Read-Modify-Write - Something that every PIC programmer should be aware of.

Whenever you write a bit on any register, including port registers, the CPU=
=20
actually reads the whole register, modifies the bit you want to set/clear a=
nd=20
then writes the whole register back. When you write a port pin it sets a la=
tch=20
on that port but when you read it, it is the actual voltage on the port pin=
=20
that is read.=20

This would all be fine in an ideal world. However, it can take some time fo=
r=20
the actual voltage on a port pin to rise from a 0 to a 1 (or vice versa). T=
his=20
is depending on the capacitive and resistive load on that pin.

Before the port pin output voltage have reach the threashold value for a=20
digital 1 on the input circuitry for that pin it will be read as a 0. So ev=
en=20
if you have set the port pin latch to a 1, it can take some time before the=
=20
port register bit for that pin will be read back as a 1.

Now, if you set two bits on the same port right after each other, the secon=
d=20
write can reset the first because the output pin for the first bit has not=
=20
settled to a high level yet (as read by the input circuitry).

There are a couple of ways to solve this.

On newer pics, each port actually has two registers, the port pin register=
=20
itself that sees the voltage on the port pin and also the latch register th=
at=20
holds the value last written to the port. The latch register is always a 1 =
when=20
it has been set to a 1 even if the voltage on the port pin hasn't reached i=
ts=20
final voltage yet (and ofcourse vice versa for a 0). Therefor you should us=
e=20
the latch register when writing individual bits to a port since a read on t=
hat=20
register always reflects what was last written.

Older pics can't read the latch register so there you have to solve the pro=
blem=20
by not writing individual bits to a port register one at a time but instead=
=20
always write all bits at once. You do this by using a shadow register (whic=
h=20
works the same way as the latch register). Set and clear individual bits to=
 the=20
shadow register and then copy this whole register to the actual port regist=
er.=20
This way, writing an individual bit to an outport, never involves a read on=
=20
that port.

/Ruben =20


> not sure i fully understand your suggestion. could you provide a a few li=
nes
> of code for me to view?
>=20
> thank you
>=20
>=20
> >> On a PIC16F88, I would use:
> >> #define LED1 PORTAbits.RA0
>=20
> > In most cases this will put you at the mercy of the Read-Modify-Write=20
> > problem. Better define a bit in a porta-shadow register, chacge that=20
> > bit, and copy it to PORTA after each change of any bit.
>=20
> > --=20
>=20
> > Wouter van Ooijen
>=20
> > -- -------------------------------------------
> > Van Ooijen Technische Informatica: www.voti.nl
> > consultancy, development, PICmicro products
> > docent Hogeschool van Utrecht: www.voti.nl/hvu
> > C++ on uC blog: http://www.voti.nl/erblog
>=20
>=20
>=20
>=20
> --=20
> Bill
> Key fingerprint =3D DB4D 251B FE8A BDCD 2BE4  E889 13F1 78D0 A386 B32B
>=20
> --=20
> http://www.piclist.com PIC/SX FAQ & list archive
> View/change your membership options at
> http://mailman.mit.edu/mailman/listinfo/piclist



=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
Ruben J=F6nsson
AB Liros Electronic
Box 9124
200 39 Malm=F6 Sweden
www.liros.se
Tel +46 40142078
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D

--=20
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
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.