On the search for SRAM chips for the Z80.

[PA0PBZ]

But hey, the schematic would work-right?

The problem is that you are making it very hard to verify what it does, compare your drawing with this:

[makerimages]

That is indeed more readable-I agree. But.. is there an Eagle library for individual logic gates?

[PA0PBZ]

I don't have/use Eagle but it looks like it is possible:

[daveshah]

The 74xx-US, 74xx-EU and 40xx libraries (should be included in your Eagle installation) have parts split up into individual logic gates. Once you have placed the part, use the invoke tool (or right click and select Invoke) to place the remaining gates and power symbol - remember not to forget the power connections!

[makerimages]

Oh really? Great!

[makerimages]

Allright, ordered the parts I need to start off with from 2 suppliers-one order is useless without the other!
Now, I wrote a small piece of asssembly code to practice with

Code: [Select]

Start:
 .ORG 000h
 LD A,5
 LD B,A
.end
.end

Using tasm301 I got this .LST file https://www.dropbox.com/s/fmtij3zcwir4szx/TRY.LST

format:

Code: [Select]

Line  Address Data Source

And now I face a challenge-at first I would like to bitbang the .LST file to the memory chip. What puzzles me is the

Code: [Select]

0003   0000 3E 05        LD A,5

line, 3E and 05 make a 16 bit value, z80 has 8 data bits-3E fits, 05 doesn`t, what I want to know is where the 05 would go. My guess is it goes to 0001 of memory. Am I right?

[makerimages]

So this is how I would bitbang it?

Code: [Select]

address                            data
0000000000         00111110
0000000001         00000101
0000000010         01000111

[makerimages]

hmm, there is still one thing. From what I know there is this indianness, beign big or small, I believe the z80 uses big, what does that mean to me, which end of address goes to a0 and which end of data goes to db0?

[grumpydoc]

Z80 is little-endian so the least significant byte of a multi-byte value goes at the lower address, so a jump to address 0x1003 will appear as

C3 03 10

in a hex dump of memory

[makerimages]

so basically I can do. a raw hex to bin conversion, add 0s in the beginning to form a byte and then just have the result going to the db from left being db0 and right end. db7, same with addresses? so the binary above is valid to enter from left to right

[grumpydoc]

left being db0 and right end. db7

Not quite

Your byte order is OK

Your bit order is correct as you have written it out.

Your bit order is wrong, however in your description above. d0 on the data bus is the least significant bit. When you write this out you will have bit 7 on the left and bit 0 on the right

IE:

address            hex    binary
                         D7    D0
                         /      \
0000000000         3E    00111110
0000000001         05    00000101
0000000010         47    01000111

[makerimages]

and address line?

[grumpydoc]

A0 is least significant

[makerimages]

so that is reversed above aswell?