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N-bit processor and ALU in MCU

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final_destination

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Hello Everyone!!

1. Can anyone explain what is the meaning of an N-bit processor?
whether its data bus width is N-bits or its the width of the internal general purpose data register?

2. How can you link a N-bit processor with processor ALU? whether its ALU width is N-bits?


Thanks in advance.
 

KlausST

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Hi,

1) Isn't the internal data bus width the same as register bit width?

2) it is a problem of the mathematical operations.
.For an 8 bit system:
An ADD of two 8 bit registers give an 8 bit result (plus carry bit). So a 8 bit register is enough for the result.
A multiply of two 8 bit registers gives 16 bit result. Therefor you usually have two 8 bit result registers.

Klaus
 

final_destination

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Hi,

1) Isn't the internal data bus width the same as register bit width?

2) it is a problem of the mathematical operations.
.For an 8 bit system:
An ADD of two 8 bit registers give an 8 bit result (plus carry bit). So a 8 bit register is enough for the result.
A multiply of two 8 bit registers gives 16 bit result. Therefor you usually have two 8 bit result registers.

Klaus

Thank you for the reply.

1. Yes internal data bus width is same as register bit width.

2.
A multiply of two 8 bit registers gives 16 bit result. Therefor you usually have two 8 bit result registers.
means the processor size and ALU size are same? if 16 bit processor ALU can process data in the range 0-65535 ?
correct me if I am wrong!

Regards,
 

KlausST

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Hi,

Code:
means the processor size and ALU size are same? if 16 bit processor ALU can process data in the range 0-65535 ?

Typically yes.

But there may be deviations.
I think I have seen 16bit x 16bit multiplication ALUs within an 8 bit controller. Then you have two 8 bit registers for each input.
And four 8bit registers as output.

Klaus
 

final_destination

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Hi,

So
1. the bit size of a processor means the ALU width?? and ALU width is the same width as the CPU's internal registers!!

2. the "size" of a processor is not really the width of its data bus? For example, the 8088 was a 8 bit bus version of the 8086, but both were considered "16 bit" processors.

Correct me if I am wrong!

Thank you,
 

KlausST

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Hi,

Afaik, the 8088 is a downsized but command compatible derivate ot the 8086.

To be code compatible it is important for an 8 bit processor to act like a 16 bit processor. I don´t know if only the external data bus is limited to 8 bit or the internal databus also.
The difference is: either it uses two access in series to 8 bit register, or one access to a 16 bit register.

Klaus
 

final_destination

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Hi Klaus,

What I am trying to prove it here is the bit size of a processor means the width of ALU not the data bus width!! is that correct definition for bit size of processor?

like 8-bit processor means - 8bit ALU inbuild
16-bit processor - 16bit ALU

or other way, An N-bit processor capable of processing N-bit width operands in ALU.

and in pic10,pic12,pic16,pic18,pic24 mcus..10,12,16,18,24 means the size width of opcode in ALU.

Correct me if I am wrong.

Thank you.
 

doraemon

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Hello!

First you should define what you call width of the ALU. Width of the input or of the output?
Usually an ALU output width is twice the input width (multply 2 16-bit numbres gives you 32).
I would say that the bit width of a processor is the width of its registers.

Dora.
 

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