Intel MCS-80/85 manual + 6 =, HOi, Ho =

For a typical calculation, see the example below.

EXAMPLE

To produce 2400 baud with the standard 6.144 MHz crystal:

2400	(6.144 X 106) + 2
	83 + 14 (HO
14 (HO	(6.144 X 106 +2)
	2400	- 83
(HO'	[(61442~O~06+ 2)_ 83J
	+ 14 = 85.5	~ 86
(HO'	8610 = 0056H
(HO	0157H = BITTIME

To determine the true data rate this parameter will produce, substitute into equation (6):

6.144 X 106 + 2

Date Rate = 83 + 14(86)

=2387 baud, which is 0.54% slow.

For 9600 baud, the same calculations will yield (HO'

=17, which is actually 0.3% slow; a sizzling 19200 baud or 38400 baud could each be generated to with- in 5% if (HU' = 6 or O! Table 9 presents the param- eters for several standard baud rates.

Notice that the resolution of the delay algorithm - the difference between bit times resulting from parameters which differ by one - is 14 machine cycles. As a result, the true bit delay produced can always manage to be within ±2.3 fJsec of the delay

desired. This guarantees that at rates up to 9600 baud, where each bit time is at least 104 fJsec wide, some value of BITTIME can be found which will be accurate to within 2.2%.

BAUD RATE IDENTIFICATION ROUTINE

The function of BRID is to compute the appropri- ate parameters BITTIME and HALFBIT. It accom- plishes this by observing the data pattern received when the space bar is pressed on the console device. Since a space character has the ASCII code 20H = 001 OOOOOB, the pattern represented back in Figure 4 is transmitted. Notice that the initial Zero level is 6 bits wide. Suppose it could be determined that this corresponds to M machine cycles. Then one bit would correspond to (M+6) machine cycles. The reason for dividing down a space several bits long is so that any distortion caused by the signal rise and fall times, or any lack of pre- cision in detecting the two transitions, will be reduced by a factor of six. Since the bit period of COUT and CIN is 83 + 14 (HO',BRID must gener- ate a value (HOi such that:

M + 6 =	83 + 14 (HOi		(7)
(HOi	(M + 6) -	83	(8)
	14
I	M	(approximately)	(9)
(HO =	84 - 6

This value can be determined by setting register pair HL to -6, then incrementing it once every 84 machine cycles during the period that the incom-

f

\:

I:,

1",1.,'.'

I

!

I

II

!,

I

Table 9

DELAY PARAMETERS FOR STANDARD BAND RATES USING 6.144 MHz CRYSTAL

TARGET	(HL)'10	(HL)' 16	(HL) or		ACTUAL	%
BAUD			BITTIME	HALFBIT	BAUD RATE
	(See Text)	(See Text)
RATE			(See Text)		PRODUCED		ERROR
110	1989	07C5	08C6	04E3	109.99	-0.006
150	1457	05Bl	06B2	0309	149.99	-0.005
300	726	0206	0307	026C	299.80	-0.068
600	360	0168	0269	01A5	599.65	-0.059
1200	177	OOBl	01B2	0159	1199.5	-0.039
2400	86	0056	0157	012C	2386.9	-0.547
4800	40	0028	0129	0115	4777.6	-0.469
~600	17	0011	0112	0109	9570.1	-0.312
19200	6	0006	0107	0104	18395.2	-4.37

A1-36