Intel MCS-80/85 manual PP-l, Pop !?, HALFPrT

I

I'

~~

I,"

POP H

POP !?

EI

RFT

INPUT ROUTINE

The console input routine uses the opposite pro- cedure; instead of moving a bit from register C to the CY, then to A7, then to SOD, CIN loads a bit from SID into A7, then moves it to CY, then into register C.

First, set up the CPU as before:

CHI PUSH H

PI

!,!\,1I B.:?

When a start bit transition arrives, the first sam- pling should not be taken until the middle of the first data bit, one and one-half bit times after the transition. Await the start bit transition, then set up the delay parameter for one-half bit time:

CIi'	RIl'l		(4)
	')PA	A	<4>
	m	(Ii	<7~.\
	LHU'	HALFPrT	'-::1t:>

Loop for one-half bit time before starting to sample data:

CI2'	[:ofF.'	L	((i:.
	mz	(12	([I)
	DCP	H	.:[:..,
	..TN:	CI2	,:.[:,':,

Wait until the middle of the next bit before sam- pling SID, then move the data bit into CY:

eB:	LHLf:t	BITTH1E	<16)
C14:	C(:P	L	;;c·)
	.m:	Cf4	<O:i
	[oCP-	H	<e')
	.m:	(:14	W)
	RIM		<4)
	PP-l		<:4>

Decrement the bit counter. If this is the ninth cycle, the 8 data bits are in register C, so quit (the first stop bit will already have been received, and be in CY):

[)CP	P	<4)
..1:	CIS	(7)

Otherwise, continue. Rotate the data bit right into register C, and repeat the cycle:

~10Y	A,C	{4>
RAR		<.f>
NOli	C.F!	{,~)
NOP	CE	<4)
..Tt1P		<113::'

(A NOP is needed to make the COUT and CIN loops exactly equal in number of machine cycles, so that each can use the same delay parameter.) Restore status and return.

CI5 POP H

E1

FH

TIMING ANALYSIS

COUT and CIN now need to be provided with parameters for BITTIME and HALFBIT. It can be seen from the above code that each routine uses 61 + D machine cycles per input or output bit, where D is the number of cycles spent in either four line delay segmen t. If (H) and (L) are the contents of the Hand L registers going into this section of code, then:

D = 22+«U-I)X 14+«H)-l)X
		[(255 X 14)+25]	(1)
If (H)' -		(H) - I, (U' == (U - I, and
		(HU' == 256 (H)' + (U'	(2)
then		22 + 14 (U'+ 3595 (H)'
D			(3)
This can be approximated by:
D	=	22 + 14 (HU'	(4)

This approximation is exact for (H)' = 0; otherwise, it is accurate to within 0.3%. Thus each loop of COUT or CIN uses a total of:

C = 61 + D = 83 + 14 (HU' machine cycles (5)

Each machine cycle uses two crystal cycles in the 8085, so the resulting data rate is:

B = cycle frequency

C

(crystal frequency) +2	(6)
83 + 14 (HU'

A1-35