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hello everybody.i can't calculate logical effort of a skewed gate.i don't understand it.please help me.i have a circuit in attachment file .thanks for reading
Logical effort of a gate is defined as the ratio of the input capacitance of the gate to the input capacitance of an inverter that can deliver the same output current.
For the Hi-skew inverter, the input cap is 5/2. An un-skewed inverter that would source the same current from the supply would have an input cap of 3. So the logical effort, h=(5/2)/3 = 5/6. In the sinking direction, the reference inverter would only have an input cap of 3/2 (1+1/2). So logical effort is (5/2)/(3/2) = 5/3.
Similarly, for the Low-skew inverter, the input cap is 2. gu=2/(3/2)=4/3; gd=2/3.
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