Inverted Pins

J

Jon Slaughter

Guest
When I see an inverted pin in a timing diagram do I need to invert it or
what?

Basically I'm looking at a pic that has overbar(MCLR). IN the timing
diagram it shows a high voltage for high. But does this mean that the
physical voltage into the pin on the device needs to be inverted?

It seems reasonable that if they went through the trouble to put the overbar
in the pin then it means its inverted with respect to the rest of the
logic... but I do not know if they already inverted it on the timing diagram
or if I'm support to invert it or if its really an internal thing and I
don't have to worry about it(i.e., just treat it like a normal pin)?

For example, one of the timing diagrams shows that Vdd goes from low to high
and then overbar(MCLR) goes from low to high momentarily. If MCLR didn't
have the overbar then I would actually have it go low to high as it shows.
But since MCLR does have the overbar it seems to imply that I should invert
the pin?

In the data sheet it says "This line is brought low to cause a Reset". So
it seems that it really means the pin is inverted and the timing diagram
does not reflect this? (or does it and the and should follow the diagram?

In general, basically when I see an inverted pin and any timing diagrams I
can essentially think of the pin as non-inverted but just put in inverter on
the pin? (or in my case I'll just flip it in the software)

(just want to make sure)

Thanks,
Jon
 
John:

The overbar on your MCLR pin indicates inverted logic. Just tie it to
Vcc (HIGH) for the chip to operate normally. When you need to reset
the chip, just pulse it LOW (typically with a pushbutton tied to
ground).

Hope this helps.
Aaron
 
On Tue, 02 Oct 2007 21:07:03 +0000, Jon Slaughter wrote:

When I see an inverted pin in a timing diagram do I need to invert it or
what?

Basically I'm looking at a pic that has overbar(MCLR). IN the timing
diagram it shows a high voltage for high. But does this mean that the
physical voltage into the pin on the device needs to be inverted?
.....
In the data sheet it says "This line is brought low to cause a Reset". So
it seems that it really means the pin is inverted and the timing diagram
does not reflect this? (or does it and the and should follow the diagram?
Click to expand...
This is called "active low". "Inversion" doesn't really have much
applicability in this context; to be "inverted", there had to have
been something to invert. The timing diagram should show you exactly
what physical signals go in and come out.

In general, basically when I see an inverted pin and any timing diagrams
I can essentially think of the pin as non-inverted but just put in
inverter on the pin? (or in my case I'll just flip it in the software)
Click to expand...
Something here might help to shed a little light on this:
http://www.google.com/search?hl=en&q=demorgan%27s+theorem

Good Luck!
Rich
 
Jon Slaughter wrote:

When I see an inverted pin in a timing diagram do I need to invert it or
what?

Basically I'm looking at a pic that has overbar(MCLR). IN the timing
diagram it shows a high voltage for high. But does this mean that the
physical voltage into the pin on the device needs to be inverted?
Click to expand...
It means the pin is ACTIVE LOW.

In other words, asserting the pin LOW causes the function (presumably master
clear in this case).

Graham
 
Jon Slaughter wrote:

In the data sheet it says "This line is brought low to cause a Reset". So
it seems that it really means the pin is inverted
Click to expand...
Think ACTIVE LOW.

Graham
 
"Jon Slaughter" <Jon_Slaughter@Hotmail.com> wrote in message
news:XpyMi.421$lD6.123@newssvr27.news.prodigy.net...
When I see an inverted pin in a timing diagram do I need to invert it or
what?

Basically I'm looking at a pic that has overbar(MCLR). IN the timing
diagram it shows a high voltage for high. But does this mean that the
physical voltage into the pin on the device needs to be inverted?

It seems reasonable that if they went through the trouble to put the
overbar in the pin then it means its inverted with respect to the rest of
the logic... but I do not know if they already inverted it on the timing
diagram or if I'm support to invert it or if its really an internal thing
and I don't have to worry about it(i.e., just treat it like a normal pin)?

For example, one of the timing diagrams shows that Vdd goes from low to
high and then overbar(MCLR) goes from low to high momentarily. If MCLR
didn't have the overbar then I would actually have it go low to high as it
shows. But since MCLR does have the overbar it seems to imply that I
should invert the pin?

In the data sheet it says "This line is brought low to cause a Reset". So
it seems that it really means the pin is inverted and the timing diagram
does not reflect this? (or does it and the and should follow the diagram?

In general, basically when I see an inverted pin and any timing diagrams I
can essentially think of the pin as non-inverted but just put in inverter
on the pin? (or in my case I'll just flip it in the software)

(just want to make sure)

Thanks,
Jon

Jon,
Click to expand...

The overbar, in this case, indicates that the function is active LOW. In
other words, a LOW on the pin causes a reset to take place. The timing
digrams should depict this as it really is and you don't have to do anything
to change them. The !MCLR (the exclamation point is another way to indicate
this) is normally held HIGH and is brought LOW, momentarily, to effect the
reset.

You wrote that, "For example, one of the timing diagrams shows that Vdd goes
from low to high and then overbar(MCLR) goes from low to high momentarily."

Which PIC and which timing diagram in the datasheet are you looking at?

Richard
 
Which PIC and which timing diagram in the datasheet are you looking at?
Click to expand...
http://www.semiconductorstore.com/pdf/newsite/microchip/PIC24FJ128GA006_PG.PDF

Page 22.

I think I got it. Thanks.

I have another question regarding the same spec though. Maybe you can take a
look at my other post about it if you end up looking at the pdf. Its on page
21 regarding Note 1.

Thanks,
Jon
 
On Tue, 02 Oct 2007 21:07:03 GMT, "Jon Slaughter"
<Jon_Slaughter@Hotmail.com> wrote:

When I see an inverted pin in a timing diagram do I need to invert it or
what?

Basically I'm looking at a pic that has overbar(MCLR). IN the timing
diagram it shows a high voltage for high. But does this mean that the
physical voltage into the pin on the device needs to be inverted?

It seems reasonable that if they went through the trouble to put the overbar
in the pin then it means its inverted with respect to the rest of the
logic... but I do not know if they already inverted it on the timing diagram
or if I'm support to invert it or if its really an internal thing and I
don't have to worry about it(i.e., just treat it like a normal pin)?

For example, one of the timing diagrams shows that Vdd goes from low to high
and then overbar(MCLR) goes from low to high momentarily. If MCLR didn't
have the overbar then I would actually have it go low to high as it shows.
But since MCLR does have the overbar it seems to imply that I should invert
the pin?

In the data sheet it says "This line is brought low to cause a Reset". So
it seems that it really means the pin is inverted and the timing diagram
does not reflect this? (or does it and the and should follow the diagram?

In general, basically when I see an inverted pin and any timing diagrams I
can essentially think of the pin as non-inverted but just put in inverter on
the pin? (or in my case I'll just flip it in the software)

(just want to make sure)

Thanks,
Jon
Click to expand...
If the reset pin is electrically active-low, but their timing diagrams
show it up for the reset state, they should be taken behind the
woodshed and whupped.

And they should be given one extra lick for using the passive voice,
as "This line is brought low to cause a Reset." A lot of the time,
it's unclear whether people are talking about levels or edges, or even
whether the damned pin is an input to the chip, or an output.

John
 
"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in message
news:3qt7g35p10r6n4ggkac1rc14bk8gtutlhh@4ax.com...
On Tue, 02 Oct 2007 21:07:03 GMT, "Jon Slaughter"
Jon_Slaughter@Hotmail.com> wrote:

When I see an inverted pin in a timing diagram do I need to invert it or
what?

Basically I'm looking at a pic that has overbar(MCLR). IN the timing
diagram it shows a high voltage for high. But does this mean that the
physical voltage into the pin on the device needs to be inverted?

It seems reasonable that if they went through the trouble to put the
overbar
in the pin then it means its inverted with respect to the rest of the
logic... but I do not know if they already inverted it on the timing
diagram
or if I'm support to invert it or if its really an internal thing and I
don't have to worry about it(i.e., just treat it like a normal pin)?

For example, one of the timing diagrams shows that Vdd goes from low to
high
and then overbar(MCLR) goes from low to high momentarily. If MCLR didn't
have the overbar then I would actually have it go low to high as it shows.
But since MCLR does have the overbar it seems to imply that I should
invert
the pin?

In the data sheet it says "This line is brought low to cause a Reset". So
it seems that it really means the pin is inverted and the timing diagram
does not reflect this? (or does it and the and should follow the diagram?

In general, basically when I see an inverted pin and any timing diagrams I
can essentially think of the pin as non-inverted but just put in inverter
on
the pin? (or in my case I'll just flip it in the software)

(just want to make sure)

Thanks,
Jon


If the reset pin is electrically active-low, but their timing diagrams
show it up for the reset state, they should be taken behind the
woodshed and whupped.

And they should be given one extra lick for using the passive voice,
as "This line is brought low to cause a Reset." A lot of the time,
it's unclear whether people are talking about levels or edges, or even
whether the damned pin is an input to the chip, or an output.
Click to expand...
I'm personally having a lot of trouble understanding the protocol from the
spec. First they introduce the different methods out of order and then
introduce the protocol out of course. They also do not explain anything in
great depth and leave, at least to me, a lot of ambiguity.

I would hope that someone writing the docs would want to put as much
specific information and examples in so that there could be no ambiguity.
Its not like its rocket science but if you don't do the exact protocol then
its not going on work. Maybe I'm just stupid though. I'm know I'm just
guessing at a few things and only about 90% sure and that makes me
uncomfortable as I need to be 99.999%. (which may or may not be their fault
but I feel it is a half ass spec(ok, maybe a quarter ass spec.))

Jon
 
On Wed, 03 Oct 2007 16:13:42 -0500, Jon Slaughter wrote:
I'm personally having a lot of trouble understanding the protocol from the
spec. First they introduce the different methods out of order and then
introduce the protocol out of course. They also do not explain anything in
great depth and leave, at least to me, a lot of ambiguity.
Click to expand...
Do you have a link to this data sheet?

Thanks,
Rich
 
"Rich Grise" <rich@example.net> wrote in message
news:pan.2007.10.04.21.57.05.595782@example.net...
On Wed, 03 Oct 2007 16:13:42 -0500, Jon Slaughter wrote:

I'm personally having a lot of trouble understanding the protocol from
the
spec. First they introduce the different methods out of order and then
introduce the protocol out of course. They also do not explain anything
in
great depth and leave, at least to me, a lot of ambiguity.


Do you have a link to this data sheet?
Click to expand...
See my other thread "Not sure about PIC ICSP Spec". It has the link there.
 
On Wed, 3 Oct 2007 16:13:42 -0500, "Jon Slaughter"
<Jon_Slaughter@Hotmail.com> wrote:


I'm personally having a lot of trouble understanding the protocol from the
spec. First they introduce the different methods out of order and then
introduce the protocol out of course. They also do not explain anything in
great depth and leave, at least to me, a lot of ambiguity.

I would hope that someone writing the docs would want to put as much
specific information and examples in so that there could be no ambiguity.
Its not like its rocket science but if you don't do the exact protocol then
its not going on work. Maybe I'm just stupid though. I'm know I'm just
guessing at a few things and only about 90% sure and that makes me
uncomfortable as I need to be 99.999%. (which may or may not be their fault
but I feel it is a half ass spec(ok, maybe a quarter ass spec.))
Click to expand...
---
Switch to Freescale.


--
JF
 
On Wed, 03 Oct 2007 13:15:19 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 02 Oct 2007 21:07:03 GMT, "Jon Slaughter"
Jon_Slaughter@Hotmail.com> wrote:

When I see an inverted pin in a timing diagram do I need to invert it or
what?

Basically I'm looking at a pic that has overbar(MCLR). IN the timing
diagram it shows a high voltage for high. But does this mean that the
physical voltage into the pin on the device needs to be inverted?

It seems reasonable that if they went through the trouble to put the overbar
in the pin then it means its inverted with respect to the rest of the
logic... but I do not know if they already inverted it on the timing diagram
or if I'm support to invert it or if its really an internal thing and I
don't have to worry about it(i.e., just treat it like a normal pin)?

For example, one of the timing diagrams shows that Vdd goes from low to high
and then overbar(MCLR) goes from low to high momentarily. If MCLR didn't
have the overbar then I would actually have it go low to high as it shows.
But since MCLR does have the overbar it seems to imply that I should invert
the pin?

In the data sheet it says "This line is brought low to cause a Reset". So
it seems that it really means the pin is inverted and the timing diagram
does not reflect this? (or does it and the and should follow the diagram?

In general, basically when I see an inverted pin and any timing diagrams I
can essentially think of the pin as non-inverted but just put in inverter on
the pin? (or in my case I'll just flip it in the software)

(just want to make sure)

Thanks,
Jon


If the reset pin is electrically active-low, but their timing diagrams
show it up for the reset state, they should be taken behind the
woodshed and whupped.
Click to expand...
That particular diagram is showing how to get the processor to enter
the ICSP (in-circuit serial programming) mode, not how to reset it.

And they should be given one extra lick for using the passive voice,
as "This line is brought low to cause a Reset." A lot of the time,
it's unclear whether people are talking about levels or edges, or even
whether the damned pin is an input to the chip, or an output.

John
Click to expand...
 
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