I am trying to create a software delay. Here is a sample program of what I am doing:
Address Data Opcode Comment
1800 06 LD, B Load register B with fix value
1801 “ “ Fixed value
1802 05 DEC, B Decrement value in register B
1803 C2 JP cc Jump to 1802 if value is not 0
1804 02 - Address XX
1805 18 - Address XX
My question is how can I calculate the required fixed value to load into register B so that the process of decrementing the value until 0 takes 2 seconds?
In my manual the time given to run the instructions is based on a 4MHz CPU but the Z80 CPU I am using has a speed of 1.8MHz. Any idea how I can calculate this? Thanks. P.S here is the decrement (DEC) and jump (JP cc) instructions from the manual:
Instruction M Cycles T states 4 MHz E.t
DEC r 1 4 1.00
JP cc 3 10 (4,3,3) 2.50
If by 1.8MHz you mean exactly 1,800,000 Hz, then to get a 2 second delay you'd need to delay for 3,600,000 T-states. Your current delay loop takes 14 T-states per iteration, which means that your initial value for B would have to be 3600000/14 == 257143, which obviously won't fit in one byte.
The greatest number of iterations that you could specify with an 8-bit register is 256, and to reach 3,600,000 T-states with 256 iterations, each iteration would have to take 14,062 T-states. That's one big loop body.
If we use a 16-bit counter things start getting a bit more manageable. At 65,536 iterations we only need 55 T-states per iteration to reach a total of 3,600,000 T-states. Below is an example of what that could look like:
; Clobbers A, B and C
ld bc,#0
1$:
bit #0,a ; 8
bit #0,a ; 8
bit #0,a ; 8
and a,#255 ; 7
dec bc ; 6
ld a,c ; 4
or a,b ; 4
jp nz,1$ ; 10, total = 55 states/iteration
; 65536 iterations * 55 states = 3604480 states = 2.00248 seconds