[email protected]2 0 1 4 Re f e re n c e :M icrop ro c e s s o r s a n d P e rip h e ra ls H ar d w ar e , So f tware, In ter f acin ga n d Ap p lic at ion s S e c o n d E d itio n Barry B. Bre y INT R ODUCT ION to MICRO COMP UTERS in ESO G U Questi on s about 8085Micr o pro ces s or withS oluti on s So lve d b y ŚŵĞƚ PD7Z Cha p t e r - 3 / Q - 5 Whatis thec l o c k c y cle tim e whwn ev er a 4 - M H z cry s t alisatt a ched t o the 8 0 8 5 A ? The cl o ck frequ enc y o fan 8 0 8 5 Ais o n e h alf t h e cry s tal frequ ency . The p erio d o f o n e T ʹcycl e isthe in v ers e o f t h e c lo ck f requ ency .At a cr y stal freq u ency o f 4 - M H z ,t h e cl o ck i s2 - M H z and c lo ck cy cle t i m e equ alt o 5 0 0 n s. Ͷ ݖܪܯʹ ൗ ൌ ݖܪܯʹͳ ݖܪܯʹൗ ൌ ͲǤͷ ݏߤൌ ͷͲͲ ݏCha p t e r - 3 / Q - 8 H o w m an y byt es o f m e m o r y c an the 8 0 8 5 Aad d r ess d i rect ly ? The 8 0 8 5 A M P Uw i th1 6 ad d ress li n es cap ab le o fadd r essing ൌ 6 5 5 3 6 (gen e ral ly kno wns as6 4 K) m em o r y lo cati o n s. Cha p t e r - 3 / Q - 1 2 Whatis th e p u rp o s e o ft h e ܦത ത ത ത sig n al? The c o n tr o lsig n alR ead( ܦത ത ത ത ) e n ab lest h e o u tpu t b u ffer , a n d datafro m thes elec t edr egi ste r are m ad e a v ail ab le o n t h e o u tpu t li n es. Thisisan act i v e l o w inp u t c o n tro l sign al u sedt o re ad da ta fro m t h e m em o r y l o cati o n who s e add ress is a v ail ab le o n add ress lin es whene v er chi p se lec t sig n alis enab le.Thissig n alis av ail a b leo n s y st em c o n tro l bu sa n d generatedb y t h e m i cro p ro ces s o r o r the o ther m as te rin t h e s y st e m such as D M A c o n tro ll er o r co - p ro ces s o r.
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Reference : Microprocessors and Peripherals Hardware, Software, Interfacing and Applications Second Edition Barry B. Brey
Chapter-5/Q-18
Develop a memory system that uses one 2716 EPROM and one 4016 RAM. Locate them anywhere
you wish. (HINT: Use incompletely specified decoding.)
2716 EPROM and 4016 RAM are 2K. We can find this dividing the last two term by two. Means that they are required totaly 4K memory locations. Therefore 8085A have 64K memory space, we must divide by 32 field. And we can take into consideration minimum field 2K in this separation.
So we can use 5 digit address bits to declare our devices’ selection bits.