x 2) - 0.3 volt drop = 2.76 volts max.
> > > >
> > > > 2.76 volts / 114 ohms = 24.2 mA !
> > > >
> > > > But in generating the audio sine wave the pin has a duty
> cycle.
> > > And,
> > > > the headphones add some resistance, maybe 16 ohms.
> > > >
> > > > At 300 Hz, with approx 40 points for each sine wave, this
pin
> is
> > on
> > > for
> > > > approx .002 second, then off .001 sec, then on again .002
sec
> > then
> > > off
> > > > .001 sec, etc. So, about 67% duty cycle.
> > > >
> > > > Is there some risk that the PIC will fail?
> > > >
> > > >
> > > >
> > > > Ransom Peek
> > > > Fort Collins CO
> >
> >
> >
> > to unsubscribe, go to http://www.yahoogroups.com and follow the
> > instructions
> >
> >
> >
> >
> >
> >
> > Yahoo! Groups Sponsor
> >
> >
> >
> > ADVERTISEMENT
> >
> >
>
<http://rd.yahoo.com/SIG=129our7br/M=295196.4901138.6071305.3001176/D
> =gr
> >
>
oups/S=1706554205:HM/EXP=1083966028/A=2128215/R=0/SIG=10se96mf6/*http
> :/c
> > ompanion.yahoo.com> click here
> >
> >
> >
> > <http://us.adserver.yahoo.com/l?
> M=295196.4901138.6071305.3001176/D=group
> > s/S=:HM/A=2128215/rand=407049941>
> >
> >
> >
> > _____
> >
> > > Service. >
> to unsubscribe, go to http://www.yahoogroups.com and follow the
> instructions >
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> ADVERTISEMENT <http://rd.yahoo.com/SIG=1297p4f5k/M=295196.4901138.6071305.3001176/D
=gr
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:/c
> ompanion.yahoo.com> click here >
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Re: Re: Phil's Driving Transistor - Is 24 mA too much for PIC I/O pin if it has a duty cyle? - Vasile Surducan - May 8 2:37:00 2004



Phil,
maybe you should read first a LED datasheet... :)
Or understand better how can be driven a NPN transistor.
Not always when is working it means that is ok.
BTW, which LED have a dropout of 100mV at 20mA ? I need 100K of those... best regards,
Vasile
http://surducan.netfirms.com On Fri, 7 May 2004, Phil wrote:

> I dunno, seems to work well for me. I have no trouble getting a
> couple 100 mV over the LED's Vf. Your approach will, of course, also
> work.
>
> --- In , "rtstofer" <rstofer@p...> wrote:
> >
> > I wouldn't put the resistor and LED in the emitter. They normally
> > get install between Vcc and the collector and the emitter is
> > grounded. Otherwise the emitter voltage rises and the base voltage
> > has to keep rising and it can never get to Vcc - Vce (sat) + Vbe
> > (sat) - in other words the collector voltage minus the drop across
> > the transistor (collector to emitter) when saturated plus the
> > voltage across the base emitter junction at saturation.
> >
> > You can get an estimate of R1 if you assume an emitter current of
> > say 20 mA and a DC current gain of say 100 (pick your own values).
> > You need base current of 200 uA (20 mA / 100). So, figure the
> > voltage across the resistor when the PIC output is logic 1 (Voh
> = ?)
> > and the transistor is saturated (Vbe = 0.7). Given that voltage
> and
> > a current of 200 uA you can determine the resistor.
> >
> > Now, you might want to lower the value by half just in case the
> > transistor doesn't quite get to saturation.
> >
> > Or you could just try 10K which is where I would start.
> >
> > --- In , "ransom peek" <ransom.peek@c...>
> > wrote:
> > > Phil - your experience of needing to use a transistor to drive
> the
> > LEDs
> > > is key. Since I have 0.3V diode drop and battery voltage varies
> > from
> > > 1.53V(new) to 1.30V(old), I also will need a transistor to drive
> > each
> > > LED.
> > >
> > >
> > >
> > >
> > >
> > > Voltage could be:
> > >
> > >
> > >
> > > (1.53V x 2) - 0.3V (diode) - 0.7V (PIC I/O drop) = 2.06V
> > >
> > > or (1.30V x 2) - 0.3V - 0.7V = 1.6V
> > >
> > >
> > >
> > >
> > >
> > > Should I use a SO2222A or MMBT3904 (surface mount) NPN
> > transistor?
> > >
> > > Is this how to do it?
> > >
> > >
> > >
> > > Collector---- +1.6V - 2.06V
> > >
> > > |
> > >
> > > PIC---/\/\/\/\/\---Base
> > >
> > > R1 |
> > >
> > > Emitter----/\/\/\/\/----LED-----GND
> > >
> > >
> > >
> > > What is value for R1?
> > >
> > > I'll need three transistors but can't find any quad smt
> transistor
> > packs
> > > in the mouser catalog. Do they exist?
> > >
> > >
> > >
> > >
> > >
> > > Ransom
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > -----Original Message-----
> > > From: Phil [mailto:phil1960us@y...]
> > > Sent: Thursday, May 06, 2004 3:40 PM
> > > To:
> > > Subject: [piclist] Re: Is 24 mA too much for PIC I/O pin if it
> has
> > a
> > > duty cyle?
> > >
> > >
> > >
> > > I agree with scott and have seen this with 3.2V battery powered
> > PICs.
> > > I was trying to power a 2.3V LED from a PIC pin. Given a 0.7
> > voltage
> > > drop from the PIC, I should have gotten 2.5V, full current and
> > > brightness from the LED with no dropping resistor. yes, I was
> > > somewhat over voltage but was PWMing the LED, same theory as you.
> > I
> > > was actually seeing, iirc, something less than 2V and the entire
> > PIC
> > > circuit pulled around 18 mA. Whether this is enough for your
> > circuit
> > > to work is hard to say. I wound up using a driver transistor for
> > the
> > > LED and it worked quite well.
> > >
> > > My guess is that you'll probably be ok - current-wise. since the
> > > chip is pretty cheap, why dont you just try it? set up a stress
> > test
> > > and see if the chip overheats or burns out. I believe the 20mA
> > limit
> > > is because of heating and thus a duty cycle that brings the
> > average
> > > below 20mA will probably be ok. Obviously, we aren't talking
> > about
> > > huge current spikes.
> > >
> > >
> > > --- In , "Scott Lee" <midl_man@y...> wrote:
> > > > This is not answering your question but pointing something out
> > that
> > > > you probably didn't realise...
> > > >
> > > > Remember that sources have internal impedance too. (And in
> this
> > > case
> > > > it is likely dynamic and changes with the amount of current
> > being
> > > > drawn by the sink.)
> > > >
> > > > Thus, I wouldn't expect the PIC to source the max rated current
> > at
> > > > the supply voltage. I would expect that as the load increases
> > the
> > > > actual voltage you see at the IO pin will drop.
> > > >
> > > > In other words, you might find that your current calculations
> > are
> > > > wrong because you won't be seeing 2.76V across the load
> resistor.
> > > >
> > > > --Scott
> > > >
> > > > --- In , "ransom peek"
> <ransom.peek@c...>
> > > > wrote:
> > > > > Using a PIC 16F688, the maximum current for an I/O pin is 20
> > mA.
> > > > Is it
> > > > > ok to exceed this if the drain has a duty cycle as follows:
> > > > >
> > > > >
> > > > > I/O pin feeds 120 ohm 5% resistor (114 ohms at low end of
> > > > tolerance).
> > > > > The resistor is part of a 6 bit binary ladder generating
> audio
> > > > direct
> > > > > feed to headphones (6 resistors to six I/O pins, then tied
> > > > together -
> > > > > this pin has the lowest value resistor, so I am concerned).
> > The
> > > > ladder
> > > > > has a 10 uf series cap (dc block) and then to a 10K pot to
> > ground.
> > > > > Center of pot to headphones.
> > > > >
> > > > > 3 volt circuit with 2N5819 schotky barrier diode, so new
> > > batteries
> > > > give
> > > > > (1.53 volts x 2) - 0.3 volt drop = 2.76 volts max.
> > > > >
> > > > > 2.76 volts / 114 ohms = 24.2 mA !
> > > > >
> > > > > But in generating the audio sine wave the pin has a duty
> > cycle.
> > > > And,
> > > > > the headphones add some resistance, maybe 16 ohms.
> > > > >
> > > > > At 300 Hz, with approx 40 points for each sine wave, this pin
> > is
> > > on
> > > > for
> > > > > approx .002 second, then off .001 sec, then on again .002 sec
> > > then
> > > > off
> > > > > .001 sec, etc. So, about 67% duty cycle.
> > > > >
> > > > > Is there some risk that the PIC will fail?
> > > > >
> > > > >
> > > > >
> > > > > Ransom Peek
> > > > > Fort Collins CO
> > >
> > >
> > >
> > > to unsubscribe, go to http://www.yahoogroups.com and follow the
> > > instructions
> > >
> > >
> > >
> > >
> > >
> > >
> > > Yahoo! Groups Sponsor
> > >
> > >
> > >
> > > ADVERTISEMENT
> > >
> > >
> >
> <http://rd.yahoo.com/SIG=129our7br/M=295196.4901138.6071305.3001176/D
> > =gr
> > >
> >
> oups/S=1706554205:HM/EXP=1083966028/A=2128215/R=0/SIG=10se96mf6/*http
> > :/c
> > > ompanion.yahoo.com> click here
> > >
> > >
> > >
> > > <http://us.adserver.yahoo.com/l?
> > M=295196.4901138.6071305.3001176/D=group
> > > s/S=:HM/A=2128215/rand=407049941>
> > >
> > >
> > >
> > > _____
> > >
> > > > Service. >
>
> to unsubscribe, go to http://www.yahoogroups.com and follow the instructions
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Re: Phil's Driving Transistor - Is 24 mA too much for PIC I/O pin if it has a duty cyle? - Phil - May 8 3:41:00 2004

Well, I'm using a classic emitter follower. Check out horowitz/hill
chapter 2. The LED I was using has a 2.3V Vf (from the data sheet
you think I didn't read), the transistor has about 0.7 V drop.
Running with 2 fresh akaline cells I get 3.2 - (2.3+.7) = 0.3. i.e.
a couple hundred mV OVER the LED's Vf as I specifically said in my
post (which I dont think you read carefully).

--- In , Vasile Surducan <vasile@s...> wrote: > Phil,
> maybe you should read first a LED datasheet... :)
> Or understand better how can be driven a NPN transistor.
> Not always when is working it means that is ok.
> BTW, which LED have a dropout of 100mV at 20mA ? I need 100K of
those... > best regards,
> Vasile
> http://surducan.netfirms.com > On Fri, 7 May 2004, Phil wrote:
>
> > I dunno, seems to work well for me. I have no trouble getting a
> > couple 100 mV over the LED's Vf. Your approach will, of course,
also
> > work.
> >
> > --- In , "rtstofer" <rstofer@p...> wrote:
> > >
> > > I wouldn't put the resistor and LED in the emitter. They
normally
> > > get install between Vcc and the collector and the emitter is
> > > grounded. Otherwise the emitter voltage rises and the base
voltage
> > > has to keep rising and it can never get to Vcc - Vce (sat) + Vbe
> > > (sat) - in other words the collector voltage minus the drop
across
> > > the transistor (collector to emitter) when saturated plus the
> > > voltage across the base emitter junction at saturation.
> > >
> > > You can get an estimate of R1 if you assume an emitter current
of
> > > say 20 mA and a DC current gain of say 100 (pick your own
values).
> > > You need base current of 200 uA (20 mA / 100). So, figure the
> > > voltage across the resistor when the PIC output is logic 1 (Voh
> > = ?)
> > > and the transistor is saturated (Vbe = 0.7). Given that voltage
> > and
> > > a current of 200 uA you can determine the resistor.
> > >
> > > Now, you might want to lower the value by half just in case the
> > > transistor doesn't quite get to saturation.
> > >
> > > Or you could just try 10K which is where I would start.
> > >
> > > --- In , "ransom peek" <ransom.peek@c...>
> > > wrote:
> > > > Phil - your experience of needing to use a transistor to drive
> > the
> > > LEDs
> > > > is key. Since I have 0.3V diode drop and battery voltage
varies
> > > from
> > > > 1.53V(new) to 1.30V(old), I also will need a transistor to
drive
> > > each
> > > > LED.
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > Voltage could be:
> > > >
> > > >
> > > >
> > > > (1.53V x 2) - 0.3V (diode) - 0.7V (PIC I/O drop) = 2.06V
> > > >
> > > > or (1.30V x 2) - 0.3V - 0.7V = 1.6V
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > Should I use a SO2222A or MMBT3904 (surface mount) NPN
> > > transistor?
> > > >
> > > > Is this how to do it?
> > > >
> > > >
> > > >
> > > > Collector---- +1.6V - 2.06V
> > > >
> > > > |
> > > >
> > > > PIC---/\/\/\/\/\---Base
> > > >
> > > > R1 |
> > > >
> > > > Emitter----/\/\/\/\/----LED-----
GND
> > > >
> > > >
> > > >
> > > > What is value for R1?
> > > >
> > > > I'll need three transistors but can't find any quad smt
> > transistor
> > > packs
> > > > in the mouser catalog. Do they exist?
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > Ransom
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > -----Original Message-----
> > > > From: Phil [mailto:phil1960us@y...]
> > > > Sent: Thursday, May 06, 2004 3:40 PM
> > > > To:
> > > > Subject: [piclist] Re: Is 24 mA too much for PIC I/O pin if it
> > has
> > > a
> > > > duty cyle?
> > > >
> > > >
> > > >
> > > > I agree with scott and have seen this with 3.2V battery
powered
> > > PICs.
> > > > I was trying to power a 2.3V LED from a PIC pin. Given a 0.7
> > > voltage
> > > > drop from the PIC, I should have gotten 2.5V, full current and
> > > > brightness from the LED with no dropping resistor. yes, I was
> > > > somewhat over voltage but was PWMing the LED, same theory as
you.
> > > I
> > > > was actually seeing, iirc, something less than 2V and the
entire
> > > PIC
> > > > circuit pulled around 18 mA. Whether this is enough for your
> > > circuit
> > > > to work is hard to say. I wound up using a driver transistor
for
> > > the
> > > > LED and it worked quite well.
> > > >
> > > > My guess is that you'll probably be ok - current-wise. since
the
> > > > chip is pretty cheap, why dont you just try it? set up a
stress
> > > test
> > > > and see if the chip overheats or burns out. I believe the
20mA
> > > limit
> > > > is because of heating and thus a duty cycle that brings the
> > > average
> > > > below 20mA will probably be ok. Obviously, we aren't talking
> > > about
> > > > huge current spikes.
> > > >
> > > >
> > > > --- In , "Scott Lee" <midl_man@y...>
wrote:
> > > > > This is not answering your question but pointing something
out
> > > that
> > > > > you probably didn't realise...
> > > > >
> > > > > Remember that sources have internal impedance too. (And in
> > this
> > > > case
> > > > > it is likely dynamic and changes with the amount of current
> > > being
> > > > > drawn by the sink.)
> > > > >
> > > > > Thus, I wouldn't expect the PIC to source the max rated
current
> > > at
> > > > > the supply voltage. I would expect that as the load
increases
> > > the
> > > > > actual voltage you see at the IO pin will drop.
> > > > >
> > > > > In other words, you might find that your current
calculations
> > > are
> > > > > wrong because you won't be seeing 2.76V across the load
> > resistor.
> > > > >
> > > > > --Scott
> > > > >
> > > > > --- In , "ransom peek"
> > <ransom.peek@c...>
> > > > > wrote:
> > > > > > Using a PIC 16F688, the maximum current for an I/O pin is
20
> > > mA.
> > > > > Is it
> > > > > > ok to exceed this if the drain has a duty cycle as
follows:
> > > > > >
> > > > > >
> > > > > > I/O pin feeds 120 ohm 5% resistor (114 ohms at low end of
> > > > > tolerance).
> > > > > > The resistor is part of a 6 bit binary ladder generating
> > audio
> > > > > direct
> > > > > > feed to headphones (6 resistors to six I/O pins, then tied
> > > > > together -
> > > > > > this pin has the lowest value resistor, so I am
concerned).
> > > The
> > > > > ladder
> > > > > > has a 10 uf series cap (dc block) and then to a 10K pot to
> > > ground.
> > > > > > Center of pot to headphones.
> > > > > >
> > > > > > 3 volt circuit with 2N5819 schotky barrier diode, so new
> > > > batteries
> > > > > give
> > > > > > (1.53 volts x 2) - 0.3 volt drop = 2.76 volts max.
> > > > > >
> > > > > > 2.76 volts / 114 ohms = 24.2 mA !
> > > > > >
> > > > > > But in generating the audio sine wave the pin has a duty
> > > cycle.
> > > > > And,
> > > > > > the headphones add some resistance, maybe 16 ohms.
> > > > > >
> > > > > > At 300 Hz, with approx 40 points for each sine wave, this
pin
> > > is
> > > > on
> > > > > for
> > > > > > approx .002 second, then off .001 sec, then on again .002
sec
> > > > then
> > > > > off
> > > > > > .001 sec, etc. So, about 67% duty cycle.
> > > > > >
> > > > > > Is there some risk that the PIC will fail?
> > > > > >
> > > > > >
> > > > > >
> > > > > > Ransom Peek
> > > > > > Fort Collins CO
> > > >
> > > >
> > > >
> > > > to unsubscribe, go to http://www.yahoogroups.com and follow
the
> > > > instructions
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > Yahoo! Groups Sponsor
> > > >
> > > >
> > > >
> > > > ADVERTISEMENT
> > > >
> > > >
> > >
> >
<http://rd.yahoo.com/SIG=129our7br/M=295196.4901138.6071305.3001176/D
> > > =gr
> > > >
> > >
> >
oups/S=1706554205:HM/EXP=1083966028/A=2128215/R=0/SIG=10se96mf6/*http
> > > :/c
> > > > ompanion.yahoo.com> click here
> > > >
> > > >
> > > >
> > > > <http://us.adserver.yahoo.com/l?
> > > M=295196.4901138.6071305.3001176/D=group
> > > > s/S=:HM/A=2128215/rand=407049941>
> > > >
> > > >
> > > >
> > > > _____
> > > >
> > > > > Service.
> >
> >
> >
> >
> > to unsubscribe, go to http://www.yahoogroups.com and follow the
instructions
> > Yahoo! Groups Links
> >
> >
> >
> >
>


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Re: Re: Phil's Driving Transistor - Is 24 mA too much for PIC I/O pin if it has a duty cyle? - Vasile Surducan - May 8 4:14:00 2004

Yaap, you've right I didn't :)
However driving leds with emiter followers should be the last valid
option.
Just because of the transistor Vbe voltage drop, Vce saturation and the
variable load voltage which is changing the transistor driving
caracteristic. Else is ok.

best,
Vasile

On Sat, 8 May 2004, Phil wrote:

> Well, I'm using a classic emitter follower. Check out horowitz/hill
> chapter 2. The LED I was using has a 2.3V Vf (from the data sheet
> you think I didn't read), the transistor has about 0.7 V drop.
> Running with 2 fresh akaline cells I get 3.2 - (2.3+.7) = 0.3. i.e.
> a couple hundred mV OVER the LED's Vf as I specifically said in my
> post (which I dont think you read carefully).
>
> --- In , Vasile Surducan <vasile@s...> wrote:
> >
> >
> > Phil,
> > maybe you should read first a LED datasheet... :)
> > Or understand better how can be driven a NPN transistor.
> > Not always when is working it means that is ok.
> > BTW, which LED have a dropout of 100mV at 20mA ? I need 100K of
> those...
> >
> >
> > best regards,
> > Vasile
> > http://surducan.netfirms.com
> >
> >
> > On Fri, 7 May 2004, Phil wrote:
> >
> > > I dunno, seems to work well for me. I have no trouble getting a
> > > couple 100 mV over the LED's Vf. Your approach will, of course,
> also
> > > work.
> > >
> > > --- In , "rtstofer" <rstofer@p...> wrote:
> > > >
> > > > I wouldn't put the resistor and LED in the emitter. They
> normally
> > > > get install between Vcc and the collector and the emitter is
> > > > grounded. Otherwise the emitter voltage rises and the base
> voltage
> > > > has to keep rising and it can never get to Vcc - Vce (sat) + Vbe
> > > > (sat) - in other words the collector voltage minus the drop
> across
> > > > the transistor (collector to emitter) when saturated plus the
> > > > voltage across the base emitter junction at saturation.
> > > >
> > > > You can get an estimate of R1 if you assume an emitter current
> of
> > > > say 20 mA and a DC current gain of say 100 (pick your own
> values).
> > > > You need base current of 200 uA (20 mA / 100). So, figure the
> > > > voltage across the resistor when the PIC output is logic 1 (Voh
> > > = ?)
> > > > and the transistor is saturated (Vbe = 0.7). Given that voltage
> > > and
> > > > a current of 200 uA you can determine the resistor.
> > > >
> > > > Now, you might want to lower the value by half just in case the
> > > > transistor doesn't quite get to saturation.
> > > >
> > > > Or you could just try 10K which is where I would start.
> > > >
> > > > --- In , "ransom peek" <ransom.peek@c...>
> > > > wrote:
> > > > > Phil - your experience of needing to use a transistor to drive
> > > the
> > > > LEDs
> > > > > is key. Since I have 0.3V diode drop and battery voltage
> varies
> > > > from
> > > > > 1.53V(new) to 1.30V(old), I also will need a transistor to
> drive
> > > > each
> > > > > LED.
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > Voltage could be:
> > > > >
> > > > >
> > > > >
> > > > > (1.53V x 2) - 0.3V (diode) - 0.7V (PIC I/O drop) = 2.06V
> > > > >
> > > > > or (1.30V x 2) - 0.3V - 0.7V = 1.6V
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > Should I use a SO2222A or MMBT3904 (surface mount) NPN
> > > > transistor?
> > > > >
> > > > > Is this how to do it?
> > > > >
> > > > >
> > > > >
> > > > > Collector---- +1.6V - 2.06V
> > > > >
> > > > > |
> > > > >
> > > > > PIC---/\/\/\/\/\---Base
> > > > >
> > > > > R1 |
> > > > >
> > > > > Emitter----/\/\/\/\/----LED-----
> GND
> > > > >
> > > > >
> > > > >
> > > > > What is value for R1?
> > > > >
> > > > > I'll need three transistors but can't find any quad smt
> > > transistor
> > > > packs
> > > > > in the mouser catalog. Do they exist?
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > Ransom
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > -----Original Message-----
> > > > > From: Phil [mailto:phil1960us@y...]
> > > > > Sent: Thursday, May 06, 2004 3:40 PM
> > > > > To:
> > > > > Subject: [piclist] Re: Is 24 mA too much for PIC I/O pin if it
> > > has
> > > > a
> > > > > duty cyle?
> > > > >
> > > > >
> > > > >
> > > > > I agree with scott and have seen this with 3.2V battery
> powered
> > > > PICs.
> > > > > I was trying to power a 2.3V LED from a PIC pin. Given a 0.7
> > > > voltage
> > > > > drop from the PIC, I should have gotten 2.5V, full current and
> > > > > brightness from the LED with no dropping resistor. yes, I was
> > > > > somewhat over voltage but was PWMing the LED, same theory as
> you.
> > > > I
> > > > > was actually seeing, iirc, something less than 2V and the
> entire
> > > > PIC
> > > > > circuit pulled around 18 mA. Whether this is enough for your
> > > > circuit
> > > > > to work is hard to say. I wound up using a driver transistor
> for
> > > > the
> > > > > LED and it worked quite well.
> > > > >
> > > > > My guess is that you'll probably be ok - current-wise. since
> the
> > > > > chip is pretty cheap, why dont you just try it? set up a
> stress
> > > > test
> > > > > and see if the chip overheats or burns out. I believe the
> 20mA
> > > > limit
> > > > > is because of heating and thus a duty cycle that brings the
> > > > average
> > > > > below 20mA will probably be ok. Obviously, we aren't talking
> > > > about
> > > > > huge current spikes.
> > > > >
> > > > >
> > > > > --- In , "Scott Lee" <midl_man@y...>
> wrote:
> > > > > > This is not answering your question but pointing something
> out
> > > > that
> > > > > > you probably didn't realise...
> > > > > >
> > > > > > Remember that sources have internal impedance too. (And in
> > > this
> > > > > case
> > > > > > it is likely dynamic and changes with the amount of current
> > > > being
> > > > > > drawn by the sink.)
> > > > > >
> > > > > > Thus, I wouldn't expect the PIC to source the max rated
> current
> > > > at
> > > > > > the supply voltage. I would expect that as the load
> increases
> > > > the
> > > > > > actual voltage you see at the IO pin will drop.
> > > > > >
> > > > > > In other words, you might find that your current
> calculations
> > > > are
> > > > > > wrong because you won't be seeing 2.76V across the load
> > > resistor.
> > > > > >
> > > > > > --Scott
> > > > > >
> > > > > > --- In , "ransom peek"
> > > <ransom.peek@c...>
> > > > > > wrote:
> > > > > > > Using a PIC 16F688, the maximum current for an I/O pin is
> 20
> > > > mA.
> > > > > > Is it
> > > > > > > ok to exceed this if the drain has a duty cycle as
> follows:
> > > > > > >
> > > > > > >
> > > > > > > I/O pin feeds 120 ohm 5% resistor (114 ohms at low end of
> > > > > > tolerance).
> > > > > > > The resistor is part of a 6 bit binary ladder generating
> > > audio
> > > > > > direct
> > > > > > > feed to headphones (6 resistors to six I/O pins, then tied
> > > > > > together -
> > > > > > > this pin has the lowest value resistor, so I am
> concerned).
> > > > The
> > > > > > ladder
> > > > > > > has a 10 uf series cap (dc block) and then to a 10K pot to
> > > > ground.
> > > > > > > Center of pot to headphones.
> > > > > > >
> > > > > > > 3 volt circuit with 2N5819 schotky barrier diode, so new
> > > > > batteries
> > > > > > give
> > > > > > > (1.53 volts x 2) - 0.3 volt drop = 2.76 volts max.
> > > > > > >
> > > > > > > 2.76 volts / 114 ohms = 24.2 mA !
> > > > > > >
> > > > > > > But in generating the audio sine wave the pin has a duty
> > > > cycle.
> > > > > > And,
> > > > > > > the headphones add some resistance, maybe 16 ohms.
> > > > > > >
> > > > > > > At 300 Hz, with approx 40 points for each sine wave, this
> pin
> > > > is
> > > > > on
> > > > > > for
> > > > > > > approx .002 second, then off .001 sec, then on again .002
> sec
> > > > > then
> > > > > > off
> > > > > > > .001 sec, etc. So, about 67% duty cycle.
> > > > > > >
> > > > > > > Is there some risk that the PIC will fail?
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > Ransom Peek
> > > > > > > Fort Collins CO
> > > > >
> > > > >
> > > > >
> > > > > to unsubscribe, go to http://www.yahoogroups.com and follow
> the
> > > > > instructions
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> > > > >
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> > > > >
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> > > > >
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> > > >
> > >
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> > > > =gr
> > > > >
> > > >
> > >
> oups/S=1706554205:HM/EXP=1083966028/A=2128215/R=0/SIG=10se96mf6/*http
> > > > :/c
> > > > > ompanion.yahoo.com> click here
> > > > >
> > > > >
> > > > >
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> > > > M=295196.4901138.6071305.3001176/D=group
> > > > > s/S=:HM/A=2128215/rand=407049941>
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> > > > >
> > > > >
> > > > > _____
> > > > >
> > > > > > Service.
> > >
> > >
> > >
> > >
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> > >
> > >
> > >
> > >
> > >
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Re: Phil's Driving Transistor - Is 24 mA too much for PIC I/O pin if it has a duty cyle? - Phil - May 8 5:04:00 2004

ok, what bad effects will I see with this? shorter battery life?
shorter component life? I've drained several batteries with this and
get pretty much the lifespan you would expect.

--- In , Vasile Surducan <vasile@s...> wrote: >
> Yaap, you've right I didn't :)
> However driving leds with emiter followers should be the last valid
> option.
> Just because of the transistor Vbe voltage drop, Vce saturation and
the
> variable load voltage which is changing the transistor driving
> caracteristic. Else is ok.
>
> best,
> Vasile
>
> On Sat, 8 May 2004, Phil wrote:
>
> > Well, I'm using a classic emitter follower. Check out
horowitz/hill
> > chapter 2. The LED I was using has a 2.3V Vf (from the data
sheet
> > you think I didn't read), the transistor has about 0.7 V drop.
> > Running with 2 fresh akaline cells I get 3.2 - (2.3+.7) = 0.3.
i.e.
> > a couple hundred mV OVER the LED's Vf as I specifically said in my
> > post (which I dont think you read carefully).
> >
> > --- In , Vasile Surducan <vasile@s...>
wrote:
> > >
> > >
> > > Phil,
> > > maybe you should read first a LED datasheet... :)
> > > Or understand better how can be driven a NPN transistor.
> > > Not always when is working it means that is ok.
> > > BTW, which LED have a dropout of 100mV at 20mA ? I need 100K of
> > those...
> > >
> > >
> > > best regards,
> > > Vasile
> > > http://surducan.netfirms.com
> > >
> > >
> > > On Fri, 7 May 2004, Phil wrote:
> > >
> > > > I dunno, seems to work well for me. I have no trouble
getting a
> > > > couple 100 mV over the LED's Vf. Your approach will, of
course,
> > also
> > > > work.
> > > >
> > > > --- In , "rtstofer" <rstofer@p...>
wrote:
> > > > >
> > > > > I wouldn't put the resistor and LED in the emitter. They
> > normally
> > > > > get install between Vcc and the collector and the emitter is
> > > > > grounded. Otherwise the emitter voltage rises and the base
> > voltage
> > > > > has to keep rising and it can never get to Vcc - Vce (sat)
+ Vbe
> > > > > (sat) - in other words the collector voltage minus the drop
> > across
> > > > > the transistor (collector to emitter) when saturated plus
the
> > > > > voltage across the base emitter junction at saturation.
> > > > >
> > > > > You can get an estimate of R1 if you assume an emitter
current
> > of
> > > > > say 20 mA and a DC current gain of say 100 (pick your own
> > values).
> > > > > You need base current of 200 uA (20 mA / 100). So, figure
the
> > > > > voltage across the resistor when the PIC output is logic 1
(Voh
> > > > = ?)
> > > > > and the transistor is saturated (Vbe = 0.7). Given that
voltage
> > > > and
> > > > > a current of 200 uA you can determine the resistor.
> > > > >
> > > > > Now, you might want to lower the value by half just in case
the
> > > > > transistor doesn't quite get to saturation.
> > > > >
> > > > > Or you could just try 10K which is where I would start.
> > > > >
> > > > > --- In , "ransom peek"
<ransom.peek@c...>
> > > > > wrote:
> > > > > > Phil - your experience of needing to use a transistor to
drive
> > > > the
> > > > > LEDs
> > > > > > is key. Since I have 0.3V diode drop and battery voltage
> > varies
> > > > > from
> > > > > > 1.53V(new) to 1.30V(old), I also will need a transistor to
> > drive
> > > > > each
> > > > > > LED.
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > > Voltage could be:
> > > > > >
> > > > > >
> > > > > >
> > > > > > (1.53V x 2) - 0.3V (diode) - 0.7V (PIC I/O drop) =
2.06V
> > > > > >
> > > > > > or (1.30V x 2) - 0.3V - 0.7V = 1.6V
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > > Should I use a SO2222A or MMBT3904 (surface mount) NPN
> > > > > transistor?
> > > > > >
> > > > > > Is this how to do it?
> > > > > >
> > > > > >
> > > > > >
> > > > > > Collector---- +1.6V - 2.06V
> > > > > >
> > > > > > |
> > > > > >
> > > > > > PIC---/\/\/\/\/\---Base
> > > > > >
> > > > > > R1 |
> > > > > >
> > > > > > Emitter----/\/\/\/\/----LED---
--
> > GND
> > > > > >
> > > > > >
> > > > > >
> > > > > > What is value for R1?
> > > > > >
> > > > > > I'll need three transistors but can't find any quad smt
> > > > transistor
> > > > > packs
> > > > > > in the mouser catalog. Do they exist?
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > > Ransom
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > > -----Original Message-----
> > > > > > From: Phil [mailto:phil1960us@y...]
> > > > > > Sent: Thursday, May 06, 2004 3:40 PM
> > > > > > To:
> > > > > > Subject: [piclist] Re: Is 24 mA too much for PIC I/O pin
if it
> > > > has
> > > > > a
> > > > > > duty cyle?
> > > > > >
> > > > > >
> > > > > >
> > > > > > I agree with scott and have seen this with 3.2V battery
> > powered
> > > > > PICs.
> > > > > > I was trying to power a 2.3V LED from a PIC pin. Given a
0.7
> > > > > voltage
> > > > > > drop from the PIC, I should have gotten 2.5V, full
current and
> > > > > > brightness from the LED with no dropping resistor. yes,
I was
> > > > > > somewhat over voltage but was PWMing the LED, same theory
as
> > you.
> > > > > I
> > > > > > was actually seeing, iirc, something less than 2V and the
> > entire
> > > > > PIC
> > > > > > circuit pulled around 18 mA. Whether this is enough for
your
> > > > > circuit
> > > > > > to work is hard to say. I wound up using a driver
transistor
> > for
> > > > > the
> > > > > > LED and it worked quite well.
> > > > > >
> > > > > > My guess is that you'll probably be ok - current-wise.
since
> > the
> > > > > > chip is pretty cheap, why dont you just try it? set up a
> > stress
> > > > > test
> > > > > > and see if the chip overheats or burns out. I believe the
> > 20mA
> > > > > limit
> > > > > > is because of heating and thus a duty cycle that brings
the
> > > > > average
> > > > > > below 20mA will probably be ok. Obviously, we aren't
talking
> > > > > about
> > > > > > huge current spikes.
> > > > > >
> > > > > >
> > > > > > --- In , "Scott Lee"
<midl_man@y...>
> > wrote:
> > > > > > > This is not answering your question but pointing
something
> > out
> > > > > that
> > > > > > > you probably didn't realise...
> > > > > > >
> > > > > > > Remember that sources have internal impedance too.
(And in
> > > > this
> > > > > > case
> > > > > > > it is likely dynamic and changes with the amount of
current
> > > > > being
> > > > > > > drawn by the sink.)
> > > > > > >
> > > > > > > Thus, I wouldn't expect the PIC to source the max rated
> > current
> > > > > at
> > > > > > > the supply voltage. I would expect that as the load
> > increases
> > > > > the
> > > > > > > actual voltage you see at the IO pin will drop.
> > > > > > >
> > > > > > > In other words, you might find that your current
> > calculations
> > > > > are
> > > > > > > wrong because you won't be seeing 2.76V across the load
> > > > resistor.
> > > > > > >
> > > > > > > --Scott
> > > > > > >
> > > > > > > --- In , "ransom peek"
> > > > <ransom.peek@c...>
> > > > > > > wrote:
> > > > > > > > Using a PIC 16F688, the maximum current for an I/O
pin is
> > 20
> > > > > mA.
> > > > > > > Is it
> > > > > > > > ok to exceed this if the drain has a duty cycle as
> > follows:
> > > > > > > >
> > > > > > > >
> > > > > > > > I/O pin feeds 120 ohm 5% resistor (114 ohms at low
end of
> > > > > > > tolerance).
> > > > > > > > The resistor is part of a 6 bit binary ladder
generating
> > > > audio
> > > > > > > direct
> > > > > > > > feed to headphones (6 resistors to six I/O pins, then
tied
> > > > > > > together -
> > > > > > > > this pin has the lowest value resistor, so I am
> > concerned).
> > > > > The
> > > > > > > ladder
> > > > > > > > has a 10 uf series cap (dc block) and then to a 10K
pot to
> > > > > ground.
> > > > > > > > Center of pot to headphones.
> > > > > > > >
> > > > > > > > 3 volt circuit with 2N5819 schotky barrier diode, so
new
> > > > > > batteries
> > > > > > > give
> > > > > > > > (1.53 volts x 2) - 0.3 volt drop = 2.76 volts max.
> > > > > > > >
> > > > > > > > 2.76 volts / 114 ohms = 24.2 mA !
> > > > > > > >
> > > > > > > > But in generating the audio sine wave the pin has a
duty
> > > > > cycle.
> > > > > > > And,
> > > > > > > > the headphones add some resistance, maybe 16 ohms.
> > > > > > > >
> > > > > > > > At 300 Hz, with approx 40 points for each sine wave,
this
> > pin
> > > > > is
> > > > > > on
> > > > > > > for
> > > > > > > > approx .002 second, then off .001 sec, then on
again .002
> > sec
> > > > > > then
> > > > > > > off
> > > > > > > > .001 sec, etc. So, about 67% duty cycle.
> > > > > > > >
> > > > > > > > Is there some risk that the PIC will fail?
> > > > > > > >
> > > > > > > >
> > > > > > > >
> > > > > > > > Ransom Peek
> > > > > > > > Fort Collins CO
> > > > > >
> > > > > >
> > > > > >
> > > > > > to unsubscribe, go to http://www.yahoogroups.com and
follow
> > the
> > > > > > instructions
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
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> > > > > > Yahoo! Groups Sponsor
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> >
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> > > > > =gr
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> > > > >
> > > >
> >
oups/S=1706554205:HM/EXP=1083966028/A=2128215/R=0/SIG=10se96mf6/*http
> > > > > :/c
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> > > > > >
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> > > > > >
> > > > > >
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