Otdfbt

My understanding of how interrupts (and more specifically, interrupt daisy-chaining) works in the Z80 is limited, to say the least, so if I get anything wrong please correct me.

Anyway, let's say I have 2 PIO chips to control 2 different peripherals (lets say, a character LCD display, and some switches, but I'm sure that's not very relevant). I've connected the IEO pin of one PIO to the CPU's INT pin, and it's IEI pin to the other PIO's IEO. I then connected that PIO's IEI pin to a 5V rail. I got this information from this image: (but instead of the three different peripheral chips they use, I just have two PIO chips.)

[from here]

Anyway, my understanding is that this kind of configuration means that if the second PIO (the one directly connected to the 5V rail) pulls its IEO pin low, the other PIO will not be able to send an interrupt. Correct?

My issue is this: How would I cause a PIO chip's IEO pin to go low? And how can I actually send an interrupt from one of these peripherals? My initial thought would be that I'd just pull the CPU's INT pin low, but that doesn't make sense the more I think about it.

(On a side note, any idea what the INTACK pin is on that diagram?)

You might want to consult the chapter for 'Interrupt Response' of the CPU manual (Chapter 8 on p.55 of the 1976 issue) and 'Interrupt Servicing' from the PIO manual (Chapter 6 on p.15 of the 1977 issue).

In general the Z80 supports 3 different interrupt modes:

• Mode 0 - like 8080, here the interrupting device must place an instrution on the bus - usually done by a 8259 interrupt controller.




• Mode 1 - All interrupts will jump to 38h (restart) and execute from there. Much like NMI




• Mode 2 - Peripherals put an interrupt service number on the bus during interrupt response, the CPU uses for an indirect call thru a table of pointers in a 256 byte page pointed to by the I register.

The picture inserted hints that you intend to use Z80s Mode 2. To make it work:

• Write some service routine for your interrupt, ending in an RETI


• Disable interrupts


• Setup a vector table in a memory page (256 byte boundry)


• Pick any vector number you like (lets say 3)


• Put the address of your service routing to that vector (xx06/xx07)


• Put the vector number into the PIO channels vector register (control word with 2^0 set)


• Set the Interrupt control word (x7) for the port to define which bits and under wich configuration will trigger an interrupt. For example 97h FFh sent to Port A will make all inputs issue interrupts as soon as they go high.


• Set interrupt Mmode 2 of the CPU


• Enable interrupts


• Enjoy whatever happens :))

A Z80 in Mode 2 is perfect suited for an interrupt driven system.

How would I cause a PIO chip's IEO pin to go low?

Err ... by waiting for an interrupt to occur, then serving it?

(Maybe I do not really understand what part of information is missing).

The short answer is that Z80 do NOT determine itself which peripheral sends an interrupt. In Zilog's framework, all compatible peripherals determine among themselves who's emitting the interrupt to the CPU this time. Or more specifically, who is sending IM2 vector on the bus during the time Z80 acknowledges the interrupt.

This kind of controlling prioritization of interrupts allows one building Z80 system not to use additional dedicated interrupt controller, as i8080 system designer had to do. However, the drawbacks are:

• extra pins every Z80-world compatible peripheral has to have.


• extra intelligence every such a peripheral has to have: particularly, they have to understand specific Z80 command (that is, RETI), whose the only purpose is to say to the peripherals the interrupt routine ends; otherwise RETI is fully equivalent to RET.


• probably the very long interrupt chains could have timing problems.