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 .
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
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.
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
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....
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.
I really understand ur explanation thank you for these vedio I need these topic it's very important views of data in database management Data user in dbms Over all system structure dbms Data model in dbms Data base language in dbms Storage management in dbms
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
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
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)
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
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!
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.
Education 4u I am asking for assembly language program like binary to ASCII code conversion, binary to BCD conversion, BCD to binary conversion,hexadecimal counter,mod 10 counter etc. Looking forward to heat from u.
What do you mean by "Holding or Keeping the address"? How 1MB space of 8086 divided into four 64KB blocks??? Four 64KB blocks=256KB right mam....how can they make 1MB?
At 15:27, the sum of four 64KB is not 1MB, its just 256KB. Can you tell something about why that didnt add up to 1MB? And I think at 09:58, the segment registers dont total to 1MB. These registers just hold base addresses of different 4 segments, which just makes it 8 bytes( for four 16 bit registers) ( +2 Bytes for IP). Is there maximum and minimum range to which each segment can address upto so that it adds to 1MB ? Great tutorial by the way madam. Thank you so much.
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 ...
it is explained wrong,it is segment register which contains the adress of segment memory,here it gets the adress then it passes the adress to memory,and collect the instruction and fetch it to instruction queue
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
Legends are watching this one day before their exam....😂 Yeah I'm one of them...🙂❤️🩹
Ultralegends see this video before one hour of exam
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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 .
No words ,perfect explanation, very clear thank you so much mam please do more videos
Really it helps a lot thank you once again mam❤
I passed exam bcz of u...thanku so much
microprocessor 8086 microprocessor architecture | Bus interface unit | part-1/2
Very good teaching amd topics are esay to learn very nice madam
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If the qstion come .
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Thanks Mam
Thanks mam
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QSTION? Or question?
two words for mam Thank you 🙏🙏🙏🙏🙏✍✍✍✍✍✍✍
Super explaination but some topics confusing
08:33 Instruction Pointer points to the current instruction not the next instruction...that job is for Program Counter. Kindly rectify
False , instruction pointer register holds the offset of the address of the next instruction to be executed Program counter == IP
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
do you have any material or theory from where i can read ?
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Excellent explaination.... Madam thank you..... explanation
U r voice is too good
Very good explanation mam 👌👌
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U r class is very good. Thank u
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.
Ok
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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
you're correct 💯
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....
Superb,,, best best video thanks 🙏 mam
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.
I really understand ur explanation thank you for these vedio
I need these topic it's very important views of data in database management
Data user in dbms
Over all system structure dbms
Data model in dbms
Data base language in dbms
Storage management in dbms
nice video mam...keep uploading vids..it helped a lot!!!!!!!
Great Explanation
In BIU the 5 segment register are not of 64 kb per segment , they are 16 bit per segment and can only contains addresses.
No it's 64 kb only. She is right, please make a note of it.
@@KomalSharma-xi5xd yeah
Actually it's 16 bit registers that contain address of 64 kb segments of memory that is outside biu
Thanks for these lectures.you are great
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Thanks maám for confusing me even further :)
Damn good teacher,! Thanks
Legends are watching one day before exam with 1.5 speed 😂
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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?
The 1MB memory is divided into 16 sub-parts each of 64KB size
Therefore,
64KB*16= 1024KB=1MB.
As per Microprocessor theory
Exactly I was thinking about that .. but yeah @Sai Priyatham Reddy was correct
Excellent explanation mam tqqq sooo much
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
Super 👌👌👌👌👌 mam
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
Easy understanding good teaching madam
Habibi come to kerla😂😂
Segment registers contain the address of corresponding memory segments they are not 64 kb big!
They are actually 16 bit registers not 64 kb
@@asherabraham3034 Yes right.
And they are then converted to 20 bit address by multiplying with 10H.
Right Brother
Podey
Also ....ip contains the offset address....might be😕
Excellent teaching medam
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Excellent madam,,,wonderful explanation...
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here, 4 segment register contains 16 segments of 64 kb, which makes total 1 MB.
not 4 segment registers. 4 segments contain the data. Segment registers are only 16 bit.
Segment registers are used to refer segments of the memory.
Sindh Uni wali hazr hoo 2024😂
Instruction pointer generates a 20bit address of the next instruction to be executed.but IP register is 16 bit..then how?
By multipying 16 bit address with 10H(Hexadecimal).
How ?
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)
64KB X 4 = 256KB, how is the total 1MB?
I was thinking that
Yaa
64×16
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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
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!
Can you point those mistakes. So the leanersy be aware of them?
nice mam😍
Very Nice Mam
madam we want notes pdf
Usefull👌
Legends are watching on day of exam
Thank u for God explain madam, I want remaining units important with answers please
Thanku so much your classes are very easy to understand
Complete memory of segment register is 1MB then how it is partitioned by 64kb???
The complete 1mb memory which the 8086 addresses is divided into 16 logical segments. Each segment thus contains 64kb of memory.
There are total 16 partitions.
4 Segment registers and 4 partition for each register.
How ?
Can u plz make a video on 80386 also plz mam
Nice video
Thanks
You are very good mam. I love you.
Segment register Is not segment from the memory.....
Segment register holds address of that location
Bc yeh kuch bhi bolti hai be
Nice madam
Very Nice
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I'm Still Searching Powerful💪 Teacher For Microprocessor Subject 😫
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Mam the theory of internal architecture of 8088 microprocessor is same or not to the internal architecture of 8086 microprocessor
Very good.
POV : exams are near 🙃
How 1MB of storage space is divided into 4 segments and each having only 64KB space??
4 segments (ES, CS, SS & DS ) with some additional space
@@education4uofficial At any given time 8086 works with only four 64KB segments withing 1 MB range.
@@ashishvidyarthi7450 she herself says whatever written in the book... Many times I have noticed that she delivers the wrong lectures....
What do you mean by the memory space of 8086 ?
Memory Space ?
@@learningexpress6224 dude you are spamming every comment
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.
Legends are watch this one day before exam.....
Me also😂
Thanks a lot mam🙏
excellent explanation. Thankyou maam
Please make videos on Assembly language program of 8085 microprocessor.
It's urgent and all my colleagues are waiting to hear from u on this topic.
already uploaded..plz visit my channel..in COA playlist..
th-cam.com/video/MGIhCGdhpjA/w-d-xo.html
Education 4u
I am asking for assembly language program like binary to ASCII code conversion, binary to BCD conversion, BCD to binary conversion,hexadecimal counter,mod 10 counter etc.
Looking forward to heat from u.
Assembly Language Program
What do you mean by "Holding or Keeping the address"?
How 1MB space of 8086 divided into four 64KB blocks???
Four 64KB blocks=256KB right mam....how can they make 1MB?
Ma'am please reply
Please explain about segment register clearly
At 15:27, the sum of four 64KB is not 1MB, its just 256KB.
Can you tell something about why that didnt add up to 1MB?
And I think at 09:58, the segment registers dont total to 1MB. These registers just hold base addresses of different 4 segments, which just makes it 8 bytes( for four 16 bit registers) ( +2 Bytes for IP).
Is there maximum and minimum range to which each segment can address upto so that it adds to 1MB ?
Great tutorial by the way madam. Thank you so much.
4 hours before exam 😂
Dsu?
@@ekbilli5074 do you mean dsuniversity... No I'm not from dsu 🙃
Please add distributed computing notes and class
Pls add
is IP and PC the same?
if you work on white board making direct connections earn more views
24/09/2023
But sound quality itna achaa hai sab clear ho jata hai
awesome
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 ...
💪🏾
The world would be better if Indian Had a comprehensive accent
It would be even better if bhikaaristanis learnt english or even better Pakistan didn't exist.. terrorists
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tnx madam you are great
bhai flip-flop ke uper kuch banao
thanks..sure..but after some time
Education 4u tumhara learning process bahat achhha hai hamare college ke sab ladke tumhare video dekhte hai.keep it bro
Flip Flop
Help me in this Question
Write a program for finding out the no of +ve, -ve and zeros from a given data array using 8086
64 kb x 4=256 kb which is not equal to 1 Mb...🙃
Mam u have good NAILS full triangular and in shape do they WORKOUTTTTT???
😂
Thanks you
position of instruction pointer
is wrong
it is explained wrong,it is segment register which contains the adress of segment memory,here it gets the adress then it passes the adress to memory,and collect the instruction and fetch it to instruction queue
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