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'PIC16F84 - RA4 Driving Problem'
2000\03\22@013640 by Werner Soekoe

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Hi

I wonder if anyone could adivse me. I made a PIC16F84 circuit using all 13
I/O pins to drive 13 LEDs, but the LED connected to RA4 doesn't light up. It
appears that the pin does provide a signal, though. When I read up some more
in the PIC16F84 datasheet, I came accross a paragraph which said that all RA
pins are CMOS drivers, except RA4 which is an "open-drain" output. Is this
why it can't deliver current? What does the "open-drain" mean?

Thanks

Werner Soekoe
spam_OUTwsoekoeTakeThisOuTspamglobal.co.za

2000\03\22@015646 by Patilano, Jr.

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If you're going to configure RA4 as output, you should tie it up first to VDD
since it is open-collector. (Use pull-up resistor).

Werner Soekoe wrote:

{Quote hidden}

2000\03\22@020733 by David Duffy

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<x-flowed>Werner Soekoe wrote:
>I wonder if anyone could adivse me. I made a PIC16F84 circuit using all 13
>I/O pins to drive 13 LEDs, but the LED connected to RA4 doesn't light up. It
>appears that the pin does provide a signal, though. When I read up some more
>in the PIC16F84 datasheet, I came accross a paragraph which said that all RA
>pins are CMOS drivers, except RA4 which is an "open-drain" output. Is this
>why it can't deliver current? What does the "open-drain" mean?

RA4 can pull down to ground but has no high-side transistor like the other
pins.
You can still use it to drive LEDs by hanging the LED & resistor between RA4
and +5V supply. (anode to +5V) This means that the logic is now inverted;
1 = off and 0 = on but that is easily catered for on the software side of
things.
Regards...

</x-flowed>

2000\03\22@021124 by w. v. ooijen / f. hanneman

picon face
> I wonder if anyone could adivse me. I made a PIC16F84 circuit using all
13
> I/O pins to drive 13 LEDs, but the LED connected to RA4 doesn't light up.
It
> appears that the pin does provide a signal, though. When I read up some
more
> in the PIC16F84 datasheet, I came accross a paragraph which said that all
RA
> pins are CMOS drivers, except RA4 which is an "open-drain" output. Is
this
> why it can't deliver current?
Exactly. It only sinks current. Connect all you LEDs and resistors between
+%V and the pins, invert what you wrte to the orts and it should work.
Wouter

2000\03\22@030531 by William Chops Westfield

face picon face
Invert all your bits and have the outputs SINK current instead of sourcing
it for the LEDs.  Then RA4 or other "open drain" outputs should behave the
same (as far as LEDs are concerned) as the other outputs.  "open drain" or
"open collector" means that there's a switching transisitor to ground, but
not one to VCC.  This allows an external pullup resistor to VCC to "pull up"
all the signals if they're all "high" (ie "disconnected"), but any one of
them can pull the line down ("wired OR" (inverted))

BillW

2000\03\22@042338 by Roland Koehler

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"open-drain" means this pin can only PULL DOWN (= current "sink" to GND),
while the other outputs can also "source" current ...
If you connect the LEDs (and current limiting resisitors) between the
outputs and VCC, and invert the output signals, everything will be ok.

Roland Koehler.

2000\03\22@091330 by James Paul

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Werner,

As you probably know, CMOS parts use FET's.  One of the
terminals on an FET is the DRAIN.  This is, power supply
wise, likened to the COLLECTOR on a BJT.  An open DRAIN
connection is just that.  The DRAIN is not connected to
the power source.  This is so that you can connect a resistor
to this pin and then to the power supply, so that you can
have a wired 'OR' connection.  Just put a pullup resistor
from 'RA4' to Vdd, and everything should work fine.  I would
use from 1K to 5K ohms for this pullup.

                                         Regards,

                                           Jim


On Tue, 21 March 2000, Werner Soekoe wrote:

{Quote hidden}

.....jimKILLspamspam.....jpes.com

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