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Added section on overclocking.
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The clock must transition, from low to high, and repeat, in a regular pattern.

It is these transitions which drive changes in the logic, not the high level. No transitions = no logic change. So without transitions, it will stop working.

This includes "extra high" voltage (it will likely be damaged.)

If only static elements are involved, then operation will appear to completely "freeze" during the entire duration of the clock removal. Restart the clock, and it will pickup exactly where it left off (like no time had passed whatsoever.)

If dynamic elements are involved (likely today, such as watchdog timers, interrupt-on-change, DRAM memory refresh, etc.) then something will likely be missed or corrupted during a long clock pause. In the case of DRAM loss, the currently running code will now be garbage, so the processor will fault in short order from trying to run garbage code.

It is the transitions of the clock which signal to the processor to "do the next step." The low-high transition or edge (legacy), and also the high-low edge (modern) now both do something.

On today's modern processors, multiple cores with multiple threads means that each clock cycle is doing hundreds, even thousands of things in parallel on each clock edge.

Clock edges must be used, because they are sequential, and these edges must be separated by some amount of time, to allow the "step" to fully propagate or complete before the next is started. "Overclocking" is increasing the clock rate higher than the normal value. The typical clock rate includes a safety margin, so it is usually possible to increase this somewhat. Increasing it causes the processor to work harder, so commonly needs a slight boost in voltage to keep operating. But increasing the clock rate too much will eventually violate these step timings, especially over all possible temperature ranges.

The clock must transition, from low to high, and repeat, in a regular pattern.

It is these transitions which drive changes in the logic, not the high level. No transitions = no logic change. So without transitions, it will stop working.

This includes "extra high" voltage (it will likely be damaged.)

If only static elements are involved, then operation will appear to completely "freeze" during the entire duration of the clock removal. Restart the clock, and it will pickup exactly where it left off (like no time had passed whatsoever.)

If dynamic elements are involved (likely today, such as watchdog timers, interrupt-on-change, DRAM memory refresh, etc.) then something will likely be missed or corrupted during a long clock pause. In the case of DRAM loss, the currently running code will now be garbage, so the processor will fault in short order from trying to run garbage code.

It is the transitions of the clock which signal to the processor to "do the next step." The low-high transition or edge (legacy), and also the high-low edge (modern) now both do something.

On today's modern processors, multiple cores with multiple threads means that each clock cycle is doing hundreds, even thousands of things in parallel on each clock edge.

The clock must transition, from low to high, and repeat, in a regular pattern.

It is these transitions which drive changes in the logic, not the high level. No transitions = no logic change. So without transitions, it will stop working.

This includes "extra high" voltage (it will likely be damaged.)

If only static elements are involved, then operation will appear to completely "freeze" during the entire duration of the clock removal. Restart the clock, and it will pickup exactly where it left off (like no time had passed whatsoever.)

If dynamic elements are involved (likely today, such as watchdog timers, interrupt-on-change, DRAM memory refresh, etc.) then something will likely be missed or corrupted during a long clock pause. In the case of DRAM loss, the currently running code will now be garbage, so the processor will fault in short order from trying to run garbage code.

It is the transitions of the clock which signal to the processor to "do the next step." The low-high transition or edge (legacy), and also the high-low edge (modern) now both do something.

On today's modern processors, multiple cores with multiple threads means that each clock cycle is doing hundreds, even thousands of things in parallel on each clock edge.

Clock edges must be used, because they are sequential, and these edges must be separated by some amount of time, to allow the "step" to fully propagate or complete before the next is started. "Overclocking" is increasing the clock rate higher than the normal value. The typical clock rate includes a safety margin, so it is usually possible to increase this somewhat. Increasing it causes the processor to work harder, so commonly needs a slight boost in voltage to keep operating. But increasing the clock rate too much will eventually violate these step timings, especially over all possible temperature ranges.

Expanded answer to address comments.
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rdtsc
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The clock must transition, from low to high, and repeat, in a regular pattern.

It is these transitions which drivetransitions which drive changes in the logic, not the high level. No transitions = no logic change. So without ittransitions, it will stop working.

This includes "extra high" voltage (it will likely be damaged.)

If only static elements are involved, then operation will appear to completely "freeze" during the entire duration of the clock removal. Restart the clock, and it will pickup exactly where it left off (like no time had passed whatsoever.)

If dynamic elements are involved (likely today, such as watchdog timers, interrupt-on-change, DRAM memory refresh, etc.) then something will likely be missed or corrupted during a long clock pause. In the case of DRAM loss, the currently running code will now be garbage, so the processor will fault in short order from trying to run garbage code.

It is the transitions of the clock which signal to the processor to "do the next step." The low-high transition or edge (legacy), and also the high-low edge (modern) now both do something.

On today's modern processors, multiple cores with multiple threads means that each clock cycle is doing hundreds, even thousands of things in parallel on each clock edge.

The clock must transition, from low to high, in a regular pattern.

It is these transitions which drive changes in the logic. No transitions = no logic change. So without it, it will stop working.

This includes "extra high" voltage (it will likely be damaged.)

The clock must transition, from low to high, and repeat, in a regular pattern.

It is these transitions which drive changes in the logic, not the high level. No transitions = no logic change. So without transitions, it will stop working.

This includes "extra high" voltage (it will likely be damaged.)

If only static elements are involved, then operation will appear to completely "freeze" during the entire duration of the clock removal. Restart the clock, and it will pickup exactly where it left off (like no time had passed whatsoever.)

If dynamic elements are involved (likely today, such as watchdog timers, interrupt-on-change, DRAM memory refresh, etc.) then something will likely be missed or corrupted during a long clock pause. In the case of DRAM loss, the currently running code will now be garbage, so the processor will fault in short order from trying to run garbage code.

It is the transitions of the clock which signal to the processor to "do the next step." The low-high transition or edge (legacy), and also the high-low edge (modern) now both do something.

On today's modern processors, multiple cores with multiple threads means that each clock cycle is doing hundreds, even thousands of things in parallel on each clock edge.

Source Link
rdtsc
  • 16.4k
  • 4
  • 33
  • 71

The clock must transition, from low to high, in a regular pattern.

It is these transitions which drive changes in the logic. No transitions = no logic change. So without it, it will stop working.

This includes "extra high" voltage (it will likely be damaged.)