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Your doubts arise from lack of understanding of the basics about synchronous sequential logic networks, of which modern CPUs and related basic blocks, like counters, are just particular examples.
As someone else already pointed out, such networks can change state only in consequence of a clock edge, i.e a clock signal transition.
I suggest you to investigate ...
answered Jan 24 at 22:39
Lorenzo Donati -- Codidact.com
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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 ...
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Usually I dislike water analogies for electricity, but...
The clock on synchronous logic is sort of like a well-water pump. The pumping action is not continuous, it happens in bursts. When you lift the pump arm up, the pump inhales a little bit of the water from down below, and when you force the pump arm down again, it expels that water upstream. Move the ...
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Clock pins on the FPGA are for clock inputs, not outputs. They have well-characterized paths to internal clock resources, such as PLLs, DCMs and the buffers for clock trees.
The clock trees themselves are routed to most of the clockable devices on the chip, mostly the FFs in the fabric and in the IOBs. There is NOT, in general, any way to connect a clock ...
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Clock signal go everywhere on the FPGA so require special fanout considerations and and must arrive at all their destinations with minimal skew. That means you cannot route clocks through regular logic fabric or through any pin on the FPGA. The clock can only go into pins dedicated to it and can only be routed on the FPGA on special clock distribution ...
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Clock stopped: if the registers and logic are static, the machine holds its state. If they’re dynamic, the state will eventually be corrupted or lost.
(Dynamic logic is used in some architectures to reduce area. Like DRAM, it needs to be cycling to refresh its state. It was more common in the NMOS era, though it still sees use in CMOS.)
Clock running beyond ...
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The clock synchronises operation of all inner components of any CPU. If it stops (or gets constantly high), the CPU should just stop too. At least 8051 microcontroller acts like that. It will definitely not "overclock" the CPU (aka. it won't work as fast as possible). All nasty side effects like clearing/changing of registers, undefined state etc. ...
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You cannot do so from native VHDL/Verilog because clock signals cannot be properly routed through the logic fabric. Too much skew. They need to go through the dedicated clock routing networks that span the entire FPGA. That means you need to use Clocking Wizard in the IP Catalog in Vivado to configure the hardware PLL or DLL.
After you finish the Clocking ...
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'minimal RMS jitter' is not a specification. How much do you want to spend? Fundamental oscillator? Overtone oscillator followed by an injection locked divider? Rubidium? Hydrogen maser?
Select the 5 MHz model and evaluate it. If it doesn't give you the performance you want, then characterise where it's deficient and improve it. Simply going for a different ...
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One method of switching crystals involves using diodes. A diode is put in series with each crystal and a voltage is used to bias the diodes on or off. RF chokes are used to keep the bias voltage from loading down the signal from the crystals.
This was done in things like police scanners and CB radios quite a bit, so you might look for schematics of those for ...
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If the end goal is produce a 5MHz clock signal with minimal RMS jitter, is it generally advantageous to select the 50MHz model and divide* it by 10 or simply select the 5MHz model?
Assuming high frequency oscillators had lower RMS jitter (which I will leave to other answers to confirm/disprove), you will have to divide the output frequency as you correctly ...
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For quartz crystal oscillators the biggest contribution to the jitter comes from random noise sources. So there should not be a direct relation between the RMS timing jitter and the oscillation frequency. NOTE: since the RMS jitter is in the orders of picoseconds or even less than 1 picoseconds, it's neglected (i.e. not considered as a limiting parameter or ...
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