Interfacing Memory in 8086 Microprocessor with Memory Chip (Problems)

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[@manolokonosko3702]

She explains everything in a way that almost anyone can understand. Great job!

[@zounedmansoor3341]

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[@kartiknagrale1479]

@@zounedmansoor3341 what?

[@manikantarobbi7252]

nice explination madam.

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[@PawanKumar-te6sp]

Ma'm In which basis u r taking odd and even in logic table.

[@hafsaafreen350]

Best explanation 💯

[@farhathabduljabbar9879]

Ece engineering 4th sem?

[@venkateswarluguddeti798]

Thank you mam very clearly explained 👍

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[@adithyasurs9044]

How can you use A0 in both address decoding logic and address line in each chip? you have to use A1 to A13 as address line for each chip

[@akankshyapanda5277]

Exactly

[@ranjanathalore9054]

Shruti Boss I was going through Microprocessor related videos and stepped onto this. So proud of you.Very well explained.

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[@niku_undhi_roi]

Thanks man 🥰

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[@prathyushakonathala6605]

Why did we take A14 - A15 as 1 for EPROM??

[@madhudmk]

Excellent way of teaching

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[@sanjaygowda5515]

superb explanation👏👌

[@ACE-ej4fu]

thanks for great explaination

[@gamemachine878]

Shouldn't the final input to ChipSelect1 be using an OR gate instead of AND?

[@shreyamishra6818]

very well explained

[@gamemachine878]

Great explanation

[@chandreshyadav2501]

best explanatio

[@NidhiMishra-rh9io]

Thank you so much

[@farhathabduljabbar9879]

Well explained!

[@bwasiookoyo1673]

Amazing

[@randomtime7475]

Good 😭.... saved me but need more sums

[@subhankarkarmakar2650]

The last part 29:29 is wrong, check the address lines, it should be from A1 to A12 and not A0 to A12, if a ROM/RAM has 4K locations how many pins will it have ? 12, mam you are trying to connect 13 pins at once, go get some practicals

[@purandarkakade9026]

Exactly same point. I am also having this doubt, how can she connect 13 address lines to the memory which is originally having only 12 bit lines??...

[@mayapatange2276]

Thank u sooo much😊

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[@ahmadsultan5419]

Cant we multiplex the address pins?

[@poolapremsai6110]

Tqqq mam

[@newdreamza1-z038]

1 ram has 4K memory locations => it should have A0-A11 lines(i.e 12 address lines) . but according to you A0-A12(13 address lines).isn't it wrong? 29:10

[@shivamjhariya7315]

newdreamz a1-z0 but in the question, it is said to interface 2 4k rom and 2 4k ram, that means 8k rom anr 8k ram, according to that 13 address lins (i.e., A0- A12) will be used.

[@saraswathips7283]

Yeah

[@subhankarkarmakar2650]

@@shivamjhariya7315 but here A0 will be used for even and odd bank selection, the address lines that should be connected will be A1 to A12 not A0 to A12 look carefully, A0 is already connected

[@jp0427]

Why in chip select line from A14 to A19 there in not all 0 in case of RAM

[@shubhamneurekar3132]

Chip select is for activating the chip If it is 0 no chip will be activated

[@abrarrazi3909]

Noise free sound

[@rakeshnipane2293]

The direction of data bus of ROM chip is in output direction only

[@dheerajkhanna7697]

you have taken two chips of each RAM and ROM. But you have taken the total memory of 8k only, which is for either 2 RAMs or 2 ROMs or each RAM and ROM. why is the total memory 8k=4k+4k and not 16k=8k+8k. ma'am its urgent, please consider replying.

[@sudiptadas1068]

8086 has an internal data bus of 16 bits or 2 bytes. Hence unlike the previous 8 bit processors you may have studied, 8086 having 16 data lines can be used to address a location in memory that contains 16 bits or 2 bytes at the same specified location (in 8085 if you were to store 2 bytes of data you would require 2 separate addresses in memory, but in 8086 you can store 2 bytes in a single memory location). However, most of the commercially available memory chips are "byte organized" and not "word organized". Hence to store 2 bytes of the data at the same specified offset within a particular segment, we would require 2 memory banks, one for storing the higher byte of data and another for storing the lower byte.

[@Manojsingh-nz5yq]

Best

[@Manojsingh-nz5yq]

Best 100\100

[@kathanpatel1896]

What will be address range if we have 2 256k EPROM in this???

[@lohithapulamarasetti9817]

256K = 2^18, So 18 address lines(A0 - A17) can be used for a 256K EPROM

[@alibehrouz9833]

you waste a lot of time plz write them faster thank you nice job

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[@snehaljosesj]

kidu

[@aishwaryaarakh8477]

Too lengthy video .....

[@vaishnavivlogs8017]

Please get proper clarity before u teach

[@shrutichavanaik5782]

Not usefull at all 👎

[@debaedits544]

8086 is a 1980s technology try to teach something new

[@lakshyarana67]

Tatti...logic lesss....

[@lasttimeifappedwas3851]

Please be fast