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'More sink current'
1998\06\18@073555 by org Hager

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Greetings!

In order to drive an IR LED, I'd like to sink currents of about 100mA for
short periods of time into 16F84 output pins. As this is not possible with
a single pin, can I just short a couple of pins to get more sink current?
Or is there some danger afflicted with it (apart from accidentally
giving different logic levels out to the connected pins ;-)?

Thanx,
Georg.

1998\06\18@074215 by David BALDWIN

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Georg Hager wrote:
>
> Greetings!
>
> In order to drive an IR LED, I'd like to sink currents of about 100mA for
> short periods of time into 16F84 output pins. As this is not possible with
> a single pin, can I just short a couple of pins to get more sink current?
> Or is there some danger afflicted with it (apart from accidentally
> giving different logic levels out to the connected pins ;-)?
>
> Thanx,
> Georg.

       It takes some of you valued pic pins! Use a BS170 mosfet instead,
because it wil disturb your internal pic supply and perhaps create a
reset.
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1998\06\18@075042 by tjaart

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Georg Hager wrote:

> Greetings!
>
> In order to drive an IR LED, I'd like to sink currents of about 100mA for
> short periods of time into 16F84 output pins. As this is not possible with
> a single pin, can I just short a couple of pins to get more sink current?
> Or is there some danger afflicted with it (apart from accidentally
> giving different logic levels out to the connected pins ;-)?

The PIC's are CMOS devices (That's why the power pins are markedVdd & Vss instead of Vcc and Vee),
so theoretically, you could.
I, however, wouldn't. I't easy enough to use a transistor for the job.

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1998\06\18@091929 by Lewis H. Cobb

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I just recently (i.e. an hour ago) ganged up four pins to "source" more
current and it seemed to have the desired effect - I would imagine the same
would be for sinking - seems to me i heard someone else on this group once
say that it was ok to do this - I'm new to pics so don't know the innerds
too well yet...

Lewis

At 01:21 PM 6/18/98 +0200, you wrote:
{Quote hidden}

1998\06\18@142810 by Thomas McGahee

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Greg,
The PIC has a max current rating per pin, AND it has a max TOTAL current
rating for the entire device. Too much current through the combined
outputs might well result in all kinds of crashes and burn outs.
Personally, I would not attempt it. Of course, if you have a PIC
you don't mind sacrificing, then experiment and see what happens.

A better solution is to use an NPN transistor with a base resistor of about
330 ohms. Connect the collector to ground (Vss), connect the cathode
of the IR LED to the collector. Connect a current limiting resistor
from the anode of the IR LED to +5. A high out from the PIC will
cause the IR LED to emit infra-red light.

Hope this helps.
Fr. Tom McGahee

----------
{Quote hidden}

1998\06\18@143153 by Martin Green

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    As long as you don't exceed the maximum source or sink current for the
    whole device, (50mA for many PIC's, 100mA for 16C/F84 I think), it is
    fine to connect multiple output pins to boost the drive capability.


    CIAO - Martin.


______________________________ Reply Separator _________________________________
Subject: More sink current
Author:  pic microcontroller discussion list <PICLISTspamspam_OUTMITVMA.MIT.EDU> at
Internet
Date:    6/18/98 1:21 PM


Greetings!

In order to drive an IR LED, I'd like to sink currents of about 100mA for
short periods of time into 16F84 output pins. As this is not possible with
a single pin, can I just short a couple of pins to get more sink current?
Or is there some danger afflicted with it (apart from accidentally
giving different logic levels out to the connected pins ;-)?

Thanx,
Georg.

1998\06\19@002749 by Mike Hamilton

picon face
I built a viscometer using two diode lasers.  I found that a setup similiar
to Tom McGahee's works great, but uses N-Chan. Enhanced FET's instead of NPN
transistors.  These use hardly any current from the pic and can switch on
and off very fast.  The FET part number is 2n7000 and can be easily ordered
from Digikey for about a $1.  In my setup I have the each laser's plus side
connected to 5V then the minus side connected to the drain and the source
connected to the ground.  The RB1 and RB2 output from the pic connects to
the gates of the FETs.  So when RB1 or RB2 goes high the particular laser
turns on.
{Original Message removed}

1998\06\19@044852 by org Hager

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On Thu, 18 Jun 1998, Thomas McGahee wrote:

> A better solution is to use an NPN transistor with a base resistor of about
> 330 ohms. Connect the emitter to ground (Vss), connect the cathode
> of the IR LED to the collector. Connect a current limiting resistor
> from the anode of the IR LED to +5. A high out from the PIC will
> cause the IR LED to emit infra-red light.

I see. But I think one could spare the base resistor by using an emitter
follower, right? I mean connecting base to PIC output, collector to +5V
and placing diode/resistor between emitter and GND. This cuts the drive
voltage for the diode/resistor combination by 0.7V, but has better
switching characteristics as the transistor is not put into saturation.
Right?

Georg.

BTW, thanks to all that contributed to this.

1998\06\19@063907 by Morgan Olsson

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At 10:11 1998-06-19 +0200, you wrote:
...
>I see. But I think one could spare the base resistor by using an emitter
>follower, right? I mean connecting base to PIC output, collector to +5V
>and placing diode/resistor between emitter and GND. This cuts the drive
>voltage for the diode/resistor combination by 0.7V, but has better
>switching characteristics as the transistor is not put into saturation.
>Right?
>
>Georg.

Excellent
/Morgan.

>BTW, thanks to all that contributed to this.
>
>
/  Morgan Olsson, MORGANS REGLERTEKNIK, SE-277 35 KIVIK, Sweden \
\  @spam@mrtKILLspamspaminame.com, ph: +46 (0)414 70741; fax +46 (0)414 70331    /

1998\06\19@081605 by Keith Howell

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Georg Hager wrote:

> I think one could spare the base resistor by using an emitter
> follower, right?

How about using a 2N7000 FET.
It has quite decent drain current.
You can drive the gate from logic levels.
Voila! No base resistor!

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