Hello I want to build an 8 bit digital to analog converter and want to know if someone can help me with the resistance values
Also I will thankfull if someone has a better design

 

The way you do it is to create an R-2R resistance ladder. Here is an article that describes the technique.

Resistor ladder - Wikipedia

 

Your best bet for the weighted resistor values are:
7.5k
15k
30k
60k (=2x30k in series, or 2x120k in parallel)
120k
240k
and then you get stuck at 6 bits
You can put 2 x 7.5k in parallel for an extra one at the MSB end and 2x240k in series for an extra one at the LSB end,
but your resistor tolerances get quite important: need to be better than 1 part in 256 or 0.4%
. . . so, you're better off with the R-2R ladder.

If you only need 4 bits, then you can get 9-way resistor networks, which although aren't very good tolerance, and reasonably well matched.

2xR in parallel
R
2xR in series
4xR in series

You can do R-2R ladders with resistor networks as well. Use two in parallel for the "R" section, and it tracks on a pcb better than doing 2 in series for the 2R sections.

 

The circuit for a resistor DAC is a solved problem. I learned it in my electronics class in 1971.

 

If you are not concerned with accuracy, try these values:

470
1k
2k
3.9k
8.2k
15k
33k
68k


in place of
500
1k
2k
4k
8k
16k
32k
64k