selecting ram bank

[Fran123]

I have a question about the addressing of the memory banks

On a cpc6128 the combinations are as follows:

-Address- 0 1 2 3 4 5 6 7
0000-3FFF RAM_0 RAM_0 RAM_4 RAM_0 RAM_0 RAM_0 RAM_0 RAM_0
4000-7FFF RAM_1 RAM_1 RAM_5 RAM_3 RAM_4 RAM_5 RAM_6 RAM_7
8000-BFFF RAM_2 RAM_2 RAM_6 RAM_2 RAM_2 RAM_2 RAM_2 RAM_2
C000-FFFF RAM_3 RAM_7 RAM_7 RAM_7 RAM_3 RAM_3 RAM_3 RAM_3

understanding that the banks are distributed as follows:

0..64k RAM_0 RAM_1 RAM_2 RAM_3
64..128k RAM_4 RAM_5 RAM_6 RAM_7

If I want to use configuration C2 (page 0, combination 2) I will have:
RAM_4 RAM_5 RAM_6 RAM_7

where the formula for each bank is:
bank 1= page*4+4 (0*4+4)=4
bank 2= page*4+5 (0*4+5)=5
bank 3= page*4+6 (0*4+6)=6
bank 4= page*4+7 (0*4+7)=7

but if I have 8 pages of 64k (512K)

page 1 0..64k RAM_0 RAM_1 RAM_2 RAM_3
page 2 64..128k RAM_4 RAM_5 RAM_6 RAM_7
page 3 128..192k RAM_8 RAM_9 RAM_10 RAM_11
page 4 192..256k RAM_12 RAM_13 RAM_14 RAM_15
page 5 256..320k RAM_16 RAM_17 RAM_18 RAM_19
page 6 320..384k RAM_20 RAM_21 RAM_22 RAM_23
page 7 384..448k RAM_24 RAM_25 RAM_26 RAM_27
page 8 448..512k RAM_28 RAM_29 RAM_30 RAM_31


How is the memory configured with the FA combination (seventh page and third combination)?
because in this case the fourth bank would be 7*4+7 = 35

Thank you!

[lightforce6128]

I had a look here: https://www.cpcwiki.eu/index.php/PAL16L8#PAL_MMR_register

Combination FA is the second row in the diagram (fourth row with headlines). So this should be:
RAM_0  RAM_1  RAM_2  RAM_27

But this is only theoretical. I do not have experience with more than 128K on the CPC. Maybe somebody else can confirm?

[McArti0]

the formula for each bank is:
Setting = #C0 + Page * 8 + 4..7
Because bit2 does not change the page number but exists and changes the configuration within the page.
The page number selection starts with bit 3 with weight 8. Each page has 8 configurations but only 4 banks of ..16kB each