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Did the original Intel peripherals use latches as input registers?
Clash Royale CLAN TAG#URR8PPP
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This is more a question to satisfy my curiosity.
When looking at the original timing diagrams for reading and writing (e.g. for the 8255 and the 8250), these signals are never expressed in function of a clock. The 8255 doesn't even have a clock input.
So my question is, is this done using level-sensitive latches?
8080
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chthon
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up vote
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down vote
favorite
This is more a question to satisfy my curiosity.
When looking at the original timing diagrams for reading and writing (e.g. for the 8255 and the 8250), these signals are never expressed in function of a clock. The 8255 doesn't even have a clock input.
So my question is, is this done using level-sensitive latches?
8080
asked 7 hours ago
chthon
1184
New contributor
chthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This is more a question to satisfy my curiosity.
When looking at the original timing diagrams for reading and writing (e.g. for the 8255 and the 8250), these signals are never expressed in function of a clock. The 8255 doesn't even have a clock input.
So my question is, is this done using level-sensitive latches?
8080
asked 7 hours ago
chthon
1184
New contributor
chthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This is more a question to satisfy my curiosity.
When looking at the original timing diagrams for reading and writing (e.g. for the 8255 and the 8250), these signals are never expressed in function of a clock. The 8255 doesn't even have a clock input.
So my question is, is this done using level-sensitive latches?
8080
8080
asked 7 hours ago
chthon
1184
New contributor
chthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
chthon
1184
New contributor
chthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
chthon
1184
New contributor
chthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
chthon
1184
asked 7 hours ago
chthon
1184
1184
New contributor
chthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
chthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
chthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
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3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Looking at the 8255 datasheet, timing is giving in reference the the RD (read) and WR (write) signals. The 8080, has READY and WRITE signals, which are described in the 8080 datasheet as
READY: The READY signal indicates to the 8080A that valid memory or input data is available on the 8080A data bus. This signal is used to synchronize the CPU with slower memory or I/O devices. If after sending an address out the 8080A does not receive a READY input, the 8080A will enter a WAIT state for as long as the READY line is low. READY can also be used to single step the CPU.
WRITE: The WR signal is used for memory WRITE or I/O output control. The data on the data bus is stable while the WR signal is active low (WR = 0).
So, yes, the timing is clockless. No, it doesn't use level-sensitive latches; instead the various latches (in the CPU and periphery) are triggered by edges on these signals or combinations that are produced by glue logic.
answered 7 hours ago
dirkt
7,0331738
There is no READY signal in 8255, so the only conclusion it was given at the diagrams for the convenience.
– lvd
3 hours ago
@lvd: The signals are calledRDandWR.
– dirkt
3 hours ago
@lvd: pin 5 and 36, negated.
– dirkt
3 hours ago
so RD stands for READ, not READY.
– lvd
3 hours ago
@lvd: but there's only a READY signal in the 8080. I have to admit I didn't look at application example circuitry.
– dirkt
3 hours ago
 |Â
show 1 more comment
up vote
3
down vote
There are no clock inputs in 8255 but that does not mean that for example /RD and /WR strobes couldn't be used as clock inputs to some internal clocked flip-flops that latch incoming data or trigger some internal state machines.
The actual answer could be found through reverse-engineering the chips in question (check visual6502.org for similar reverse-engineering projects).
answered 3 hours ago
lvd
1,942315
add a comment |Â
up vote
1
down vote
Many I/O devices, especially historically, had bus timing entirely controlled by transitions on chip-select and read/write lines. Even devices like UARTs which need to have clocks for various purposes would often have bus timing which was independent of that clock, thus allowing them to be used on a wide range of CPUs with different clocking systems.
The biggest difficulty with this approach is something called metastability, which wasn't thoroughly understood in the 1980s. If a device on the bus initiates a read of UART's status almost the precise moment an incoming byte is completed, it would be equally acceptable for the device to report a zero in the "is data ready" bit [the data wasn't ready when the request was received] or a one [the data arrived just in time to be detected].
Unfortunately, there's a third possibility: the device might drive the data bus with a logic level somewhere between a high and low signal, or might start to drive it with a low level and switch to a high level sometime in the middle of a cycle. While a CPU might regard that situation as equivalent to either outputting zero or outputting 1, it might also end up in a weird state, especially if different parts of the CPU grab the signal at slightly different times. Many products in the 1980s didn't pay much attention to such issues, and would have occasional weird failures as a result.
The normal way to prevent problems like this is to use a device called a "double synchronizer". This will ensure that an input transition that occurs near a clock edge will end up being regarded as having been either before it or after it, but at the expense of not being able to say what the signal level currently is--merely what it used to be two cycles ago.
Unfortunately, trying to use a double-synchronizer in a UART without requiring the use of a bus clock would require doing that some bus-initiated action occur before a cycle that's supposed to report UART status, and have the latter action report what the status was when the former action occurred. For example, one could specify that each read of the status register reports the value it had at the previous read request, so code wanting current data would need to read the register twice consecutively. I don't think I've ever seen a UART actually work that way.
The other possibility is to require the use of a bus clock. If serial transmission and reception use a different clock, a double synchronizer would still be necessary, but delaying data-readiness indications by two bus clocks is likely to be less irksome than requiring the use of two separate read events to capture it.
answered 1 hour ago
supercat
5,338532
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Looking at the 8255 datasheet, timing is giving in reference the the RD (read) and WR (write) signals. The 8080, has READY and WRITE signals, which are described in the 8080 datasheet as
READY: The READY signal indicates to the 8080A that valid memory or input data is available on the 8080A data bus. This signal is used to synchronize the CPU with slower memory or I/O devices. If after sending an address out the 8080A does not receive a READY input, the 8080A will enter a WAIT state for as long as the READY line is low. READY can also be used to single step the CPU.
WRITE: The WR signal is used for memory WRITE or I/O output control. The data on the data bus is stable while the WR signal is active low (WR = 0).
So, yes, the timing is clockless. No, it doesn't use level-sensitive latches; instead the various latches (in the CPU and periphery) are triggered by edges on these signals or combinations that are produced by glue logic.
answered 7 hours ago
dirkt
7,0331738
There is no READY signal in 8255, so the only conclusion it was given at the diagrams for the convenience.
– lvd
3 hours ago
@lvd: The signals are calledRDandWR.
– dirkt
3 hours ago
@lvd: pin 5 and 36, negated.
– dirkt
3 hours ago
so RD stands for READ, not READY.
– lvd
3 hours ago
@lvd: but there's only a READY signal in the 8080. I have to admit I didn't look at application example circuitry.
– dirkt
3 hours ago
 |Â
show 1 more comment
up vote
4
down vote
accepted
Looking at the 8255 datasheet, timing is giving in reference the the RD (read) and WR (write) signals. The 8080, has READY and WRITE signals, which are described in the 8080 datasheet as
READY: The READY signal indicates to the 8080A that valid memory or input data is available on the 8080A data bus. This signal is used to synchronize the CPU with slower memory or I/O devices. If after sending an address out the 8080A does not receive a READY input, the 8080A will enter a WAIT state for as long as the READY line is low. READY can also be used to single step the CPU.
WRITE: The WR signal is used for memory WRITE or I/O output control. The data on the data bus is stable while the WR signal is active low (WR = 0).
So, yes, the timing is clockless. No, it doesn't use level-sensitive latches; instead the various latches (in the CPU and periphery) are triggered by edges on these signals or combinations that are produced by glue logic.
answered 7 hours ago
dirkt
7,0331738
There is no READY signal in 8255, so the only conclusion it was given at the diagrams for the convenience.
– lvd
3 hours ago
@lvd: The signals are calledRDandWR.
– dirkt
3 hours ago
@lvd: pin 5 and 36, negated.
– dirkt
3 hours ago
so RD stands for READ, not READY.
– lvd
3 hours ago
@lvd: but there's only a READY signal in the 8080. I have to admit I didn't look at application example circuitry.
– dirkt
3 hours ago
 |Â
show 1 more comment
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Looking at the 8255 datasheet, timing is giving in reference the the RD (read) and WR (write) signals. The 8080, has READY and WRITE signals, which are described in the 8080 datasheet as
READY: The READY signal indicates to the 8080A that valid memory or input data is available on the 8080A data bus. This signal is used to synchronize the CPU with slower memory or I/O devices. If after sending an address out the 8080A does not receive a READY input, the 8080A will enter a WAIT state for as long as the READY line is low. READY can also be used to single step the CPU.
WRITE: The WR signal is used for memory WRITE or I/O output control. The data on the data bus is stable while the WR signal is active low (WR = 0).
So, yes, the timing is clockless. No, it doesn't use level-sensitive latches; instead the various latches (in the CPU and periphery) are triggered by edges on these signals or combinations that are produced by glue logic.
answered 7 hours ago
dirkt
7,0331738
Looking at the 8255 datasheet, timing is giving in reference the the RD (read) and WR (write) signals. The 8080, has READY and WRITE signals, which are described in the 8080 datasheet as
READY: The READY signal indicates to the 8080A that valid memory or input data is available on the 8080A data bus. This signal is used to synchronize the CPU with slower memory or I/O devices. If after sending an address out the 8080A does not receive a READY input, the 8080A will enter a WAIT state for as long as the READY line is low. READY can also be used to single step the CPU.
WRITE: The WR signal is used for memory WRITE or I/O output control. The data on the data bus is stable while the WR signal is active low (WR = 0).
So, yes, the timing is clockless. No, it doesn't use level-sensitive latches; instead the various latches (in the CPU and periphery) are triggered by edges on these signals or combinations that are produced by glue logic.
answered 7 hours ago
dirkt
7,0331738
edited 3 hours ago
answered 7 hours ago
dirkt
7,0331738
answered 7 hours ago
dirkt
7,0331738
answered 7 hours ago
dirkt
7,0331738
7,0331738
There is no READY signal in 8255, so the only conclusion it was given at the diagrams for the convenience.
– lvd
3 hours ago
@lvd: The signals are calledRDandWR.
– dirkt
3 hours ago
@lvd: pin 5 and 36, negated.
– dirkt
3 hours ago
so RD stands for READ, not READY.
– lvd
3 hours ago
@lvd: but there's only a READY signal in the 8080. I have to admit I didn't look at application example circuitry.
– dirkt
3 hours ago
 |Â
show 1 more comment
There is no READY signal in 8255, so the only conclusion it was given at the diagrams for the convenience.
– lvd
3 hours ago
@lvd: The signals are calledRDandWR.
– dirkt
3 hours ago
@lvd: pin 5 and 36, negated.
– dirkt
3 hours ago
so RD stands for READ, not READY.
– lvd
3 hours ago
@lvd: but there's only a READY signal in the 8080. I have to admit I didn't look at application example circuitry.
– dirkt
3 hours ago
There is no READY signal in 8255, so the only conclusion it was given at the diagrams for the convenience.
– lvd
3 hours ago
There is no READY signal in 8255, so the only conclusion it was given at the diagrams for the convenience.
– lvd
3 hours ago
@lvd: The signals are called
RD and WR.– dirkt
3 hours ago
@lvd: The signals are called
RD and WR.– dirkt
3 hours ago
@lvd: pin 5 and 36, negated.
– dirkt
3 hours ago
@lvd: pin 5 and 36, negated.
– dirkt
3 hours ago
so RD stands for READ, not READY.
– lvd
3 hours ago
so RD stands for READ, not READY.
– lvd
3 hours ago
@lvd: but there's only a READY signal in the 8080. I have to admit I didn't look at application example circuitry.
– dirkt
3 hours ago
@lvd: but there's only a READY signal in the 8080. I have to admit I didn't look at application example circuitry.
– dirkt
3 hours ago
 |Â
show 1 more comment
up vote
3
down vote
There are no clock inputs in 8255 but that does not mean that for example /RD and /WR strobes couldn't be used as clock inputs to some internal clocked flip-flops that latch incoming data or trigger some internal state machines.
The actual answer could be found through reverse-engineering the chips in question (check visual6502.org for similar reverse-engineering projects).
answered 3 hours ago
lvd
1,942315
add a comment |Â
up vote
3
down vote
There are no clock inputs in 8255 but that does not mean that for example /RD and /WR strobes couldn't be used as clock inputs to some internal clocked flip-flops that latch incoming data or trigger some internal state machines.
The actual answer could be found through reverse-engineering the chips in question (check visual6502.org for similar reverse-engineering projects).
answered 3 hours ago
lvd
1,942315
add a comment |Â
up vote
3
down vote
up vote
3
down vote
There are no clock inputs in 8255 but that does not mean that for example /RD and /WR strobes couldn't be used as clock inputs to some internal clocked flip-flops that latch incoming data or trigger some internal state machines.
The actual answer could be found through reverse-engineering the chips in question (check visual6502.org for similar reverse-engineering projects).
answered 3 hours ago
lvd
1,942315
There are no clock inputs in 8255 but that does not mean that for example /RD and /WR strobes couldn't be used as clock inputs to some internal clocked flip-flops that latch incoming data or trigger some internal state machines.
The actual answer could be found through reverse-engineering the chips in question (check visual6502.org for similar reverse-engineering projects).
answered 3 hours ago
lvd
1,942315
answered 3 hours ago
lvd
1,942315
answered 3 hours ago
lvd
1,942315
answered 3 hours ago
lvd
1,942315
1,942315
add a comment |Â
add a comment |Â
up vote
1
down vote
Many I/O devices, especially historically, had bus timing entirely controlled by transitions on chip-select and read/write lines. Even devices like UARTs which need to have clocks for various purposes would often have bus timing which was independent of that clock, thus allowing them to be used on a wide range of CPUs with different clocking systems.
The biggest difficulty with this approach is something called metastability, which wasn't thoroughly understood in the 1980s. If a device on the bus initiates a read of UART's status almost the precise moment an incoming byte is completed, it would be equally acceptable for the device to report a zero in the "is data ready" bit [the data wasn't ready when the request was received] or a one [the data arrived just in time to be detected].
Unfortunately, there's a third possibility: the device might drive the data bus with a logic level somewhere between a high and low signal, or might start to drive it with a low level and switch to a high level sometime in the middle of a cycle. While a CPU might regard that situation as equivalent to either outputting zero or outputting 1, it might also end up in a weird state, especially if different parts of the CPU grab the signal at slightly different times. Many products in the 1980s didn't pay much attention to such issues, and would have occasional weird failures as a result.
The normal way to prevent problems like this is to use a device called a "double synchronizer". This will ensure that an input transition that occurs near a clock edge will end up being regarded as having been either before it or after it, but at the expense of not being able to say what the signal level currently is--merely what it used to be two cycles ago.
Unfortunately, trying to use a double-synchronizer in a UART without requiring the use of a bus clock would require doing that some bus-initiated action occur before a cycle that's supposed to report UART status, and have the latter action report what the status was when the former action occurred. For example, one could specify that each read of the status register reports the value it had at the previous read request, so code wanting current data would need to read the register twice consecutively. I don't think I've ever seen a UART actually work that way.
The other possibility is to require the use of a bus clock. If serial transmission and reception use a different clock, a double synchronizer would still be necessary, but delaying data-readiness indications by two bus clocks is likely to be less irksome than requiring the use of two separate read events to capture it.
answered 1 hour ago
supercat
5,338532
add a comment |Â
up vote
1
down vote
Many I/O devices, especially historically, had bus timing entirely controlled by transitions on chip-select and read/write lines. Even devices like UARTs which need to have clocks for various purposes would often have bus timing which was independent of that clock, thus allowing them to be used on a wide range of CPUs with different clocking systems.
The biggest difficulty with this approach is something called metastability, which wasn't thoroughly understood in the 1980s. If a device on the bus initiates a read of UART's status almost the precise moment an incoming byte is completed, it would be equally acceptable for the device to report a zero in the "is data ready" bit [the data wasn't ready when the request was received] or a one [the data arrived just in time to be detected].
Unfortunately, there's a third possibility: the device might drive the data bus with a logic level somewhere between a high and low signal, or might start to drive it with a low level and switch to a high level sometime in the middle of a cycle. While a CPU might regard that situation as equivalent to either outputting zero or outputting 1, it might also end up in a weird state, especially if different parts of the CPU grab the signal at slightly different times. Many products in the 1980s didn't pay much attention to such issues, and would have occasional weird failures as a result.
The normal way to prevent problems like this is to use a device called a "double synchronizer". This will ensure that an input transition that occurs near a clock edge will end up being regarded as having been either before it or after it, but at the expense of not being able to say what the signal level currently is--merely what it used to be two cycles ago.
Unfortunately, trying to use a double-synchronizer in a UART without requiring the use of a bus clock would require doing that some bus-initiated action occur before a cycle that's supposed to report UART status, and have the latter action report what the status was when the former action occurred. For example, one could specify that each read of the status register reports the value it had at the previous read request, so code wanting current data would need to read the register twice consecutively. I don't think I've ever seen a UART actually work that way.
The other possibility is to require the use of a bus clock. If serial transmission and reception use a different clock, a double synchronizer would still be necessary, but delaying data-readiness indications by two bus clocks is likely to be less irksome than requiring the use of two separate read events to capture it.
answered 1 hour ago
supercat
5,338532
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Many I/O devices, especially historically, had bus timing entirely controlled by transitions on chip-select and read/write lines. Even devices like UARTs which need to have clocks for various purposes would often have bus timing which was independent of that clock, thus allowing them to be used on a wide range of CPUs with different clocking systems.
The biggest difficulty with this approach is something called metastability, which wasn't thoroughly understood in the 1980s. If a device on the bus initiates a read of UART's status almost the precise moment an incoming byte is completed, it would be equally acceptable for the device to report a zero in the "is data ready" bit [the data wasn't ready when the request was received] or a one [the data arrived just in time to be detected].
Unfortunately, there's a third possibility: the device might drive the data bus with a logic level somewhere between a high and low signal, or might start to drive it with a low level and switch to a high level sometime in the middle of a cycle. While a CPU might regard that situation as equivalent to either outputting zero or outputting 1, it might also end up in a weird state, especially if different parts of the CPU grab the signal at slightly different times. Many products in the 1980s didn't pay much attention to such issues, and would have occasional weird failures as a result.
The normal way to prevent problems like this is to use a device called a "double synchronizer". This will ensure that an input transition that occurs near a clock edge will end up being regarded as having been either before it or after it, but at the expense of not being able to say what the signal level currently is--merely what it used to be two cycles ago.
Unfortunately, trying to use a double-synchronizer in a UART without requiring the use of a bus clock would require doing that some bus-initiated action occur before a cycle that's supposed to report UART status, and have the latter action report what the status was when the former action occurred. For example, one could specify that each read of the status register reports the value it had at the previous read request, so code wanting current data would need to read the register twice consecutively. I don't think I've ever seen a UART actually work that way.
The other possibility is to require the use of a bus clock. If serial transmission and reception use a different clock, a double synchronizer would still be necessary, but delaying data-readiness indications by two bus clocks is likely to be less irksome than requiring the use of two separate read events to capture it.
answered 1 hour ago
supercat
5,338532
Many I/O devices, especially historically, had bus timing entirely controlled by transitions on chip-select and read/write lines. Even devices like UARTs which need to have clocks for various purposes would often have bus timing which was independent of that clock, thus allowing them to be used on a wide range of CPUs with different clocking systems.
The biggest difficulty with this approach is something called metastability, which wasn't thoroughly understood in the 1980s. If a device on the bus initiates a read of UART's status almost the precise moment an incoming byte is completed, it would be equally acceptable for the device to report a zero in the "is data ready" bit [the data wasn't ready when the request was received] or a one [the data arrived just in time to be detected].
Unfortunately, there's a third possibility: the device might drive the data bus with a logic level somewhere between a high and low signal, or might start to drive it with a low level and switch to a high level sometime in the middle of a cycle. While a CPU might regard that situation as equivalent to either outputting zero or outputting 1, it might also end up in a weird state, especially if different parts of the CPU grab the signal at slightly different times. Many products in the 1980s didn't pay much attention to such issues, and would have occasional weird failures as a result.
The normal way to prevent problems like this is to use a device called a "double synchronizer". This will ensure that an input transition that occurs near a clock edge will end up being regarded as having been either before it or after it, but at the expense of not being able to say what the signal level currently is--merely what it used to be two cycles ago.
Unfortunately, trying to use a double-synchronizer in a UART without requiring the use of a bus clock would require doing that some bus-initiated action occur before a cycle that's supposed to report UART status, and have the latter action report what the status was when the former action occurred. For example, one could specify that each read of the status register reports the value it had at the previous read request, so code wanting current data would need to read the register twice consecutively. I don't think I've ever seen a UART actually work that way.
The other possibility is to require the use of a bus clock. If serial transmission and reception use a different clock, a double synchronizer would still be necessary, but delaying data-readiness indications by two bus clocks is likely to be less irksome than requiring the use of two separate read events to capture it.
answered 1 hour ago
supercat
5,338532
answered 1 hour ago
supercat
5,338532
answered 1 hour ago
supercat
5,338532
answered 1 hour ago
supercat
5,338532
5,338532
add a comment |Â
add a comment |Â
chthon is a new contributor. Be nice, and check out our Code of Conduct.
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