8086 microprocessor architecture | Bus interface unit | part-1/2

architecture of 8086 microprocessor with diagram

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QSTION? Or question?

[@venkateswararaoaddanki6420]

Mam its very easy for understanding and reading. I watch all your videos on microprocessor. Thank you very much. I want a help for preparing 80386 and 80486 microprocessor intro, architecture, processor model and programming model. Its one compulsory question from one unit. Please keep videos for these topics. Please mam. I subscribed your channel and referred to my friends.

[@kovidsingh2558]

Note that the 8086 does not work the whole 1MB memory at any given time. However, it works only with four 64KB segments within the whole 1MB memory. we can anywhere position that 64 kb segments in 1mb memory .

[@Allinone-ry8dr]

Hello mam I like your videos but one thing need to be correct , that since cpu don't take access all 1 mb memory at a time so it use segmentation process and will segment it to 64kb each to make it easy for access and then there are 4 segment register inside the cpu which are useful to contain address of the corresponding segmented memory they are addressing and they hold 16 bit each as we know 8086 contains 20bit address so it add 4 zeroes on pre side of the segment register like 0000 then 16 bit address....I hope you understand guys I have been confused too but now good with it

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[@GNRLOGICS]

Excellent explaination.... Madam thank you..... explanation

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Easy understanding good teaching madam

[@anujasebastian8034]

These are the segment registers which holds the address of the segments in the memory. These segment registers in BIU are 16- bit registers. Memory is completely different entity and not in the BIU. The 1 MB memory is segmented into 4 segments each of 64KB (CS,DS,SS,ES) together with other portions not for the user....

[@suryaroyals8128]

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[@Sushant6543]

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[@abhaykumar4848]

thank you, mam... Everything is fine but one thing you mixed up here that Segment register and Segment memory is complete two different things. segment register is of 16 bit which holds the starting or source address of the corresponding segmented memory location. N that segment memory is of 1MB 64 kb each.

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microprocessor 8086 microprocessor architecture | Bus interface unit | part-1/2

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this video help me so much,in order to get high experience from microprocessor, please guys ,when you have comfortable condition attach some video on my email and make videos on assembly language,thanks guys

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[@shinejoseph6656]

Dear Madam, Thank you for uploading the video n I appreciate your presentation skills but there are some mistakes in the concept while you delivered it. Kindly look into that. Thank you. Have a great future in online teaching!

[@harishrajaaruntheeban8896]

Can you point those mistakes. So the leanersy be aware of them?

[@epictetus__]

Thanks maám for confusing me even further :)

[@hitech1918]

Thank u for God explain madam, I want remaining units important with answers please

[@Hari_creations05]

Clg valla kanna mire Baga ceptunnaru

[@MuhammadBilal-we8tt]

In BIU the 5 segment register are not of 64 kb per segment , they are 16 bit per segment and can only contains addresses.

[@KomalSharma-xi5xd]

No it's 64 kb only. She is right, please make a note of it.

[@sakshh16]

@@KomalSharma-xi5xd yeah

[@elvisedison1741]

Actually it's 16 bit registers that contain address of 64 kb segments of memory that is outside biu

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[@jittususanmathew468]

in order to increase execution speed and fetching speed, 8086 segments the memory. Its 20-bit address bus can address 1MB of memory, it segments it into 16 64kB segments. 8086 works only with four 64KB segments within the whole 1MB memory.

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taught*

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24/09/2023

[@ajjunikhila]

Tq so much

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Thanks Mam

[@ashok20599]

Please explain about segment register clearly

[@TutorialTechie]

POV : exams are near 🙃

[@mallesh8399]

Ur explanation is good but ur explaining wrongly about segment registers . Segment registers can't store 64kb of data. Those can store only base address of that particular segments.segments are different from segment registers. Segments can store 64 kb of data, not registers

[@LakshmiD-oi8wr]

Nice madam

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Very Nice

[@AMARKUMAR-ch8rl]

Thanks mam

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Please add all microprocessor and microcontrollers in one playlist madam

[@A-L-L-in-O-N-E]

Within the 1 MB of memory space the 8086 defines four 64K-byte memory blocks called the code segment, stack segment, data segment, and extra segment. Each of these blocks of memory is used differently by the processor. This all the memory segments .... The four segment registers (CS, DS, ES, and SS) are used to "point" at location 0 (the base address) of each segment

[@rahulravindran322]

Segment registers contain the address of corresponding memory segments they are not 64 kb big!

[@asherabraham3034]

They are actually 16 bit registers not 64 kb

[@AkashDubey1706]

@@asherabraham3034 Yes right. And they are then converted to 20 bit address by multiplying with 10H.

[@learningexpress6224]

Right Brother

[@silentcat4292]

Podey

[@shaifaliverma153]

Also ....ip contains the offset address....might be😕

[@c4yourselfyt]

08:33 Instruction Pointer points to the current instruction not the next instruction...that job is for Program Counter. Kindly rectify

[@ravinderkaur124]

Mam the theory of internal architecture of 8088 microprocessor is same or not to the internal architecture of 8086 microprocessor

[@user-tv6hv5jg6z]

Internal registers are not of 64K however they are 16 bit wide. the 64K portion lies in the memory not in the inside registers. Inside registers hold the Base address of that corresponding segment

[@shivaPrasad-vg9dp]

Thank you madam

[@reshmaben988]

Nice video

[@education4uofficial]

Thanks

[@lokeshparab]

How this 1mb is divided into 64kb each block? Because 1mb = 1024kb then we divided by 4 we get 256kb for each then how 64kb for each block?

[@saipriyathamreddy2126]

The 1MB memory is divided into 16 sub-parts each of 64KB size Therefore, 64KB*16= 1024KB=1MB. As per Microprocessor theory

[@varshareddy8907]

Exactly I was thinking about that .. but yeah @Sai Priyatham Reddy was correct

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Tq 🌚😍

[@GHANTAAproductions]

here, 4 segment register contains 16 segments of 64 kb, which makes total 1 MB.

[@technicult0]

not 4 segment registers. 4 segments contain the data. Segment registers are only 16 bit.

[@technicult0]

Segment registers are used to refer segments of the memory.

[@Ani_tech04]

But sound quality itna achaa hai sab clear ho jata hai

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not perfect, but the best video out there on the topic.

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[@shaxz1514]

Segment register Is not segment from the memory..... Segment register holds address of that location

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Assembly Language Program

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Can u plz make a video on 80386 also plz mam

[@dr.shahbazkhan1206]

Please check the video at 15:30. Segment registers only contains addresses of these segments in external memory. There is no memory in these segments. Correct me if i am wrong. Thanks

[@coolsai]

you're correct 💯

[@aurkom]

64KB X 4 = 256KB, how is the total 1MB?

[@georgegeoffrick288]

I was thinking that

[@akhileshyadav9862]

Yaa

[@gopikrishna8969]

64×16

[@akhiladevibv9265]

Instruction pointer generates a 20bit address of the next instruction to be executed.but IP register is 16 bit..then how?

[@AkashDubey1706]

By multipying 16 bit address with 10H(Hexadecimal).

[@learningexpress6224]

How ?

[@pinoylyrics867]

multiply it by 10H. If u convert 10H to binary, it will be 10000. Obviously, You've got additional 4 digits. 1111111111111111(16bits) * 10000 = 11111111111111110000(20bits)

[@saminashameem3149]

Please add distributed computing notes and class

[@saminashameem3149]

Pls add

[@abdellahbelhaddad7791]

Hhhhhhhhhhh thank you verry much madam

[@abhirajkj5094]

The segment register have 1mb memory and it segmented as 4 each segment have a memory of 64 kb then ., 4*64=256 kb then the rest of the memory from the 1 mb is not mentioned in ur section ...

[@CCNA2024]

well

[@naanummalayali6520]

I can understand this more i read

[@kirangk5428]

Complete memory of segment register is 1MB then how it is partitioned by 64kb???

[@sravani1191]

The complete 1mb memory which the 8086 addresses is divided into 16 logical segments. Each segment thus contains 64kb of memory.

[@AkashDubey1706]

There are total 16 partitions. 4 Segment registers and 4 partition for each register.

[@learningexpress6224]

How ?

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Legends are watching on day of exam

[@Loalen]

Thanks Teacher