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7:40 it should be 2^16= 65536 registers. since 8085 microprocessor has 8 bit register so total bits= 65536*8 bits = 65536*8/8 bytes. therefore memory size is 64kb.
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design a memory system consisting of 8K byte ROM and 16Kbyte RAM spaces.The ROM starts at address 0000H while the RAM starts at 8000H.If the CPU address is 16bit ,list all possible ways in which ROM and RAM spaces can be accessed.Assuming individual ROM and RAM chips to be of size 4K bytes each,design the full memory system..............can u plz help me by showing how to do this??
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Thank you mam very nice gide & very nice best information memory chips interfacing microprocessor teaching video.👍
brilliant mam great job
i passed my test because of you thank you very much
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Thank you so much. I understand this topic better because of you. Keep up the good work ma'am.
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7:40 it should be 2^16= 65536 registers. since 8085 microprocessor has 8 bit register so total bits= 65536*8 bits = 65536*8/8 bytes. therefore memory size is 64kb.
you are just awesomee ma'am
thank you so much for your effortss
great work
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Superb explaination regarding this topic mam. Thank you so much madam🙏
thank you so much 😍🌸
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U did a very good job mam.. Thank you
Thanks a lot mam 😊
thank you very much mam
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Thank you mam
Best explanation of this topic 😊
wow you are awesome !!..
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Can you please upload lectures about arm microprocessor
At 20:06, It's AD0 to AD7 and u wrote it as A0 to A7 and that is an address bus not data bus, D0 to D7 is data bus , just don't teach wrong!!
tu ch...... hain tujhe pata nahi hain magar tu hain
design a memory system consisting of 8K byte ROM and 16Kbyte RAM spaces.The ROM starts at address 0000H while the RAM starts at 8000H.If the CPU address is 16bit ,list all possible ways in which ROM and RAM spaces can be accessed.Assuming individual ROM and RAM chips to be of size 4K bytes each,design the full memory system..............can u plz help me by showing how to do this??
At 6:30, its 65536 bytes not bits
shouldn't it be S0 and S1?
thank u mam
thank you maam