firashelou:
i didn’t really understand what do you mean 
Yes, I had the same feeling. In theory he is right, but I am struggling to figure out how to implement it.
So we define S1, S2, S3, S4 as bit 0, bit 1, bit 2, bit 3, we define R1 as 1k, R2 as 2k, R3 as 4k, R4 as 8k.
This would give equivalent resistances of
0 0 0 1 1000.00
0 0 1 0 2000.00
0 0 1 1 666.67
0 1 0 0 4000.00
0 1 0 1 800.00
0 1 1 0 1333.33
0 1 1 1 571.43
1 0 0 0 8000.00
1 0 0 1 888.89
1 0 1 0 1600.00
1 0 1 1 615.38
1 1 0 0 2666.67
1 1 0 1 727.27
1 1 1 0 1142.86
1 1 1 1 533.33
Which is clearly useless. So reversing the order of value for R1-R4 gives R1 as 8k, R2 as 4k, R3 as 2k, R4 as 1k and thus equivalents of :-
0 0 0 1 8000.00
0 0 1 0 4000.00
0 0 1 1 2666.67
0 1 0 0 2000.00
0 1 0 1 1600.00
0 1 1 0 1333.33
0 1 1 1 1142.86
1 0 0 0 1000.00
1 0 0 1 888.89
1 0 1 0 800.00
1 0 1 1 727.27
1 1 0 0 666.67
1 1 0 1 615.38
1 1 1 0 571.43
1 1 1 1 533.33
Which now at least runs in chronological order…
So now choosing an arbitrary value for R5 of 2k would give analogue values of :-
0 0 0 1 204
0 0 1 0 341
0 0 1 1 438
0 1 0 0 511
0 1 0 1 568
0 1 1 0 613
0 1 1 1 651
1 0 0 0 682
1 0 0 1 708
1 0 1 0 730
1 0 1 1 750
1 1 0 0 767
1 1 0 1 782
1 1 1 0 795
1 1 1 1 807
Maybe Paul__B can pick up from here… I can’t figure it out beyond here…
Regards,
Graham
