Banks size in 8051

Question

I read a book about intel 8051 in which the author says, 8051 has three banks 00h to 1Fh, each bank has 8 registers and each bank is of 8 bytes. ?

Now I am confused what does he mean by each bank is of 8 bytes when each bank has 8 registers each 8 bytes wide. Kindly guide me

Regards

Answer 1

bank is of 8 bytes when each bank has 8 registers each 8 bytes wide

A register is 8 bits wide, and not 8 bytes.

Answer 2

Also, look at the Chapter 14 Figure 3 Memory Spaces chart here: (http://www.the8051microcontroller.com/select-figures) Hopefully, it will make the picture clearer.

Answer 3

In the 8051, there are 4 bank registers B0 to B3. Their memory address locations are
B0 - 00H - 07H
B1 - 08H - 0FH
B2 - 10H - 17H
B3 - 18H - 2FH

The default bank is B0. Each bank is 8 bytes. In each bank, there are 8 Registers which are 1 byte each R0 - R7. Each register is 1 byte that is 8 bits.

The banks can be switched by using the PSW (Processor Status Word) Register. To sum it up,
Each register is 8 bits(1 byte) R0 - R7
Each bank is 8 bytes B0 - B3