Reply-To: "John B"
From: "John B"
References: <email@example.com> <6nnu9.5168$6F4.firstname.lastname@example.org>
Subject: Re: designing COM port interface without UART
X-Newsreader: Microsoft Outlook Express 6.00.2600.0000
Date: Sat, 26 Oct 2002 12:42:09 GMT
NNTP-Posting-Date: Sat, 26 Oct 2002 05:42:09 PDT
Organization: EarthLink Inc. -- http://www.EarthLink.net
"John B" wrote in message
> > For even parity, the parity bit will be a logic 1 (negative) if the
> > number of "1" data bits in the byte is odd, in order to make the total
> > number of "1" bits even. (and vice-versa for odd parity).
> OK, I got that. Also, I see that a parity bit follows immediately after
> last data bit, in the frame. However, what if there is "no parity"? Does
> the "parity bit" disappear, and the stop bit immediately follow after the
> last data bit, in the frame? Or does the parity bit become another stop
> bit; i.e, always a "space"? (Sorry if this sounds like a stupid question,
> but I'd hate to miss this.)
parity bit disappears when not used.
It seems this and other issues were answered in the article:
As in other cases of technical confusion, it all becomes easier to
understand if the history of the technology is discovered. I appreciate the
story about the teletype, and how fault-discovery concerns dictated that the
current should flow during quiescent operation. This was called a "marking"
state. If the line should have broken, then this "break" would have become
immediately apparent by the "space" state associated with 0 mA of current