Q3. A processor is connected to a RAM with a capacity of 2M and a ROM with a capacity of 40K. The number of address lines for the processor is 32, and the number of data lines is 8. Calculate the following: a. The address range of the RAM and the number of locations within the memory. b. The address range of the ROM and the number of locations within the memory
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Question: I have two eproms connected The address map is set for the eprom to be 8000hex to FFFFh I have 16 address lines connected to allow 8000h up to FFFFh I have low bank and high bank eprom with d0 to d7 on low and d8 to d15 on high The 8086 has address 8000h and reads low byte and high byte instruction It then moves on next memory cycle to 8002 and reads low and high But how ? The address on high bank is 8000h but the same address 8000h is on low bank too Should the high bank address not be 8000h and the low bank address be 8001h ? It seems to be using the same address on both EPROMs ?
17:30 - The bidirectional bus of data on ROM's is wrong. It should be only an entering data bus, since ROM means "READ ONLY MEMORY". It doesn't write data.
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Sir ,why did you select OR gates for chip selection,i think that both or gates 1 for any value of A0 and Bhe' .since they are connected a16-a19 and those are 1 for EPROM
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my question is "should we consider 'A0' as a line address" , I mean in the EPROM we have 16 address lines (A1-A16), so The Start of EPROM will be equal to "E0000h" ????!
I teach at Karbala University The topics you explained are all excellent and very understandable You are aware of the situation in Iraq and the lack of resources. If you have the books available, can you send me a pdf copy?
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Q3. A processor is connected to a RAM with a capacity of 2M and a ROM with a capacity of 40K. The number of address lines for the processor is 32, and the number of data lines is 8. Calculate the following:
a. The address range of the RAM and the number of locations within the memory.
b. The address range of the ROM and the number of locations within the memory
Gajab padhaya sir. Main sab samajh gaya . lessgoo
wah
zabardast
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dimag ka bhosda hogya bhai mere tereko kaise samjha
your videos are very helpful sir thankyou very much.
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Question:
I have two eproms connected
The address map is set for the eprom to be 8000hex to FFFFh
I have 16 address lines connected to allow 8000h up to FFFFh
I have low bank and high bank eprom with d0 to d7 on low and d8 to d15 on high
The 8086 has address 8000h and reads low byte and high byte instruction
It then moves on next memory cycle to 8002 and reads low and high
But how ?
The address on high bank is 8000h but the same address 8000h is on low bank too
Should the high bank address not be 8000h and the low bank address be 8001h ?
It seems to be using the same address on both EPROMs ?
I think you should see the explanation of interfacing again.
Sir I think for 64 kb eprom 32 kb is available so address lines should be 15 not 16 same case with ram
@@tiieej3950 😂
17:30 - The bidirectional bus of data on ROM's is wrong. It should be only an entering data bus, since ROM means "READ ONLY MEMORY". It doesn't write data.
See it properly. With ROM I have provided unidirectional lines and with RAM it is bidirectional lines.
@@EngineeringFunda You should see it properly. The diagram is wrong and so are you.
@@Mynhassty ice on the cake
Sir, Thanks for the video. Sir can you please make some problem solving videos for this topic.
Master of interfacing ❤
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excellent lecture sir
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Thank U Sir ❤️🙏
Your Appreciations, care and share matters a lot to me. #EnginneringLove.
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Sir ,why did you select OR gates for chip selection,i think that both or gates 1 for any value of A0 and Bhe' .since they are connected a16-a19 and those are 1 for EPROM
sir perfect ho aap
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17:14 it should be A15-A1 and A16-A1 address lines
@@akashmajumdar-b1h it is perfectly correct
Sir 8085 me ROM is starting from 0000H and 8086 me RAM is starting from 00000H ?
my question is "should we consider 'A0' as a line address" ,
I mean in the EPROM we have 16 address lines (A1-A16), so The Start of EPROM will be equal to "E0000h"
????!
What you are saying is difficult to implement
Coding in memory is done using complier
@@EngineeringFunda ice on the cake
Hello, Possible book to take problems from
In 1st video, I have already mentioned books used by me.
I teach at Karbala University
The topics you explained are all excellent and very understandable
You are aware of the situation in Iraq and the lack of resources. If you have the books available, can you send me a pdf copy?
Is memory interfacing in 8088 microprocessor will remain the same as 8086??
Yes
Finally crystal cleae
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ice on the cake
Hi,Possible name of the book
In 1st video of playlist, it is already mentioned.
Won't the one byte of memory in odd bank get wasted if we store only 8 bit data for some reason
For that you have to my video based on memory segmentation of 8086, it will clear your doubts.