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'[PIC]: Using 2 outputs in parallel for higher curr'
2002\11\06@180319 by Daniel Rubin

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I know the output current limit per pin on the PICs is 20 ma... what if
I combine 2 output pins to supply 40 ma of current?  Can it be done and
more importantly should it be done?  If so would you just tie the 2
pins together in the circuit?

Many thanks
- Dan

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2002\11\06@180942 by Tony Nixon

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Daniel Rubin wrote:
>
> I know the output current limit per pin on the PICs is 20 ma... what if
> I combine 2 output pins to supply 40 ma of current?  Can it be done and
> more importantly should it be done?  If so would you just tie the 2
> pins together in the circuit?
>
> Many thanks
> - Dan



It could be done, but you run the risk of opposing logic states existing
between the pins causing a short.

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2002\11\06@190745 by Stef

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Daniel Rubin wrote:

>I know the output current limit per pin on the PICs is 20 ma... what if
>I combine 2 output pins to supply 40 ma of current?  Can it be done and
>more importantly should it be done?  If so would you just tie the 2
>pins together in the circuit?
>
If you only need current in one direction, use 2 shottkey diodes.
Stef Mientki

{Quote hidden}

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2002\11\06@192134 by Steve Ruse

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Of course you could put a diode on each output, though I don't really think
that is the best design. Using a transistor to jump the current up from a
single pin is probably the best way if you have to have more current. If its
is just something you are playing with (doesn't have to be extremely
dependable), tie them together & see what happens. If you write your code so
that those two pins should ALWAYS match, I can't imagine that you would ever
have a problem with opposing logic states.

Steve

{Original Message removed}

2002\11\06@230634 by Daniel Rubin

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The main reason for doing this is that there is no room any additional
components! I need to drive a 35-40 ma coil and I would like to use nothing
more than two pins from the PIC12F629.  There really is no room for an
additional smd transistor let alone 2 diodes.  I already have a back-emf
diode across the coil.

If I always write to GPIO making sure the paired outputs are always the
same using MOVWF (as opposed to BCF/BSF) would I have any problems?  Are
people concerned with the internals of the PIC and output pins are toggled
based on register changes or are they just concerned with user code that
might not keep the paired pins in sync?  I can certainly eliminate the later.

- Dan

At 06:19 PM 11/6/2002 -0600, you wrote:
>Of course you could put a diode on each output, though I don't really think
>that is the best design. Using a transistor to jump the current up from a
>single pin is probably the best way if you have to have more current. If its
>is just something you are playing with (doesn't have to be extremely
>dependable), tie them together & see what happens. If you write your code so
>that those two pins should ALWAYS match, I can't imagine that you would ever
>have a problem with opposing logic states.
>
>Steve
>
>{Original Message removed}

2002\11\06@233751 by Josh Koffman

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Perhaps just to be super safe, instead of toggling the pins on and off,
toggle them on and then tristate them. That way you'd be sure you
wouldn't have one up and one down.

Just an idea.

Josh
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Daniel Rubin wrote:
> If I always write to GPIO making sure the paired outputs are always the
> same using MOVWF (as opposed to BCF/BSF) would I have any problems?  Are
> people concerned with the internals of the PIC and output pins are toggled
> based on register changes or are they just concerned with user code that
> might not keep the paired pins in sync?  I can certainly eliminate the later.

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2002\11\07@014317 by thys

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I don't think it is a good idea to parallel the pins.
What about start-up?
I will at least make space for some diodes, some thing like a BAV70LT1 OR
BAW56LT, depending on sink or source. These are two diodes in a sot23 common
cathode or anode.

Thys

{Original Message removed}

2002\11\07@020142 by Daniel Rubin

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Josh,

Thanks for your input... What do you mean by tristate?  Is there another
state besides on and off?  Are you saying I do the following:

disable outputs
write W to GPIO
enable outputs

If that is what you are referring to I am not sure that is possible as this
output is PWM and the extra 2 instructions per toggle would cut my possible
rate.


- Dan

At 12:06 AM 11/7/2002 -0600, you wrote:
{Quote hidden}

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2002\11\07@023119 by Scott Dattalo

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On Fri, 8 Nov 2002, Daniel Rubin wrote:

> Josh,
>
> Thanks for your input... What do you mean by tristate?  Is there another
> state besides on and off?  Are you saying I do the following:
>
> disable outputs
> write W to GPIO
> enable outputs
>
> If that is what you are referring to I am not sure that is possible as this
> output is PWM and the extra 2 instructions per toggle would cut my possible
> rate.


Dan,

This kind of thing is done (at least I've done it) to boost the current
drive of a logic gate. You probably have a resistor in series with the
line already to limit the current. If so, replace it with two resistors
with twice the resistance. Tie one end of the resistors together (and to
your load) and tie the other ends to the two I/O pins. If the I/O's
momentarily drive in opposite states, then the current will be limited.

But, if there is no series resistor to limit the current (then I guess the
question would be how are you limiting the current?) then there's a chance
the I/O's will experience significant loading if they're driving in
opposite states. The duration of these illegal states is presumably quite
low. I.e. it only happens at the edges. I strongly doubt this will damage
the I/O drivers, since it's energy and not current that kills. In other
words, If you left the i/o's in this illegal state for a *long* time, then
you may damage them (I doubt it though - pic I/O's are notoriously
robust).

The only caveat is that you want to advoid RMW (read-modify-write)
operations on the I/O port. There's a good chance that the RMW operation
can cause a bit a error. To advoid this, add a shadow register and copy
that to the I/O port. You mention GPIO so I assume you're using the
12c508. If you were using the 18f's you could write directly to the LATch
registers.


BTW, if you're using PWM on the 12c508, check out:

http://www.dattalo.com/technical/software/pic/pwm256.txt

It was originally written for the 12-bit cores. Since first written, I've
used it on the 14-bit cores, so there may be some 14-bit dependcy that has
crept in.

---
<straying off topic (for this thread, but not for the list...)>

With the 18f's you can toggle an LED with something like:

   BTG  LATA,0     ; bit toggle a LATch bit.

Isn't that cute!

Scott

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2002\11\07@094013 by Josh Koffman

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Yes, you got it. I didn't know you were doing PWM though. As I said in
my original post, it was just an idea that struck me as I was reading
your post :)

Josh
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Daniel Rubin wrote:
{Quote hidden}

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2002\11\07@180448 by Dwayne Reid

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At 11:06 PM 11/7/02 -0500, Daniel Rubin wrote:

>If I always write to GPIO making sure the paired outputs are always the
>same using MOVWF (as opposed to BCF/BSF) would I have any problems?  Are
>people concerned with the internals of the PIC and output pins are toggled
>based on register changes or are they just concerned with user code that
>might not keep the paired pins in sync?  I can certainly eliminate the later.

This sort of thing is done on a regular basis.  So long as you ensure that
the pins are always the same as each other, you should have no
problems.  Just write GPIO all at once instead of using individual bsf/bcf
commands and you should be fine.

The easy way to check for this is to put a small value resistor in series
with either the Vss or Vdd pin of the pic and monitor the peak current with
a scope.  Disconnect the load and toggle the pins.  The peak current should
remain the same whether the pins are connected together or not.

dwayne

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2002\11\08@074928 by Steve Russell

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Call me old-fashioned, but I don't think that you should just connect
the pins together as many here seem to suggest. You should use the
resistor technique described by Scott Dattalo. Without doing so, you
have no way of knowing how the current will be divided by the two ports
and you could exceed the spec on one of them.

Steve.

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2002\11\08@085133 by Morgan Olsson

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Hej Daniel Rubin. Tack för ditt meddelande 05:06 2002-11-08 enligt nedan:
> There really is no room for an
>additional smd transistor let alone 2 diodes.  I already have a back-emf
>diode across the coil.

Remove the back-emf diode, and use a transistor in emitter follwer configuration. (that will also dissipate the back EMF)  =No more components than today.  You will get 1V drop across the tranny though.

Another idea
Hmmm have i seen a Mosfet with internal zener...?  Or select one that withstand the actual back EMF voltage.

/Morgan

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2002\11\08@085831 by Dave Tweed

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Steve Russell <raicheaEraseMEspam.....YAHOO.COM> wrote:
> source=http://www.piclist.com/postbot.asp?id=piclist\2002\11\07\023119a
>
> Call me old-fashioned, but I don't think that you should just connect
> the pins together as many here seem to suggest. You should use the
> resistor technique described by Scott Dattalo. Without doing so, you
> have no way of knowing how the current will be divided by the two ports
> and you could exceed the spec on one of them.

OK, you're old-fashioned! :-)

Seriously, the pin drivers are MOSFETs, which generally don't need ballast
resistors when operated in parallel. If one starts to warm up relative to
the other because of excess current, its resistance goes up and the current
is diverted the other way.

--------

To another poster who was wondering about "open-collector" (really
open-drain) outputs: You can simulate this, within the voltage limits
established by the protection diodes, by setting the output data
permanently to 0 and turning the tristate bit on and off.

-- Dave Tweed

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2002\11\08@101143 by Daniel Rubin

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If I use a transistor I'll have to add a resistor.  Anyone know of any
transistors (in a SMD package) that don't require a resistor between the
PIC and it's base?


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2002\11\08@102438 by Stef Mientki

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Daniel Rubin wrote:

> If I use a transistor I'll have to add a resistor.  Anyone know of any
> transistors (in a SMD package) that don't require a resistor between the
> PIC and it's base?

use a FET, it doesn't require the base resistor
Stef Mientki

>

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2002\11\08@104108 by Olin Lathrop

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> > If I use a transistor I'll have to add a resistor.  Anyone know of any
> > transistors (in a SMD package) that don't require a resistor between
the
> > PIC and it's base?
>
> use a FET, it doesn't require the base resistor

Neither does a bipolar transistor in emitter follower configuration, which
I thought the original suggestion was about.


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2002\11\08@104117 by Dave Tweed

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Daniel Rubin <EraseMEdanspamDESIGNDEVICES.COM> wrote:
> Morgan wrote:
> >Remove the back-emf diode, and use a transistor in emitter follwer
> >configuration. (that will also dissipate the back EMF)  =No more
> >components than today.  You will get 1V drop across the tranny though.
>
> If I use a transistor I'll have to add a resistor.  Anyone know of any
> transistors (in a SMD package) that don't require a resistor between the
> PIC and it's base?

Emitter follower means that the load goes between the emitter lead and
ground (assuming NPN here); no resistor in base or collector required.

When the transistor cuts off, the coil will drive the emitter below ground,
drawing current through the base and collector until the energy is
dissipated. The base current will come from the low-side pin driver MOSFET
and/or the protection diode -- you may need to make sure this current
doesn't cause any problems for the PIC.

You could also configure a PNP as a low-side switch, but a similar issue
arises when the output goes high.

-- Dave Tweed

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2002\11\08@105117 by Alan B. Pearce

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>If I use a transistor I'll have to add a resistor.  Anyone
>know of any transistors (in a SMD package) that don't
>require a resistor between the PIC and it's base?

Yes do a search for "digital transistor", typically a single npn transistor
with a base-emitter resistor and a series base resistor to the outside
world. Designed for exactly this sort of use. All this comes in a SOT-23
package IIRC.

Made by Rohm and Vishay/Telefunken among others IIRC. If you have an RS
components CD look for digital transistor, I know they have carried them in
the past.

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2002\11\09@084826 by Spehro Pefhany

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At 10:10 AM 11/9/02 -0500, you wrote:
>If I use a transistor I'll have to add a resistor.  Anyone know of any
>transistors (in a SMD package) that don't require a resistor between the
>PIC and it's base?

Yes, you could use a MOSFET or there are so-called "digital transistors"
that include base resistors or a resistor from the to emitter and
another from the base to the "base" lead. Check Panasonic, Rohm etc.
If you're in the USofA or Canada, Digikey carries the Panasonic ones.

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2002\11\09@091943 by Steve Russell

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Ok - old-fashioned, but not that old-fashioned :0)

I'm aware that paralleling MOSFETs is common in relatively high-power
applications, but at low drain currents (low=a few 10's of mA), the
temp coefficient changes sign.

Looking at the PIC12C508A spec (happened to have that one around) the
curves for current against temp do show +ve coefficient all the way
down to zero. The output devices will be pretty well matched, I
imagine, so you'll probably get away with it - even though there's no
guarantee that the current will be shared equally.

Steve.


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2002\11\09@135547 by Dave Tweed

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Steve Russell <RemoveMEraicheaspam_OUTspamKILLspamYAHOO.COM> wrote:
> source=http://www.piclist.com/postbot.asp?id=3Dpiclist\2002\11\08\085831a
>
> I'm aware that paralleling MOSFETs is common in relatively high-power
> applications, but at low drain currents (low=a few 10's of mA), the
> temp coefficient changes sign.

But remember that "high power" and "low drain currents" are relative to
the size of the MOSFET in question (current density is the key, not just
total current), and the PIC pin drivers are much smaller than the typical
discrete power MOSFET.

-- Dave Tweed

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2002\11\09@154414 by Steve Russell

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Good point - I guess that's why the PIC curves stay +ve...

Steve.


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2002\11\10@070731 by Roman Black

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Daniel Rubin wrote:
>
> If I use a transistor I'll have to add a resistor.  Anyone know of any
> transistors (in a SMD package) that don't require a resistor between the
> PIC and it's base?
>
> >Remove the back-emf diode, and use a transistor in emitter follwer
> >configuration.


Hi Daniel, using two paralleled PIC outputs
should be fine for your coil (provided you keep
the back-emf diode).

The PIC outputs are MOSFET type and will share
the current fairly well.

This is a reliable setup from a hardware view as
the main problem with outputs is that the input
latch gets set/clr due to voltages on the pin
itself, or output latches get state-change due
to large PSU glitches, but as they are tied together
the ins/outs are also tied together and the two
pins *should* always be the same state.

BUT, software could be a problem, always write to
the port as a complete byte, and be very careful
with any AND/XOR etc on that port that you don't
change one bit by mistake.

My preference would be to use SOME value of resistor
even if your current needs mean the resistors are
a low value, this will probably save your PIC when/if
you get the PSU glitches etc. In conjunction with
this I would set the pins constantly, (every few mS
if possible) so that if the PSU glitches and a pin
state gets trashed it will get fixed in a very short
time. Paranoia can be good. :o)
-Roman

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