CS222 Lab Lecture: Introduction to the MPF-I and Z80
Need:
1. MPF-I's to sign out
2. Transparencies: Overall block-diagram of MPF-I
Z80 Architecture
Z80 Instruction set (4)
Z80 Memory
Z80 IO Timing
Excerpt from Z80 Timing Table
Z80 pinouts
Z80 M1 cycle timing
Z80 INT timing
Z80 NMI timing
Z80 BUSRQ timing
MPF-I circuit diagrams (3)
74LS138 diagram
3. Handouts: Overall block-diagram of MPF-I
Monitor, editor, and assembler commands
Z80 Architecture;
Z80 Instruction set (4)
4. Copy of Z80 Microprocessor Programming and Interfacing
I. Overview of the MPF-I hardware
A. The MPF-I is a one-board microcomputer, built around a Z80 8-bit
microprocessor.
B. Show one to the class. Each box will contain:
1. MPF-I unit itself
2. Power supply
3. Three manuals: User guide, monitor source code listing, experiment
manual. Of these, the user guide will be most helpful.
4. Each team will be issued an MPF-I, for which they will be responsible.
(You may take your manuals out of the lab overnight for study, but you
are responsible for returning them.)
5. We also have some additional boards that connect to the MPF-I. These
must be shared:
a. Two printers
b. Three EPROM programmers
c. Two speech synthesizers (not used in any required labs, but
available for optional use if desired.)
C. The MPF-I consists of a Z80 CPU, a ROM containing a monitor, editor, and
two assemblers (all in less than 8K!), 4K of RAM, a 48 character keyboard,
and a 20-character display.
D. Overall block-diagram of MPF-I. (TRANSPARENCY, HANDOUT)
1. Note: we will look at detailed schematics later in the lecture.
2. Show chips on board.
System bus (40 bits)
===================================================================> Connector
|| || || || || || || to other
------- ------- ------- ------- ------- ------- ------ boards
| Z80 | | 8K | |open | | 2K | | 2K | |3 IO | |3 IO |
| CPU | | ROM | |slot | | RAM | | RAM | |ports| |ports|
| | | | |for | | | | | | | | |
| | | | |EPROM| | | | | | | | |
| | | | | | | | | | | | | |
| | |0000-| |2000-| |F000-| |F800-| | | | |
| | |1FFF | |3FFF | |F7FF | |FFFF | | | | |
------- ------- ------- ------- ------- ------- -------
| | | |
Keyboard ____| |__ ___| |
monitor, | | | |
Editor, Keyboard Display Cassette
Assembler, interface
Utility & speaker
Routines
E. Note that Memory addresses are assigned as follows:
1. 0000-1FFF (hex) are the monitor, editor, assemblers etc.
2. 2000-3FFF (hex) are dedicated to slot U3 on the board, into which one
can insert an EPROM containing a user-written program or data (as we
shall do later.)
3. F000-FFFF (hex) are the on-board RAM. These addresses are important
to note, because programs we write will be stored here.
4. The remaining addresses are available for off-board expansion - such
as the dissassembler that is built into the printer board.
II. Overview of the MPF-I built-in software
A. When the MPF-I is powered up, control passes to the monitor at memory
address 0. The monitor then displays the start up message MPF-I PLUS.
Once you start entering commands, however, the monitor prompt becomes
an upside down 7 followed by a carat (SHOW).
B. You should think of the monitor as playing a role similar to the command
language interpreters on the VAX. The upside-down 7 + carat prompt is
analogous to the $ prompt of DCL.
C. When at monitor command level, you can enter one of the other packages
such as the editor or assembler. The commands for so doing are
summarized in the top box of the handout. NOTE WELL: These are control
keys, not letter keys - e.g. to re-enter the monitor, type CONTROL-Q.
(CONTROL-Q functions like CONTROL-C on the VAX). In addition to
control-Q, there are two other control characters we want
to note: Control-P toggles the printer on and off, and Control-G toggles
the keyclick bell on and off. (SHOW).
D. For our early work, we will primarily be using the editor and the
two-pass assembler. Note that the former has two entry points:
CONTROL-E to enter the editor for creating a new program, and CONTROL-R
to re-enter it to edit an existing program. We will discuss the details
of using these two facilities first, before going on to talk about other
monitor commands.
1. When you first enter the editor using CONTROL-E, it will ask you for
the starting address of a memory region it can use as a text buffer.
For now, pressing return to accept its default value will be the
simplest solution.
2. The editor has basically two modes of operation: command mode, in
which you issue commands to move around in the text, make changes
etc., and input mode, in which you type new text.
a. When you enter the editor to create a new program you are thrown
into INPUT mode immediately.
b. To enter the input mode otherwise, type I return.
c. To leave input mode, enter a blank line.
3. GO OVER OTHER COMMANDS ON HANDOUT. NOTE THAT ALL CAN BE ENTERED AS
FIRST LETTER.
4. To leave the editor, type Q in command mode.
5. To enter the assembler, type CONTROL-A. Like the editor, it will
begin by prompting you for RAM addresses for the program it is to
assemble, for its symbol table, and for the object code. For now,
just press return in response to the prompts to accept the default.
6. The assembler makes two passes over the text. During pass 1, it
reports errors. An error will be reported as a line number, a
memory address where its code is going, and an error code - eg.
3. FC04 *I*. When you have noted the error, press return and the
assembler will go on - possibly to find further errors!
7. You may use the editor G command to go to the offending line number
in order to try to figure out your error.
E. When at monitor command level you also have immediately available to you
ODT-like functions. These are summarized on the rest of the first page
of the handout. NOTE that these commands are invoked by typing the
letter specified - not by control characters. (Very confusing - you
should probably write in CTRL before the Q, E, R, A, L, D, B, and C in
the top box.) Also, note that while the commands in the first box are
executed immediately upon pressing the specified control character, the
remaining commands require you to press the letter then press return.
(NOTE: SEE MPF-I USER'S MANUAL CHAPTER 4 FOR FURTHER INFO.)
1. The R command allows you to display or alter the registers.
a. If you type R followed by a register name, the specified register
is displayed. If it is an 8-bit register, its partner is displayed
with it.
b. If you just type R, the AF and BC pairs are displayed.
c. To look at different registers, you can press the up or down
arrow keys. The down-arrow steps through the registers (2 16-bit
pairs at a time) in this order:
AF and BC
DE and HL
A'F' and B'C' (displayed as A.F. and B.C.)
D'E' and H'L'
IX and IY
SP and PC
PC and I
d. To alter a register, one types R, followed by the register name,
followed by a colon, followed by the new value (either 2 or 4
hex digits) followed by return.
2. The M command allows you to display or alter memory locations. (Of
course, you can only examine existing locations, and you can only
alter locations in RAM - F000 to FFFF.)
a. To examine a location, type M followed by a 4-digit address and
return. MPF-I displays 4 consecutive bytes starting at that
address.
b. As with registers, the uparrow and downarrow keys may be used to
step forward or backward through the range of addresses.
c. To alter a location, type M followed by its address followed by
a colon and a new value. You can alter several consecutive bytes
at once by entering new values (as pairs of hex digits) separated
by spaces.
d. M followed by an address followed by a . followed by a second
address can be used to dump a range of memory to the printer. (Of
course, the printer must be connected.)
3. The I command allows you to insert one or more bytes of data, with
the rest of memory contents being shifted up to make room for it.
a. Type I, followed by the address AFTER WHICH the new data is to go,
followed by one or more data bytes.
b. All bytes up to FE00 will be shifted up to make room for the newly
inserted bytes. Of course, this means that some bytes will be
shifted off the end of memory. (The address space above FE00 is
reserved for monitor scratchpad work and is not altered) Note: FE00
is the default limit. You can specify a different limit if you
wish.
4. The D command allows you to delete a byte of data, with
the rest of memory contents being shifted down to make room for it.
A zero is filled into the vacated byte. Again, memory above FE00 is
untouched unless you specify a different limit.
5. The G command allows you to start program execution. Just entering G
followed by return will start you at the current PC value; however,
you can enter a different starting address after the G if you wish.
6. The B command allows you to set a breakpoint at a specified address.
Only one breakpoint may be active at a time.
7. The S command allows you to single step a program. S by itself
executes the instruction pointed to by the PC; or you can type S
followed by an address.
8. The W command allows you to write to cassette tape. You type W,
followed by the start address of the memory range you wish to write,
followed by a space, followed by the end address, followed by a
space and a 4 character filename. Be sure the recorder is running
before pressing return.
9. The L command, followed by a filename, reads a file back from tape
to the location in memory from which it was written.
10. The J command causes MPF-I to calculate for you the relative address
needed in a relative jump instruction at a specified address jumping
to a specified absolute address.
III. Z80 CPU architecture
A. The Z80 was developed by a team of engineers who formerly worked for
Intel, the company that developed the first microprocessor. At the
time, Intel's most powerful chip was the 8080 (the chip on which many
home computers - including all that run CP/M - were built). Intel
had begun an internal effort to develop an improved 8-bit microprocessor.
Evidently, the team that ultimately developed the Z80 wanted to go
farther from the original 8080 design than did corporate management. The
result was that they left Intel to form Zilog corporation and to develop
their own chip. (Intel went on to release an improved version of the
8080 known as the 8085; but the changes were mostly in terms of external
connections such as power supply requirements rather than in terms of
internal architecture.)
B. One of the design goals of the Z80 was that it should be able to run
software written for the 8080. This causes some peculiarities in the
instruction set, as we shall see.
C. The Z80 is a pure 8-bit microprocessor, which means that its internal
registers and data paths (as well as its external data bus) are 8 bits
wide. Since 16-bit integers are the minimal size for many applications,
this means that many ordinary integer arithmetic operations
must be done in two separate 8-bit steps. However, addresses are 16 bits
long; therefore, the internal data paths for addressing are 16 bits
wide. We shall see that provision is made at several points to join
two 8-bit registers together to form a 16-bit pointer to a memory
location. TRANSPARENCY OF STRUCTURE - GO OVER.
1. Note external connections: 8-bit data bus, 16-bit address bus, plus
13 control lines. Together, these form one PHYSICAL BUS and two
LOGICAL BUSSES: a memory bus and an IO bus. The two logical busses
use the same data and address busses, but different control lines.
The memory bus is used for all instruction and data fetches, and the
IO bus by the IN and OUT instructions.
2. We will discuss details of the bus structure later.
D. The Z80 is also basically a one-accumulator machine, which means that for
most arithmetic and logical operations the A register (the accumulator)
contains one of the source operands and receives the result of the
operation. Again, this contrasts with the VAX, which is a general
register machines on which any of the 16 general registers can
participate fully in arithmetic and logical operations.
E. The 8080 has 8 8-bit registers and 2 16-bit registers which are carried
over directly into the architecture of the Z80:
1. The 16-bit PC contains the address of the next instruction to be
executed (as on the VAX); however, it is not a general register
and cannot be used for any other purpose as the PC on the VAX can.
2. The 16-bit SP is a stack pointer; again, it is not a general register
and cannot be used for anything else.
3. The 8-bit A register is the accumulator, and participates in most
arithmetic and logical operations.
4. The 8-bit F register plays a role similar to that of the PSW on the
VAX. Actually, only 5 of its bits are used - the other 3 have
undefined values. These bits serve as condition codes to reflect the
results of various arithmetic and logical operations.
5. The 8-bit registers H and L are seldom used as 8-bit registers. More
often they are paired to form a 16-bit register that is used to point
to a memory location. Many memory-reference instructions require that
the address of the item to be fetched or stored be in HL. (This
corresponds roughly to mode 6 - register deferred - addressing on the
VAX - but only this one register can be used.)
6. The 8-bit registers B,C,D and E provide temporary storage for
intermediate results of operations, and can also be paired up (BC,
DE) to form 16-bit registers that can be used as pointers to memory
cells.
7. In some contexts the F register can be paired with A to form a 16
bit AF register. This is primarily done when pushing registers on
a stack - they are pushed 16 bits at a time, so A and F are pushed
together. (Of course, they cannot be used together as a pointer to
memory as the other pairs can!)
F. The Z80 more than doubled the number of bits of registers of the 8080.
1. In place of a single set of registers A,B,C,D,E,F,H,L, the Z80 has two
sets - one designated A', B' etc. However, only one set can be in use
at any time. The only operations possible on the alternate set is an
exchange operation which swaps the two register sets wholesale. The
primary use of this feature is to allow operating system components
(such as interrupt handlers) to be able to use registers without
destroying application-program data. (Switching register sets is much
faster than pushing all the registers on the stack.)
2. The Z80 has two index registers - IX and IY - which allow for indexed
addressing similar to VAX displacement modes A,C,E. (The so-called
index mode on the VAX is different). There was no provision for this
on the 8080.
3. The Z80 has an I register that is used to provide more sophisticated
interrupt vectoring (more on this later) and an R register to provide
for automatic refresh of dynamic memory (again, more on this later.)
4. For our purposes, the IX and IY will be of considerable use; however,
we will defer use of the others until an appropriate time later in the
course.
IV. Overview of the Z80 instruction set
A. We have noted that the Z80 architecture is derived from that of the 8080.
This causes some peculiar patterns in the instruction set:
1. The 8080 uses instructions with one byte (8-bit) opcodes, possibly
followed by 1-2 bytes of immediate data or direct address - yielding
an instruction length of 1-3 bytes:
op__code
or op__code imm_data
or op__code imm_data imm_data
or op__code address_ address_
2. Theoretically, the 8080 could have 256 different operations. However,
it only has 244, leaving 12 of the possible opcodes unused.
3. The Z80 uses 8 of the unused 8080 opcodes for instructions not
available on the 8080 - making a total of 252 instructions with a
one-byte opcode. The remaining 4 codes serve as "windows" to an
extended instruction set with opcodes of 2 or 3 bytes plus possibly
1-2 bytes of immediate data or address, producing instructions up to 4
bytes long:
_window_ op__code
_window_ op__code imm_data
_window_ op__code imm_data imm_data
_window_ op__code address_ address_
_window_ op__code op__code
_window_ op__code op__code imm_data
4. Our approach will be to begin our discussion with the one-byte
opcodes. Then we move on to the 2-3 byte extensions.
5. SHOW TRANSPARENCY OF Z80 1-BYTE INSTRUCTIONS; HAND OUT
B. NOP 0000 0000; HALT 0111 0110
C. The 8-bit LD group: Similar to VAX MOVB instruction. (Note: the VAX
uses one instruction with a variety of modes/registers; Z80 uses a similar
approach but lists each as a separate instruction.) Also, note that the
flags (condition codes) are unaffected (contrast with the VAX)
1. Register addressing mode: 0100 0000 .. 0111 1111
a. Note that, in contrast to the VAX, the first register named
is the destination and the second is the source. This is true
both in assembly language and in machine language.
b. Note pattern to format: 01dddsss
000 = B; 001 = C; 010 = D; 011 = E; 100 = H; 101 = L;
110 = (HL) (8080 mnemonic = M); 111 = A.
c. Note use of register indirect (HL) to address memory. To move data
to/from a memory location using these instructions, one must first
place the 16-bit address in the HL pair (using other instructions
to be covered shortly.)
d. Slot that would be LD (HL) (HL) is HALT instead.
2. 8-bit immediate loads (two-byte instruction) 00___110 _value__
3. Further addressing options with A only:
a. Register (other than HL) indirect: 000__010
b. Absolute address: 0011_010 _addrLSB _addrMSB (3 byte instruction)
[Note byte reversed format: least significant byte of address is
first, then most significant. This is true across the board for
16-bit values.]
D. 8-bit arithmetic/logical group: Note that destination operand is always A.
This may be implicit or explicit in the assembly language syntax.
(Explicit mention of A is needed in ADD, ADC, SBC because there are
versions of these for 16-bit register pairs as we shall see later.)
These instructions do set the flags (condition codes).
1. Register-mode addressing: 1000 0000 .. 1011 1111
Encoding pattern: 10fffsss (fff = function)
a. 000 ADD - 2's complement; can also be used for pair of BCD
digits in conjunction with DAA (see discussion below).
b. 001 ADC - adds a number plus carry from previous operation.
(Like ADWC on VAX.)
c. 010 SUB - again can be 2's complement or BCD.
d. 011 SBC - like ADC
e. 100 AND, 101 OR, 110 XOR - bitwise operation
f. 111 CP = compare; A-source is used to set flags but isn't
stored.
2. Immediate mode addressing: 11fff110 _value__
E. 8-bit increment (00___100); decrement (00___101)
1. Note encoding: middle three bits specify register as with LD.
2. As on the VAX, these do set the flags.
F. 8-bit miscellaneous: 00___111
/------>/
1. RLCA - C <- 7 <--- 0
/<------/
2. RRCA - 7 ---> 0 - > C
/----------->/
3. RLA - C <- 7 <--- 0
/<-----------/
4. RRA - 7 ---> 0 -> C
5. DAA (A register): This can be done immediately after an ADD or SUB
involving the A register, and corrects the A register value and carry
to what it should be if the values added were interpreted as a pair
of BCD digits instead of an 8-bit binary number.
a. Consider what would happen if a pair of BCD digits were added as
if binary - e.g.
11 + 22 = 33 - ok
14 + 66 = 7A - wrong - should be 80 (the first nibble should have
carried into the second instead of becoming
A)
18 + 68 = 80 - wrong - should be 86 (the internibble carry occurred,
but the low nibble is 6 too low)
etc.
b. The Z80 flags include a half-carry flag H, which is set by ADD/SUB
to record the inter-nibble carry. The DAA instruction uses this
flag, together with the contents of the A register, to make a
correction so that the ADD/SUB comes out correctly for BCD. For
example, in the above cases:
11 + 22 = 33 - H is not set, both nibbles of A are valid BCD digits,
so DAA would do nothing
14 + 66 = 7A - H is not set, but low nibble of A is not a valid BCD
digit. DAA automatically adds 06 (to compensate for
the 6 unused codes), yielding the correct result 80.
18 + 68 = 80 - H is set, so even though both nibbles are valid BCD
digits DAA automatically adds 06, yielding the
correct result 86.
(Note: DAA uses the C flag in conjunction with the upper nibble in
a similar way.)
c. Thus, to add two 2-digit BCD numbers (say one in A, one in B), code:
ADD A,B
DAA
To add two 4-digit BCD numbers (say one in B,C, other in D,E, with
result in B,C), code:
LD A,C
ADD A,E
DAA
LD C,A
LD A,B
ADC A,D
DAA
LD B,A
d. DAA works with subtraction as well as addition. The N flag in the
CPU records whether the last operation was add or subract, and the
DAA uses this to adjust accordingly.
6. CPL (1's complement A)
7. SCF, CCF: Set, clear carry. This can be used before operations like
RLA, RRA, ADC or SBC to control (initial) value of the C flag.
G. 16-bit operations
1. Immediate loads to register pairs 00rr0001 _LSBdata_ _MSBdata_
[Note byte-reversed format again]
2. HL load/store with direct memory address: 0010_010 _LSBaddr_ _MSBaddr_
3. Add register pair to HL: 00rr1001
4. Increment/decrement register pair: 00rrf011
5. Load SP from HL 11111001
H. Conditional/unconditional transfers of control
1. As on the VAX, the Z80 sets certain condition code bits as a
result of various operations, and these can then be tested for
conditional jumps and other operations. The condition codes are called
flags, and the register containing them is the F register, laid out
as follows:
S Z x H x P/V N C
Meaning of the flags:
a. S = sign bit of result. (Referred to in mnemonics below as P ->
result was >= 0, therefore S=0; M -> result was < 0 therefore S=1).
(Similar to VAX N bit. Note that Z80 has an N bit but it
has a very different meaning.)
b. Z = result was 0. (Same as VAX Z bit)
c. P/V = parity or overflow. This flag is used for two different
functions depending on the operation just performed:
i. After a bitwise logical instruction such as AND, OR, XOR, this
flag is set if the result had even parity, and cleared if it had
odd. (The VAX has no equivalent condition code bit.)
ii. After an arithmetic operation such as ADD the flag will be set to
denote overflow. (Same as VAX V bit.)
(Mnemonics use PE = parity even or PO = parity odd. There is no
mnemonic for overflow; note that JP PE,__ is equivalent to jump if
overflow occurred.)
d. C = carry out of arithmetic operation such as ADD or bit lost on a
shift. (Similiar to VAX C bit.)
e. H and N are set on ADD/SUB type operations to retain half carry for
use with DAA, and to record whether the operation was an ADD (N=0)
or SUB (N=1).
2. Conditional jumps: 11ccc010 + two-byte direct address
001cc000 + one-byte relative offset (Z80 unique)
Note assembler syntax for conditional form: JP cc, dest or JR cc, dest -
where cc is one of the mnemonics:
NZ -> Z flag clear (result <> 0)
Z -> Z flag set (result = 0)
NC -> C flag clear
C -> C flag set
PO -> parity odd: P/V flag clear
PE -> parity even: P/V flag set
P -> S flag clear: result >= 0
M -> S flag set: result <0
Note also that JR allows only some of these possibilities. This is
because JP was an 8080 instruction, while JR is unique to the Z80.
There were not enough one-byte opcodes left to give JR the same
flexibility as JP.
3. Unconditional jumps: 11000011 + two-byte direct address
11101001 : jump to address in HL
00011000 + one-byte relative offset (Z80 unique)
4. CALLs (return address pushed on stack): 11ccc100 or 11001101 + 2-byte
direct address).
Note that these can be unconditional or conditional as with jumps.
(A conditional JSR)
5. Subroutine returns (pop stack): 11ccc000 or 11001001. These can be
unconditional or conditional as with jumps.
6. Resets (software interrupts) 11___111. We will discuss these later.
7. DJNZ + one byte relative address: unique to Z80
(Decrements B - jumps if B<>0): 00010000
I. Stack operations, exchanges:
1. 16-bit pushes: 11rr0001 - where rr designates one of the pairs
AF, BC, DE, HL
2. 16-bit pops: 11rr0101 - rr as above
3. 16-bit exchanges: 00001000 (unique to Z80) - exchanges two versions
of AF
11011001 (unique to Z80) - exchanges two versions of
remaining registers
11100011 - exchanges HL register with top item on
stack (memory cell pointed to by SP)
11101011 - exchanges DE register pair with HL pair
J. Input output:
1. OUT: 11010011 + port
2. IN: 11011011 + port
3. Interrupt enable disable: 1111_011
K. Two byte opcodes with prefix = CB (all unique to Z80) (TRANSPARENCY)
1. Extended shift/rotate operations: CB + 00______. On the 8080, only
the A register could be shifted/rotated. This group allows
shifts/rotates to be done on other registers or a memory cell pointed
to by HL, plus adds three types of shift not found on 8080.
a. RLC, RRC: circular with 8-bits of register, CF also set (cf
RLCA, RRCA above). Note that this group includes an operation on A,
but for this register use the one-byte instructions RLCA, RRCA.
b. RL, RR rotate treating A + CF as 9 bits - see RLA, RRA above
c. SLA, SRA: arithmetic shifts - propagate sign on right shift, shift
in zero on left shift; CF always captures bit lost. (If needed on
A register, must use this form since there is no one-byte form.)
SLA: C <- 7 <--- 0 <- 0
/---/
SRA: /-> 7 ---> 0 -> C
d. SRL: logical right shift: shift in zero to sign; CF captures bit
lost.
0 -> 7 ---> 0 -> C
2. Bit test, reset, set operations: CB + 01______, 10______, 11______
a. Note format: oprrrbbb.
b. On BIT, the bit of register rrr specified by bbb is tested, and the
Z bit is set if tested bit was 0 (not set).
c. SET sets bit bbb of register rrr to 1, and RES resets it to 0.
d. On all of these, bit 7 is the MSB (sign), bit 0 the LSB.
L. Two byte opcodes with first byte = DD, FD: allow use of IX, IY
1. DD + op: TRANSPARENCY - compare to basic one-byte instructions and
note that IX occurs where HL occurs there, <IX+disp> where (HL)
occurs.
a. The <IX+disp> form allows for a form of indexed-mode addressing
(cf VAX modes A,C,E). The effective address for the
operation is formed by fetching a byte from the instruction stream
and adding it to IX to form an address. This is very useful for
stepping through tables, etc.
b. Note that IX is only accessible through these multi-byte opcodes.
Recall that IX (and IY) were not part of the original 8080
architecture.
2. DD + CB + op: TRANSPARENCY-compare to two-byte beginning with CB;
again IX fills HL slots.
3. FD + op, FD + CB + op are similar to the above, save they use IY.
M. Two byte opcodes with first byte = ED (TRANSPARENCY)
1. Extended IN, OUT with port number in C; data going to/from any register
ED + 01___000, 01___001.
(8080 IN, OUT instructions - also present on Z80 - transfer data
to/from A and require port number as second byte of instruction)
2. 16-bit ADC, SBC allows 32+bit operations: ED + 01___010: C bit is set
by carry/borrow out of one 16-bit add/subtract and then can be brought
into another.
3. Direct addressing 16-bit loads, stores for register pairs:
ED + 01___011
4. NEGate register A (2's comp): ED + 01000100
5. Control for more sophisticated Z80 interrupt mechanism (more on this
later): ED + 0100_101, 01___110, 01__0111.
6. "Nibble" oriented rotates: ED + 0110_111. These rotate a nibble
between the A register and the memory location pointed to by HL.
a. RLD: A Memory cell pointed to by HL
high <-- low <--- high <--- low <--
nibble | nibble nibble nibble |
| |
----------------------------------------------
(unchanged)
b. RRD: A Memory cell pointed to by HL
high --> low ---> high ---> low --
nibble | nibble nibble nibble |
| |
----------------------------------------------
(unchanged)
7. Block transfer instructions: ED + 10______:
a. Common to all is the use of the BC pair as a byte counter, and
HL and possibly DE as string pointers.
b. The "non-R" forms perform a single byte transfer, adjusting the
pointers (HL and possibly DE) up ("I" form) or down ("D"), then
decrement BC.
c. The "R" form repeats the operation of transfer until BC becomes 0 -
i.e. it loops on the non-R form.
d. LDxx allows movement of a string of bytes from a location pointed
to by HL to a location pointed to by DE.
e. CPxx allows comparison of a string of bytes at a location pointed to
by HL to A. Comparison will terminate on the first match to the
value in A, or when BC becomes 0 ("R" form).
f. INxx and OTxx allow transfer of a string of bytes pointed to by HL
to/from an IO port specified by C. For these operations, B alone is
used as the byte counter.
V. Calculating Z80 instruction timing and program run times
A. In our previous work in programming, we have sometimes dealt with the
matter of the execution times of programs from the standpoint of
efficiency - seeking to write a program that runs in the shortest
possible time.
B. In the case of microprocessors, however, we frequently encounter
situations in which our concern is with accuracy: knowing exactly how
much time a given program will take to run. This is because the
interface between the microprocessor and the outside world sometimes
depends on precise timing relationships for the sending and receiving of
information.
C. Lab #9 is intended to get you into the matter of timing by having you
program the MPF-I to behave as a stopwatch. Therefore, you will have to
determine the exact amount of time needed for the execution of each
section of the program you write.
D. On most microprocessor systems, timing is controlled by an external clock
which is an oscillator that generates pulses at a fixed rate. Since all
instruction times are a multiple of the clock period, this clock is
normally crystal-controlled to ensure accuracy. On the MPF-I, the clock
frequency is 1.79 mhz. This is about 30% less than the maximum
permissible frequency for the chip used - 2.5 mhz. (Note: other versions
of the Z80 can use clock frequencies up to 4.5 mhz.) Using a less than
maximal clock frequency helps to ensure reliable system performance.
E. Given a knowledge of the clock frequency, one need only figure out the
number of clock pulses needed for the execution of a given instruction to
determine its execution time. This information is obtained from tables
supplied by the microprocessor manufacturer.
1. For example, the instruction LD A,B requires four clock pulses to
execute, and therefore takes 4/1.79*10-6 sec-1 = 2.234637 microseconds.
(On a Z80 with a 4.5 mhz clock it would take 4/4.5 * 10-6 sec-1 =
.8888889 microseconds.)
2. We will see how to use the necessary tables shortly.
F. What are the factors which give rise to the particular execution timing
of a particular instruction? To answer this, we must examine how the Z80
actually executes an instruction, as well as the protocol for the Z80's
communication with the outside world:
1. The execution time for a given instruction is composed of a number of
major cycles, called M-cycles in the Z80 documentation. Normally,
each M cycle involves the transfer of one byte of data between the Z80
and either memory or an IO port. Sometimes, however, an M cycle is
dedicated to internal processing within the chip itself.
a. For example, consider the instruction
ADD A,B - add the contents of register B to register A
This instruction is 1 byte long (just the opcode) and involves
no further access to memory since all operands are in registers.
Executing it, therefore, involves one major (M) cycle - that
needed to fetch the instruction in the first place.
b. Again, consider the instruction:
LD A,(nn) - load the A register with the byte contained at memory
address nn.
This instruction occupies 3 bytes: 1 for the opcode and 2 for the
address nn. Fetching this instruction therefore involves reading
3 bytes from memory. In addition, having fetched the instruction
it becomes necessary to go to the designated memory address to fetch
a value to load into A. This requires a 4th memory access. Thus,
this instruction requires 4 major (M) cycles:
M1 - fetch the opcode byte
M2 - fetch the low-order byte of the address word
M3 - fetch the high-order byte of the address word
M4 - fetch the operand
2. Each M cycle, in turn, requires anywhere from 3 to 6 clock pulses or T
states for completion. For example, in the above M1 requires 4 T
states and the remaining cycles require 3 each, for a total of 13 T
states for the entire instruction.
3. To understand the relationship between M cycles and T states, we need
to consider what the Z80 has to do to transfer a byte of data to/from
memory or an IO port.
a. The clock for the Z80 is a square wave. Each T state corresponds
to a complete cycle of this square wave. Thus, there are two
clock transitions per T-state.
i. The rising transition signals the end of one T-state and the
start of the next.
ii. The falling transition occurs in the middle of the T-state.
b. During the course of a data transfer, the Z80 generates a sequence
of control signals that serve to synchronize the processor and the
memory or external device. Each change in a control signal
coincides with one of the transitions of the clock.
c. Example: Memory read:
TRANSPARENCY - MEMORY READ/WRITE SEQUENCES (From Carr p29)
i. To initiate a memory read, the Z80 places a 16-bit address on the
____ __
address bus and asserts the control signals MREQ and RD (memory
request and read.) Note that these signals are "low true": their
asserted state is 0V, not 5V.
- The address is put on the bus at the start of the first T-state
of the M-cycle, on the rising transition of the clock.
- The control lines are asserted in the middle of the T-state,
on the falling clock transition. This allows time for the
address to "settle" on the bus so that no device will
misinterpret it when the control signals are asserted.
ii. The two control signals remain asserted for a total of two
clock times - until the middle of the third T-state. This
allows time for the memory to recognize the command and
respond by putting the data on the bus. The Z80 actually
reads the data during the first half of the third T-state,
just before deasserting the control signals.
iii. Note that the address actually remains on the bus until the
end of the third T-state. This is done so that the address
bus does not change value during the time that the control
signals are still asserted, lest the wrong memory see an
address in transition and the control signals asserted and
respond when it shouldn't.
iv. Actually, this diagram assumes that the memory is able to get
its data up on the bus not more than 1-1/2 clock periods after
the request. Some memories may not be fast enough to do this.
To allow such memories to function, the Z80 has a control line
____
called WAIT (low true). A slow memory may assert this to force
the Z80 to inject one or more wait states (Tw) before capturing
the data on the third and final T-state. The timing tables we
will use assume no wait states will occur (which is true for the
MPF-I.)
d. Example: memory write
e. Example: I/O read write
TRANSPARENCY - I/O READ/WRITE SEQUENCES (from Carr page 30)
f. Example: Opcode fetch - see Carr p. 27. The opcode fetch (M1)
cycle is a special variant of the memory read which takes an
additional T-state to allow for memory refresh. We will come
back to this later.
G. To find out the number of clock pulses needed to execute a given
instruction, one can consult an appropriate table.
1. Two places to look are
MPF-I user's Guide appendix C, beginning with page C14.
Z80 Microprocessor Programming and Interfacing, appendix A.
2. In either case, the place to look is the column headed "Number of T
states".
EXAMPLE: TRANSPARENCY SHOWING M CYCLES AND T STATES FOR A NUMBER OF
INSTRUCTIONS FROM THE 8-BIT LOAD GROUP
3. Example: timing calculation for the following program segment:
LD C,A 1 M cycle, 4 T-states
LD B,0 2 M cycles, 7 T-states
LD A,(BC) 2 M cycles, 7 T-states
4. Three cautions are in order, however:
a. Be sure you look at the right instruction. In the first example,
LD A,B was an 8-bit load in which both source and destination were
registers. The number of T states, and hence the timing, is
radically different for other variants of LD - for example:
LD A,(HL) takes 7 T states - almost twice as long - since it needs to
make an additional reference to memory to get one operand.
LD SP,HL takes 6 T states, since it is manipulates 16 bits.
LD SP,IX takes 10 T states, since it is a 16 bit operation whose
opcode is also two bytes long.
b. For some conditional jump instructions, the number of T states
depends on whether or not the jump is taken. The same also holds
for high-power looping instructions. For example, for JR NZ,(nn):
i. If the jump is NOT taken, only 7 T states are needed.
ii. If it is taken, another major cycle and 5 more T states (12
total) are needed to compute the relative jump address
internally - since this requires a 16-bit add to the current PC
value.
In reading the table, you will find two consecutive entries for
such instructions - one of the branch is taken and one if it is
not.
c. Understand that the times given in the table are best case times.
Under certain circumstances, an instruction can take longer. Two
things can prolong an instruction:
i. If the memory chips in use are not able to read or write data
quickly enough, additional wait T states will be needed whenever
a reference to memory is made. The memory chip causes this by
asserting the control line WAIT during the T state after that
producing MREQ. The Z80 inserts TW wait states until the memory
releases WAIT. (TRANSPARENCY) (This is not a problem on the
MPF-I)
ii. IO operations can be delayed if the IO port selected cannot
respond immediately. Notice that in the IO timing one TW wait
state is always inserted; the port can request more by asserting
WAIT. (TRANSPARENCY)
In either case, however, the instruction time will be some even
multiple of the clock period - the delay is effected by adding wait
states to the execution cycle.
H. When writing programs in which exact timing is crucial you must balance
both halves of every alternate path in your program. For example,
suppose the following decision structure occurred in the middle of a
segment whose exact time is critical:
IF A = B THEN
C := 0
1. If timing were not a concern, one could code this as:
CP B
JR NZ,(SKIPCLR)
LD C,0
SKIPCLR
The time for this segment would be:
A = B: CP = 4 + JR (not taken) = 7 + LD = 7 - total 18
A <>B: CP = 4 + JR (taken) = 12 = total 16
2. To balance both sides, one might code as follows:
CP B
JR NZ,(SKIPCLR)
LD C,0
JR (ENDIF)
SKIPCLR JR Z,(ENDIF) ; Cannot possibly be true since we came here on NZ
JR Z,(ENDIF)
ENDIF
The time would now be:
A = B: CP = 4 + JR (not taken) = 7 + LD = 7 + JR = 12-total 30
A <> B: CP = 4 + JR (taken) = 12 + 2 JR's that can't be taken =
2*7 - total 30
3. One way to handle such problems is to flowchart the program, computing
the total number of T-states on each branch of the flowchart and
making sure they are equal.
4. Obviously, one only does this if knowing the exact time of a segment
is more important than keeping its size and time down! (As, for
example, in simulating a clock.)
VI. Z80 Pinouts
A. Introduction
1. Thus far, we have looked at the Z80 primarily from a software
standpoint: how one goes about programming it.
2. Now, we begin looking at the Z80 from a hardware standpoint: how one
interfaces other devices to it.
3. We will do two things to accomplish this goal:
a. First, we will look at the pinouts and control signals of the Z80.
b. Second, we will look in detail at the MPF-I circuit diagrams to see
how the MPF-I designers interfaced various devices to the bus.
B. As we have noted previously, the Z80 has 40 pins: an 8-bit data bus,
a 16-bit data bus, 14 control signals, and +5V and ground power
connections.
TRANSPARENCY: Z80 PINOUTS
C. The 8-bit data bus consists of lines D0..D7, where D0 is the least
significant bit. This bus is unique on the Z80 in that it is
bidirectional (note the double-headed arrows.) The Z80 can send a byte
of data to an external device over the bus, and an external device can
send a byte of data to the Z80 over the bus.
1. This bus is used both for data transfers to/from memory and for
IO operations.
2. The Z80 connects to these pins via tri-state gates. When the Z80
is sending data over the bus, it drives each bit to either 5V or
ground as appropriate. When it is receiving data over the bus, it
presents a high-impedance load so that the sending device can drive
it to the appropriate level. The Z80 will also "float" the bus
(present a high impedance load to it) when it is allowing a direct
memory access device to use the bus to communicate with memory, as
we shall discuss later.
3. External devices that can write to the bus must also connect to it
via tri-state gates so that they do not interfere with the Z80 or
other devices. Most chips that are meant to connect directly to
microprocessor data busses (e.g. memories) have these tri-state
gates built in.
D. The 16 bit address bus consists of lines A0..A15, where A0 is the least
significant bit. This is a one-way bus - Z80 to outside world.
1. For memory operations, the Z80 places a 16-bit address on the bus.
2. For IO operations, the Z80 places an 8-bit port number on bits
A0..A7. The high order 8 bits are not used for IO.
3. Though the bus is one-directional, the Z80 connects to it via
tri-state gates. This is so the Z80 can "float" the bus to allow
direct memory access devices to take it over and address memory
themselves.
____ __ __ ____
E. The output lines MREQ, RD, WR, and WAIT are used together for memory
operations.
1. Note that these signals are low true: the asserted (true) state
is represented by 5V and the unasserted state by 0V. This is the
meaning of the bar over the mnemonics for the signals.
2. When no bus operation is taking place, then, these lines are at
5V, which looks to external devices like a logic 1, though the
meaning is "false". Memories look for a 0V condition (that looks
like logic 0) in order to respond.
____ __ __ ____
3. Note that MREQ, RD, and WR are outputs from the Z80 while WAIT is
____ __ __
an input. MREQ, RD, and WR are connected to the bus via tri-state
gates; when the Z80 is allowing a DMA device to control the address
and data lines for direct transfer to/from memory, it also floats
these lines to allow the device to control them as well. (The DMA
device will follow the same protocol in controlling these lines
that the Z80 would.) This however, does not make them input lines -
when they are "floated", the Z80 ignores them.
4. To do a memory read, the Z80:
a. Places a 16 bit address on A0..A15.
b. "Floats" D0..D7.
____ __
c. After allowing time for the bus to settle, it asserts MREQ and RD.
d. The appropriate memory chip must now look up the requested byte
and place it on the data bus. The Z80 will capture the data about
2 T-states after it asserts the request lines, and will then
deassert the request lines. If the memory cannot respond this
____
quickly, it must assert WAIT (pull it to 0V) to cause the Z80 to
insert extra Tw wait states before capturing the data.
e. Finally, the Z80 will take the address off the bus.
REVIEW TRANSPARENCY: MEMORY TIMING (READ)
5. To do a memory write, the Z80:
a. Places a 16 bit address on A0..A15.
b. Places a byte of data on D0..D7.
____ __
c. After allowing time for the bus to settle, it asserts MREQ and WR.
d. The appropriate memory chip must now read the data byte from the
bus and store it. If it cannot do so in the allowed time, it
____
must assert WAIT to request one or more Tw wait states.
e. After about 2 T-states (plus any requested wait states), the Z80
will deassert the control signals, while still leaving address and
data valid. (The memory may use this transition to actually
trigger the data transfer.)
f. Finally, the Z80 will take the address and data off the bus.
REVIEW TRANSPARENCY: MEMORY TIMING (WRITE)
____ __ __ ____
F. The control signal IORQ is used in conjunction with RD, WR and WAIT for IO
transfers.
1. IO reads and writes follow much the same protocol as memory reads
____ ____
and writes, except that the control line IORQ is used in place of MREQ.
This distinction is what causes IO devices to respond, rather than
memories.
2. Another distinction between IO operations and memory operations is
the use of an 8 bit port number on A0..A7 in place of a 16 bit
address using the entire address bus.
____
3. IO devices may use WAIT just as memories can to request additional
time. A distinction between IO and memory operations is that all
IO operations include at least one Tw wait state automatically.
REVIEW TRANSPARENCY: IO TIMING
__
G. The M1 signal is an active low output that is asserted (0 Volts) when
the Z80 is in the opcode fetch (M1) major cycle of an instruction, and
is unasserted (5 Volts) at all other times. In the case of multi-byte
__
opcodes (window byte + additional byte), M1 is low for each byte
fetched.
TRANSPARENCY: M1 CYCLE (INSTRUCTION FETCH)
____
H. The RFSH output is used as part of a very nice feature of the Z80:
automatic refresh for dynamic memory. We will discuss this later when
we talk about memory systems. (Note: this is not of interest on the
MPF-I, since it uses static RAM, not dynamic RAM.)
NOTE ON M1 CYCLE TRANSPARENCY
___ ___
I. The INT and NMI lines are active low inputs to the Z80 that allow an
external device to initiate an iterrupt. We will discuss the details
of interrupts later in the course. For now, we note the following:
___
1. The INT line provides a very flexible interrupt mechanism whereby
the interrupting device can cause the Z80 to execute a specific
interrupt service routine for that device. This kind of interrupt
can be enabled and disabled under software control by using the EI
and DI instructions. In particular, interrupt service routines
normally run with interrupts disabled so that the servicing of one
interrupt will not be interrupt by another request.
___
2. The NMI line provides a high priority interrupt mechanism that always
executes an interrupt service routine at a specific address in memory:
___ ___
0066H. Unlike INT, interrupts requested via NMI cannot be disabled -
a non-maskable interrupt request will always be honored at the end
of the currently-executing instruction. Moreover, if requests are
___ ___ ___
pending on both INT and NMI, the NMI request will have priority. This
provides a way of guaranteeing fast response to certain kinds of
external event, such as impending power failure.
___ ___
3. In the case of INT (but not NMI) the Z80 will acknowledge the
interrupt request when it is able to do so, allowing the external
device to provide further information as to the action to take.
____ __
The Z80 acknowledges an interrupt by asserting both IORQ and M1 at
the same time. This is a cue for the interrupting device to place
additional data on the bus.
TRANSPARENCY: INT TIMING
____
For now, the importance of this is that IORQ does not always mean that
an IO transfer is being done, and IO devices cannot therefore look
____
only at the address bus and IORQ to see if they are being addressed.
In particular:
____ __
IORQ and M1 indicates that an interrupt is being acknowledged
____ __
IORQ and RD indicates that a read from the port addressed by
A0..A7 is being done
____ __
IORQ and WR indicates that a write to the port addressed by
A0..A7 is being done
(We will see the effects of this when we look at the MPF-I's circuit
diagrams.)
_____ _____
J. BUSRQ and BUSAK are input lines to the Z80 that provide a way for
direct-memory-access devices to gain control of the system bus.
1. Certain kinds of IO devices have data transfer rates that approaches
the maximum speed of the memory system. For example, a floppy disk
may have a transfer rate of 250K bytes/sec - or one byte every
4 microseconds. (Hard disks are even faster.) On a 2.5 mhz Z80,
each M-cycle takes almost 2 microseconds; so the floppy disk would
require a transfer every other M-cycle. This is probably too fast
for data transfer under program control (though it might be possible
using the INDR instruction.) The controllers for such devices
typically transfer data directly from the device to memory (or vice
versa) without CPU intervention.
_____
2. The BUSRQ line can be asserted by a DMA controller to indicate that
it needs to use the system bus for a data transfer. At the end of
the current M-cycle, the Z80 will:
a. Float its data, address, and control lines so that the external
device can control them.
_____
b. Assert BUSAK, to let the device know that it can use the bus.
_____
c. When the device is through with the bus, it releases BUSRQ to allow
the Z80 to resume using the bus.
TRANSPARENCY: BUSRQ TIMING
3. We will come back to this later.
____
K. The HALT output from the Z80 is asserted when the processor is halted -
i.e. it has executed the HALT instruction. When the CPU is halted, it
stops fetching and executing instructions, but it will respond to
___ ___ ____
interrupts (NMI always, INT only if enabled.) On some systems, HALT
___
is wired to NMI; thus, if a halt instruction is executed either
accidentally or deliberately an immediate non-maskable interrupt will
occur, restarting the processor.
_____
L. The RESET line provides a mechanism for orderly system start up when
power is first applied to a system.
1. When the power is first turned on to any digital system, the various
flip flops in the system will be in a random, unpredictable state.
This will also be true of the data stored in semiconductor RAM.
2. Many systems have a circuit that functions when power is first
applied, asserting a system reset line for a few milliseconds or
so. This reset line is then applied to all of the devices in the
system, including the CPU, to initialize them.
_____
3. When the RESET input to the Z80 is asserted, the Z80's program counter
is set to 0, and it begins executing whatever program is loaded there.
Thus, Z80-based systems typically reserve their lowest memory
addresses for ROM, with a system initialization program beginning at
location 0.
_____
4. On the MPF-I, RESET is asserted under two conditions:
a. At power-up.
b. When the reset button on the keyboard is pressed.
M. The clock input (O/) must be connected to a source of square waves
of the appropriate frequency (e.g. 2.5 or 4.5 mhz).
N. +5V and ground connections complete the Z80 pinouts.
VII. MPF-I circuit diagrams
A. As a concrete example of how various devices can be interfaced to the
Z80 bus, we consider the circuit diagrams for the MPF-I. These are
five pages long in all, in appendix D of the MPF-I User's Manual.
1. Sheet 1 is the main circuit diagram
2. Sheets 2,3 detail display and tape recorder circuits
3. Sheet 4 details the keyboard circuit
4. Sheet 5 details the power supply circuits
B. Sheet 1 - the main circuit TRANSPARENCY
1. Clock:
a. An oscillator at upper left-hand corner, consisting of two
inverters, two resistors, and a crystal, produces a 3.58 mhz
raw clock.
b. This is applied to the clock input of two D flip flops
(U11 - a 74LS74 dual D flip-flop). The lower of these is part of
the reset circuit and will be considered later.
c. The upper D flip flop has its Q' output wired to its D input.
Thus, its state changes on every clock pulse. What is the effect
of this? (ASK)
d. It divides the clock frequency by 2 - the flip flop changes state
3.58E6 times per second, which means that it goes through a
complete cycle of state changes (lo to hi to lo again) 1.59E6
times per second - yielding the desired 1.59 mhz square wave clock
for the Z80.
e. This signal is applied to the O/ input on the Z80, and is also
made available on the system bus (though none of the other MPF-I
devices use it.)
2. Reset:
a. The other half of the U11 dual D flip-flop is used for system
reset. It gets its clock from the same oscillator as described
above. This means that its output be updated to reflect the
current value of its input 3.58E6 times per second.
_____
b. The Q output of this flip flop is connected to the RESET input of
the Z80. Thus, a 0 in the flip flop will cause a system reset.
The normal operating condition of the system is for this flip
flop to contain a 1.
c. The D input to this flip flop is normally held high by a 10K
resistor to +5V. There are two ways it can be pulled low (putting
the flip flop in the zero state and resetting the system):
i. The reset switch on the keyboard can ground the D input.
ii. There is a 4.7 uf capacitor from the D input to ground. When
the system is powered down, this capacitor is discharged.
At power up, it takes a while for this capacitor to charge up
through the 10K resistor. (The time constant is about 47 ms)
Thus, the D input to the flip flop is held low during the first
few milliseconds after the system is powered up, forcing a
power-up reset of the Z80 as desired. As the capacitor charges
up, its value eventually crosses the threshold recognized as
the 0 to 1 transition by the flip flop, at which point the
flip flop goes to 1 and stays there for normal operation,.
d. The Q' output of this flip flop is connected to the reset inputs
of the 8255 IO ports shown on sheet 2 (which we will discuss later).
Unlike the Z80, these devices require an active HIGH reset signal;
thus, when Q is low and the Z80 is being reset, Q' will be high
and the ports will also be reset, as desired.
____ _____ ___
3. WAIT, BUSRQ, and INT:
a. These inputs are not used by the MPF-I. Since they are active
low, they are tied to +5V by 10K pullup resistors to prevent
spurious signalling.
b. They are connected to the system bus connector so that external
devices can use them. Although it is not shown in the diagram, the
_____ _____
BUSAK output is also put on the bus so that devices using BUSRQ can
know when the Z80 has responded to their request.
___
4. NMI: This input is used by the MPF-I as part of a circuit to implement
a single step/program breakpoint feature in the monitor.
a. One nice debugging feature of the monitor is the S command. This
will prompt the user for an address
/_S_\ =
will execute the instruction at that address, and will then return
to the monitor prompt, allowing the user to examine registers etc.
Repeated uses of this command make it possible to step through a
program one instruction at a time.
b. Implementing a feature like this requires hardware support. On
many CPU's, this takes the form of a flag in the condition codes
which the user can set to cause the CPU to trap after each
instruction. However, the Z80 lacks this feature.
c. To implement single stepping, the MPF-I uses one bit of one of its
IO ports as a break signal (active low). This line is normally
held high; the monitor outputs a 0 to this bit as part of the
setup for stepping the program. (The logic for this is on sheet 2,
which we will look at later.)
_____
d. This break signal from the port is applied to the reset input of a
74LS90 divide-by-10 counter. When it is high (the normal state),
the counter is held in the zero state. Because of the inverter
___ ___
between the counter output and the NMI, this holds NMI high
(inactive.)
_____
e. When break goes low, the counter starts running. The counter clock
__
is connected to the M1 output of the Z80, so it counts M1 states -
i.e. instructions executed (one M1 state for opcode fetch per
instruction.)
f. With the connections shown, the counter functions as a divide-by-5
counter - i.e. its input and output waveforms are:
__ _ _ _ _ _ _ _ _ _ _ _
input: |_| |_| |_| |_| |_| |_| |_| |_| |_| |_| |_| |_
___
output: __________________| |_________
Since the output of the counter is inverted before being applied to
___ ___
NMI, this means that NMI will go low - and an interrupt will be
triggered, on the fifth M1 cycle after the monitor sets the bit in
the port. The software is arranged, of course, so that this
instruction will be the user instruction to single step (the other
___
4 being monitor code); since NMI is asserted during its opcode
fetch, the interrupt is actually taken after it finishes execution
as desired. (Of course, the monitor must quickly reset the break
bit to 1 to prevent the interrupt service routine from itself
being interrupted!)
____
5. HALT: This output is connected through a driver transistor to a red
____
LED. When the CPU is running (HALT deasserted or high), the LED will
be lit; should a halt instruction be executed the LED will go out.
6. The memory system.
a. The on-board memory consists of three chips, plus a slot for a
fourth: 1-2 ROMs and 2 RAMs. (We actually only have on ROM on
our MPF-I's, the other slot being a BASIC option we did not buy.
However, we will discuss the hardware as if the ROM were there.)
b. Each of the memory chips is organized into 8 bit bytes; therefore,
each connects to all 8 bits of the data bus D0..D7.
c. Each of the ROM's is an 8K chip; therefore, each must receive the
low order 13 bits of the address bus: A0..A12. The RAM's are only
2K each, so they only get 11 bits: A0..A10. These low order
address bus bits are decoded on the chip to access the correct
byte.
d. The high-order bits of the address bus are decoded externally to
__ __
the memories, and are used to derive the CS and OE enable inputs to
them. These active-low inputs are used by the memories to decide __
whether or not they should respond to activity on the bus;only if CS
is low, will the chip respond. The decoder circuitry also looks
____
at the MREQ control line, since the memory chips should only respond
during a memory operation, regardless of the state of the address
bus. The desired truth tables for the various chip selects is as
follows:
____ __
Slot Assigned addresses MREQ A15..A11 CS
U2 0000-1FFFH 0 000xx 0
0 all others 1
1 xxxxx 1
U3 2000-3FFFH 0 001xx 0
0 all others 1
1 xxxxx 1
U4 F000-F7FFH 0 11110 0
0 all others 1
1 xxxxx 1
U5 F800-FFFFH 0 11111 0
0 all others 1
1 xxxxx 1
__
These CS values could be derived by using NAND gates - e.g. for U2:
A15 __|\o____ ________
|/ |__| \
A14 __|\o_______| \ __
|/ | )o_____ CS
A13 __|\o_______| )
____ |/ __| /
MREQ__|\o____| |_______/
|/
e. However, the MPF-I uses far fewer chips by using a 74LS138 1 out of
8 decoder chip to handle memory decoding.
TRANSPARENCY: 74LS138 diagram and truth table
i. The 74LS138 has 6 inputs and 8 outputs.
ii. The outputs are active low. At most 1 of the 8 outputs
can be low at any one time. Note that this is convenient for use
with memory or IO chips whose chip select input is active low.
(In fact, the memories are made this way BECAUSE decoders like
the 74LS138 have active low outputs. The decoders, in turn, have
active low outputs because they are built from NAND gates.)
iii. Three of the inputs are decoded to select one of the 8 outputs.
iv. The remaining three inputs are enables, which are anded together;
i.e. all must have the proper value for the selected output line
to go low.
- Two of the enables are active low: unless they are both low,
none of the outputs is low.
- One of the enables is active high: unless it is high, none
of the outputs is low.
f. The memory decoder 74LS138 on the MPF-I (U6) is connected as
follows:
i. Its active high enable input is tied to +5V, and its two
____
active low enables are both tied to MREQ. Thus, its selected
output is enabled whenver a memory operation is being called
for.
ii. Its three decoded inputs are connected to address bus bits
A15 .. A13. Thus, the first output line is selected when the
high order three bits of the address are 000; this is routed
__ __
to CS on the ROM in U2. The second line goes to CS on the ROM
in U3; this is selected when the high order bits of the address
are 001. The last output line is low when the high order bits
of the address bus are 111, which means that RAM is being
addressed; however, more decoding is needed to select a
specific RAM chip.
g. The 74LS138 output that is selected when RAM is addressed goes
as an enable to another decoder that decodes two more bits of the
address bus (A12..A11) to select the correct RAM chip. The
network of jumpers at the right hand side allows for connections
to be changed if different size RAMs are used or a RAM is put
in the ROM slot U3.
__
h. The RAM chips have an additional control input WE, which the __
selected chip uses to decide whether to do a read or a write. WE
is a low true write enable; when it is low, the chip does a write,
__ __
otherwise it does a read. Note that WE is simply connected to WR
on the Z80 bus; if a read operation is in progress the chip need
__ __
only know that WR is not asserted - it need not actually see RD
asserted.
7. The IO system.
a. The MPF-I has a total of 43 bits of IO to/from various devices. These
are provided for by two 8255 chips, each of which has 3 eight-bit ports
for a total of 48 bits. (Five are unused.) Each 8255 interfaces
to the data bus and the two low-order address bus bits, the latter
being used to select one of the 3 ports or the chip's 8-bit control
register. (We will look at the 8255 later when we get to parallel IO.)
Each 8255 also has a low true chip select, whose derivation we
___ ___
consider now. (Note off sheet connections for CS1 and CS2.)
b. One of the 8255's is assigned port addresses 80-83, and the other
90-93. Since port numbers are eight bits, and the chip itself decodes
two bits to select from among the ports on the chip, the external
circuitry must decode 6 address bits for chip select (A7..A2).
c. Also, as with memory we only want to enable an IO chip when an IO
operation is in progress. ____ __
In this case, however, we must look at two control lines: IORQ and M1.
____
This is because IORQ is not only asserted during an IO operation, but
__
also during an interrupt acknowledge; however, in the latter case M1 is
also asserted. Thus, an for an IO operation to be selected it must be
____ __
the case that IORQ is low and M1 is HIGH.
d. It would seem, then, that 8 lines must be decoded to derive chip
select for the 8255's: 6 address lines (A7..A2) and two control lines.
However, the MPF-I uses a single 74LS138 (U7) that only decodes A7..A4
plus the control lines. (A3 and A2 are ignored.) This means that
the 8255's will actually respond to any addresses in the ranges
80..8F or 90..9F. (This is not a problem, since 256 ports is far
more than the number needed. This is a common trick to save
decoder logic.)
__ ____
i. M1 is connected to the active high enable, and IORQ to one active
low enable. This guarantees that the IO chips will only respond
to IO operations, as required.
ii. Address bus bit A7 is inverted and connected to the other active
low enable. This guarantees that the chips will only respond to
port addresses >= 80.
iii. Finally, bits A6..A4 are connected to the selection inputs, and
the output lines corresponding to 000 and 001 are used to drive
___ ___
CS1 and CS2, causing the 8255's to respond to IO operations on
port addresses of the form 1000 xxxx and 1001 xxxx as required.
C. Sheet 2 - the IO ports TRANSPARENCY
D. Sheet 4 - the keyboard matrix TRANSPARENCY
Copyright ©1999 - Russell C. Bjork