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which one consumes less power?

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eexuke

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Dear all,
If I have a DFF with value 1 store in it,when I update its value at clock posedge, input new value "1" or "0" consumes less power? Or just equal?
Many thanks!
 

I feel any state change 1-0 or 0-1 will result in a dynamic power consumption but maintaining the same state will not atleast theoritically consume power.
 

any dff whih has a transition of "1" to "0" or " o" to "1" consume dynamic power. and the dynamic power consumption is the most important part of cmos circuit.
"0"to"0 does not consume dynamic power but static,
which occupies less compared to dynamic!
 

Maintainig the same state does not consume much power. Hoever if your VCC is lower than specified, say 3.1 instead of 5, then the transistors shall not be saturated but shall be in the active region and hence offer greate resistance and hence higher power consumptio. I think the same shall go on for CMOS as well.

Has anyone thought...Does an elivator consume more power traveling up than down? It is more-or-less the same. Ther is a counter weight for the elivator.
 

You can use clock gating in order to disable DFF's operation (and thus prevent the DFF from consuming power) when it's going to keep it's value unchanged, i.e it's state is 0 (or 1) and the new input value is 0 (or 1) too.
 

Hi,
When an o/p is 1 and again 1 is applied as the i/p then there will be no dynamic power consumption. Dynamic power dissipation is the main source of power disipation in CMOS. So a 0 on i/p means the o/p has to change from 1 to 0, so it consumes more power.
 

The power consumed while swithing from 0 to 1 or 1 to 0 is approx same (depenson the sizes of the tr.)
while maintaining the same state logic results in only the leakage power consumption.
 

I think value 1 input will consume less power,

because DFF's internal logic needn't

change, but for input 0, the dff's internal

logic will change, and internal node

will be charged or discharged, consume

greater power consumption.


cmos power consumption formula: P = C*V*V*f

(C node capacitance; V power supply voltage; f operation frequency).


best regards




eexuke said:
Dear all,
If I have a DFF with value 1 store in it,when I update its value at clock posedge, input new value "1" or "0" consumes less power? Or just equal?
Many thanks!
 

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Not open for further replies.

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