Interrupt Behaviour of the Z80 CPU
by Achim Flammenkamp
Interrupt timing
Only at the end of an instruction execution, except a NOP in case HALT, a LDD in case LDDR, a OUTI in case OTIR, etc., the CPU checks for an interrupt request.
Thus the INT-pin should be active for at least 23 clock ticks because some IX
resp. IY instruction last so long.
Now an interrupt is accepted if INT-pin is low(active) and the interrupt flip flop
IFF1 is set or the NMI-pin is respectively was active during the just performed instruction.
If this is the case, the following happens:
Immediately IFF1 is reset to 0 and if INT-pin is low also IFF2 is reset to 0.
Then the behavior depends whether it is a non maskable interrupt (NMI) or a
maskable interrupt (INT). In the later case the interrupt mode (IM) of the CPU is
also decisive:
- NMI
It takes 11 clock cycles to get to #0066:
  -  M1 cycle: 5 T states to do an opcode read and decrement SP
  
 -  M2 cycle: 3 T states write high byte of PC to the stack and decrement SP
  
 -  M3 cycle: 3 T states write the low byte of PC and jump to #0066.
 
   
 - INT and interrupt mode 0 set
In this mode, timing depends on the instruction put on the bus.
The interrupt processing last 2 clock cycles more
than this instruction usually needs.
Two typical examples follow:
a RST n on the data bus, it takes 13 cycles to get to 'n':
-  M1 cycle: 7 ticks
acknowledge interrupt and decrement SP
 -  M2 cycle: 3 ticks
write high byte and decrement SP
 -  M3 cycle: 3 ticks
write low byte and jump to 'n' 
 
With a CALL nnnn on the data bus, it takes 19 cycles:
-  M1 cycle: 7 ticks
acknowledge interrupt
 -  M2 cycle: 3 ticks
read low byte of 'nnnn' from data bus
 -  M3 cycle: 3 ticks
read high byte of 'nnnn' and decrement SP
 -  M4 cycle: 3 ticks
write high byte of PC to the stack and decrement SP
 -  M5 cycle: 3 ticks
write low byte of PC and jump to 'nnnn'. 
 
 - INT and interrupt mode 1 set
It takes 13 clock cycles to reach #0038:
 -  M1 cycle: 7 ticks
 acknowledge interrupt and decrement SP
  -  M2 cycle: 3 ticks
 write high byte of PC onto the stack and decrement SP
  -  M3 cycle: 3 ticks
 write low byte onto the stack and to set PC to #0038.
 
 - INT and interrupt mode 2 set
It takes 19 clock cycles to get to the interrupt routine:
 -  M1 cycle: 7 ticks
 acknowledge interrupt and decrement SP
  -  M2 cycle: 3 ticks
 write high byte of PC onto stack and decrement SP
  -  M3 cycle: 3 ticks
 write low byte onto the stack
  -  M4 cycle: 3 ticks
 read low byte from the interrupt vector
  -  M5 cycle: 3 ticks
 read high byte from bus and jump to interrupt routine 
 
 
Some remarks:
-  "acknowledge interrupt" means pin IORQ gets together with M1-pin
active/low. Furthermore the CPU adds 2 wait states to the normal machine cycle.
 - Quoted remark:
- If a LD A,I or LD A,R (which copy IFF2
to the P/V flag) is interrupted, then the P/V flag is reset, even if
interrupts were enabled beforehand.
 - If interrupts are disabled when a EI instruction is
interrupted, then the interrupt will not occur until after the
instruction following the EI, as when IFF1 is sampled
during the one and only machine-cycle of the EI, it will be reset.
 
 - An INT will let the INT-pin low(active) until it is indirectly
reseted by an RETI or other external hardware. In contrast a signal
at the NMI-pin will be recognized when its state going from high to low!
Thus it is not a level trigger signal, but a pulse which is stored internally
by the CPU until the next M1 cycle.
 
Reset Timing
One can consider a reset (RESET-pin gets low) as a very special interrupt.
If this occurs, the following happens in one machine cycle:
It takes 3 clock cycles:
- IFF1 and IFF2 as well as interrupt mode is set to 0
 - PC is set to 0, I and R registers are reset also.
 - SP is set to 0xffff as well as the A and the F register is set to 0xff.
 
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